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4.7 — Structs

There are many instances in programming where we need more than one variable in order to represent an object. For example, to represent yourself, you might want to store your name, your birthday, your height, your weight, or any other number of characteristics about yourself. You could do so like this:

However, you now have 6 independent variables that are not grouped in any way. If you wanted to pass information about yourself to a function, you’d have to pass each variable individually. Furthermore, if you wanted to store information about someone else, you’d have to declare 6 more variables for each additional person! As you can see, this can quickly get out of control.

Fortunately, C++ allows us to create our own user-defined aggregate data types. An aggregate data type is a data type that groups multiple individual variables together. One of the simplest aggregate data types is the struct. A struct (short for structure) allows us to group variables of mixed data types together into a single unit.

Declaring and defining structs

Because structs are user-defined, we first have to tell the compiler what our struct looks like before we can begin using it. To do this, we declare our struct using the struct keyword. Here is an example of a struct declaration:

This tells the compiler that we are defining a struct named Employee. The Employee struct contains 3 variables inside of it: a short named id, an int named age, and a double named wage. These variables that are part of the struct are called members (or fields). Keep in mind that Employee is just a declaration -- even though we are telling the compiler that the struct will have member variables, no memory is allocated at this time. By convention, struct names start with a capital letter to distinguish them from variable names.

Warning: One of the easiest mistakes to make in C++ is to forget the semicolon at the end of a struct declaration. This will cause a compiler error on the next line of code. Modern compilers like Visual Studio 2010 will give you an indication that you may have forgotten a semicolon, but older or less sophisticated compilers may not, which can make the actual error hard to find.

In order to use the Employee struct, we simply declare a variable of type Employee:

This defines a variable of type Employee named joe. As with normal variables, defining a struct variable allocates memory for that variable.

It is possible to define multiple variables of the same struct type:

Accessing struct members

When we define a variable such as Employee joe, joe refers to the entire struct (which contains the member variables). In order to access the individual members, we use the member selection operator (which is a period). Here is an example of using the member selection operator to initialize each member variable:

As with normal variables, struct member variables are not initialized, and will typically contain junk. We must initialize them manually.

In the above example, it is very easy to tell which member variables belong to Joe and which belong to Frank. This provides a much higher level of organization than individual variables would. Furthermore, because Joe’s and Frank’s members have the same names, this provides consistency across multiple variables of the same struct type.

Struct member variables act just like normal variables, so it is possible to do normal operations on them:

Initializing structs

Initializing structs by assigning values member by member is a little cumbersome, so C++ supports a faster way to initialize structs using an initializer list. This allows you to initialize some or all the members of a struct at declaration time.

In C++11, we can also use uniform initialization:

If the initializer list does not contain an initializer for some elements, those elements are initialized to a default value (that generally corresponds to the zero state for that type). In the above example, we see that frank.wage gets default initialized to 0.0 because we did not specify an explicit initialization value for it.

C++11/14: Non-static member initialization

Starting with C++11, it’s possible to give non-static (normal) struct members a default value:

Unfortunately, in C++11, the non-static member initialization syntax is incompatible with the initializer list and uniform initialization syntax. For example, in C++11, the following program won’t compile:

Consequently, in C++11, you’ll have to decide whether you want to use non-static member initialization or uniform initialization. Uniform initialization is more flexible, so we recommend sticking with that one.

However, in C++14, this restriction was lifted and both can be used. If both are provided, the initializer list/uniform initialization syntax takes precedence. In the above example, Triangle x would be initialized with length and width 2.0. In C++14, using both should be preferred, as it allows you to declare a struct with or without initialization parameters and ensure the members are initialized.

We talk about what static members are in chapter 8. For now, don’t worry about them.

Assigning to structs

Prior to C++11, if we wanted to assign values to the members of structs, we had to do so individually:

This is a pain, particularly for structs with many members. In C++11, you can now assign values to structs members using an initializer list:

Structs and functions

A big advantage of using structs over individual variables is that we can pass the entire struct to a function that needs to work with the members:

In the above example, we pass an entire Employee struct to printInformation(). This prevents us from having to pass each variable individually. Furthermore, if we ever decide to add new members to our Employee struct, we will not have to change the function declaration or function call!

The above program outputs:

ID:   14
Age:  32
Wage: 24.15

ID:   15
Age:  28
Wage: 18.27

A function can also return a struct, which is one of the few ways to have a function return multiple variables.

This prints:

The point is zero

Nested structs

Structs can contain other structs. For example:

In this case, if we wanted to know what the CEO’s salary was, we simply use the member selection operator twice: myCompany.CEO.wage;

This selects the CEO member from myCompany, and then selects the wage member from within CEO.

You can use nested initializer lists for nested structs:

Struct size and data structure alignment

Typically, the size of a struct is the sum of the size of all its members, but not always!

Consider the Employee struct. On many platforms, a short is 2 bytes, an int is 4 bytes, and a double is 8 bytes, so we’d expect Employee to be 2 + 4 + 8 = 14 bytes. To find out the exact size of Employee, we can use the sizeof operator:

On the author’s machine, this prints:

The size of Employee is 16

It turns out, we can only say that the size of a struct will be at least as large as the size of all the variables it contains. But it could be larger! For performance reasons, the compiler will sometimes add gaps into structures (this is called padding).

In the Employee struct above, the compiler is invisibly adding 2 bytes of padding after member id, making the size of the structure 16 bytes instead of 14. The reason it does this is beyond the scope of this tutorial, but readers who want to learn more can read about data structure alignment on Wikipedia. This is optional reading and not required to understand structures or C++!

Accessing structs across multiple files

Because struct declarations do not take any memory, if you want to share a struct declaration across multiple files (so you can instantiate variables of that struct type in multiple files), put the struct declaration in a header file, and #include that header file anywhere you need it.

Struct variables are subject to the same rules as normal variables. Consequently, to make a struct variable accessible across multiple files, you can use the extern keyword to do so.

Final notes on structs

Structs are very important in C++, as understanding structs is the first major step towards object-oriented programming! Later on in these tutorials, you’ll learn about another aggregate data type called a class, which is built on top of structs. Understanding structs well will help make the transition to classes that much easier.

The structs introduced in this lesson are sometimes called plain old data structs (or POD structs) since the members are all data (variable) members. In the future (when we discuss classes) we’ll talk about other kinds of members.

Quiz

1) You are running a website, and you are trying to keep track of how much money you make per day from advertising. Declare an advertising struct that keeps track of how many ads you’ve shown to readers, what percentage of ads were clicked on by users, and how much you earned on average from each ad that was clicked. Read in values for each of these fields from the user. Pass the advertising struct to a function that prints each of the values, and then calculates how much you made for that day (multiply all 3 fields together).

2) Create a struct to hold a fraction. The struct should have an integer numerator and an integer denominator member. Declare 2 fraction variables and read them in from the user. Write a function called multiply that takes both fractions, multiplies them together, and prints the result out as a decimal number. You do not need to reduce the fraction to its lowest terms.

Quiz Answers

1) Show Solution

2) Show Solution

4.8 -- The auto keyword
Index
4.6 -- Typedefs and type aliases

225 comments to 4.7 — Structs

  • Zidane

    #include
    using namespace std;

    struct Fraction
    {
    unsigned short nNumerator;
    unsigned short nDenominator;
    };

    Fraction GetFractionData()
    {
    Fraction Fra;
    static int c=1; // ‘c’ is a Fraction Identifier
    cout<<"Fraction "<<c++<<endl;
    cout<>Fra.nNumerator;
    cout<>Fra.nDenominator;

    return Fra;
    }

    void PrintMultiplyFractions(Fraction x, Fraction y)
    {
    cout<<x.nNumerator<<"/"<<x.nDenominator<<
    " * "<<
    y.nNumerator<<"/"<<y.nDenominator<<
    " = "<<
    (static_cast(x.nNumerator * y.nNumerator)/(x.nDenominator * y.nDenominator))<<
    endl;
    }

    int main()
    {
    PrintMultiplyFractions(GetFractionData(),GetFractionData());
    return 0;

    }

  • Zidane

    //This is how I did it. Please comment πŸ˜‰

    #include
    using namespace std;

    struct Adverts
    {
    unsigned short nAdsShown;
    float fClickRate;
    float fAveEarnPerClick;
    };

    Adverts GetAdsData()
    {
    Adverts Ads;
    cout<>Ads.nAdsShown;
    cout<>Ads.fClickRate;
    cout<>Ads.fAveEarnPerClick;
    return Ads;
    }

    void PrintAdsResult(Adverts Ads)
    {
    cout<<"Total of adverts shown:\t\t"<<Ads.nAdsShown<<endl;
    cout<<"Click through rate:\t\t"<<Ads.fClickRate<<endl;
    cout<<"Average earnings per click:\t"<<Ads.fAveEarnPerClick<<endl;
    cout<<"Total earnings for the day:\t$"<<Ads.nAdsShown * Ads.fClickRate * Ads.fAveEarnPerClick<<endl;
    }

    int main()
    {
    PrintAdsResult(GetAdsData());
    return 0;

    }

  • Speeds03

    This is how I did it. It worked perfectly. But correct me if I should have done it another way. By the way, I pre-set the values for the variables because we aren’t really getting any user data to actually calculate a result.

    #include “stdafx.h”
    #include

    using namespace std;

    struct MoneyPerDay
    {
    int nAdShown; // 45 Ads shown each month
    long float fAdsClicked; // Amount of Ads clicked by users
    long float fAmountPerClick; // Each Ad click revenue is $.10
    long float nUsersPerDay;

    };

    void PrintMoneyEarnedPerDay(MoneyPerDay sMoneyPerDay)
    {
    cout << "Number of users per day: " << sMoneyPerDay.nUsersPerDay << endl;
    cout << "Number of Ads shown: " << sMoneyPerDay.nAdShown << endl;
    cout << "Number of Ads clicked: " << sMoneyPerDay.fAdsClicked << endl;
    cout << "Amount of money per Ad click: " << sMoneyPerDay.fAmountPerClick << endl;
    }

    int main()
    {
    MoneyPerDay sMonday;
    sMonday.nUsersPerDay = 10780;
    sMonday.nAdShown = 45;
    sMonday.fAdsClicked = 20;
    sMonday.fAmountPerClick = .10;

    MoneyPerDay sTuesday;
    sTuesday.nUsersPerDay = 20678;
    sTuesday.nAdShown = 45;
    sTuesday.fAdsClicked = 20;
    sTuesday.fAmountPerClick = .10;

    // Prints stats for monday
    cout << "Monday stats-- " << endl;
    PrintMoneyEarnedPerDay(sMonday);
    cout << "The result is: $" << sMonday.nUsersPerDay / sMonday.fAdsClicked * sMonday.fAmountPerClick << endl;

    cout << " " << endl;

    // Prints stats for tuesday
    cout << "Tuesday stats-- " << endl;
    PrintMoneyEarnedPerDay(sTuesday);
    cout << "The result is: $" << sTuesday.nUsersPerDay / sTuesday.fAdsClicked * sTuesday.fAmountPerClick << endl;

    cout << " " << endl;

    return 0;
    }

    • codeez

      Long float is kind of an oxymoron I think? VS2012 gives warnings:

      nonstandard extension used : long float c:user

      If you meant double the size of float then use ‘double’, but seeing as your data doesn’t exceed a float though, just the float would’ve done. πŸ™‚

      You didn’t need the blanks here: cout << " " << endl; // just do: cout << endl;

      Keep coding. πŸ™‚

  • Blackout

    This is what I did, I didn’t use structs because I wasn’t sure how, but it works.

  • KanedaSyndrome

    My results:

    main.cpp

    #include <iostream>
    #include "functions2.h"
    using namespace std;

    advertising campaign1;
    fraction frac1, frac2;

    int main()
    {
    campaign1 = getData(campaign1);
    printResults(campaign1);

    cout << endl << "NEXT PART OF THE EXERCISE!" << endl;

    cout << "First fraction" << endl;
    frac1 = getFrac(frac1);
    cout << "Second fraction" << endl;
    frac2 = getFrac(frac2);
    fracProduct(frac1, frac2);

    cin.get(); //for halting the program
    cin.ignore(); //for halting the program
    return 0;
    }

    functions2.cpp

    #include <iostream>
    #include "functions2.h"
    using namespace std;

    advertising getData(advertising x)
    {
    cout << "Input the number of adds in the campaign: ";
    cin >> x.numberAdds;
    cout << "Input the hit rate of the campaign: ";
    cin >> x.hitRatio;
    cout << "Input the average revenue per hit: ";
    cin >> x.avgRevenuePerHit;
    return x;
    }

    void printResults(advertising x)
    {
    float result = x.numberAdds*x.hitRatio*x.avgRevenuePerHit;
    cout << endl << "The campaign has resulted in a total revenue of: " << result << endl;
    }

    fraction getFrac(fraction x)
    {
    cout << "Input a numerator: ";
    cin >> x.numerator;
    cout << "Input a denominator: ";
    cin >> x.denominator;

    return x;
    }

    void fracProduct(fraction x, fraction y)
    {
    float result = static_cast<float>(x.numerator*y.numerator)/(x.denominator*y.denominator);

    cout << "Multiplying the two fractions yields: " << result << endl;
    }

    functions2.h

    #ifndef FUNCTIONS2_H
    #define FUNCTIONS2_H

    struct advertising
    {
    int numberAdds;
    float hitRatio;
    float avgRevenuePerHit;
    };

    struct fraction
    {
    int numerator;
    int denominator;
    };

    advertising getData(advertising x);
    void printResults(advertising x);
    fraction getFrac(fraction x);
    void fracProduct(fraction x, fraction y);

    #endif

    ~KanedaSyndrome

  • DrSuse

    I had trouble with :

    void Multiply(Fraction sF1, Fraction sF2)

    because in section 2.1 Basic Addressing and Variable Declaration, it states:

    “The first mistake is declaring each variable as int (or whatever type it is) in sequence. This is not a bad mistake because the compiler will complain and ask you to fix it.

    int nValue1, int nValue2; // wrong (compiler error)

    int nValue1, nValue2; // correct”

    where instead I typed (something like) :
    void Multiply(Fraction sF1, sF2)
    and got compiler errors all over the pace.

    Am I to assume it’s different for function parameters?

    • codeez

      “Am I to assume it’s different for function parameters?”.

      Up to this point in the lessons, I’ve also noticed this hasn’t been explained, but you are right about it for sure!

      • Alex

        Yes, it works differently for function parameters, and it’s inconsistent with the way you declare variables.

        I’m not sure why it was done this way.

        • Andy356

          It’s probably due to the way function declarations are designed. We use

          for declaring, so making the function definition symmetrical would make sense. But then again, keeping it symmetrical with variable definitions would make more sense, since it’s done far more often than function declaration.

  • Mindstormscreator

    Just thought I’d share my code like some people are:

    First exercise, in main.cpp:
    #include <iostream>
    #include "main.h"

    using namespace std;

    int main()
    {
    cout << "Hi there! Um, please enter the following:\n" <<
    "-- Total ads shown to users\n" <<
    "-- Percent of ads clicked\n" <<
    "-- Average profit per ad" << endl;
    Advertising sAds;
    cin >> sAds.nAdsShown >>
    sAds.fPercentClicked >>
    sAds.fAverageProfit;
    sAds.fPercentClicked /= 100; // Convert from percent to decimal so next calculation can occur
    cout << "Total profit today: $" <<
    (
    sAds.nAdsShown *
    sAds.fPercentClicked *
    sAds.fAverageProfit
    ) << endl;

    }

    First exercise, main.h:
    #ifndef MAIN_H
    #define MAIN_H

    struct Advertising
    {
    int nAdsShown;
    float fPercentClicked;
    float fAverageProfit;
    };

    #endif

    Second exercise, main.cpp:
    #include <iostream>
    #include "main.h"

    using namespace std;

    int main()
    {
    Fraction sMyFrac, sYourFrac;
    cin >> sMyFrac.nNumer >> sMyFrac.nDenom;
    cin >> sYourFrac.nNumer >> sYourFrac.nDenom;
    Fraction sProduct = multFractions(sMyFrac, sYourFrac);
    printFraction(sProduct);
    }

    Fraction multFractions(Fraction sX, Fraction sY)
    {
    Fraction sProduct;
    sProduct.nNumer = sX.nNumer * sY.nNumer;
    sProduct.nDenom = sX.nDenom * sY.nDenom;
    return sProduct;
    //return (new Fraction{sX.nNumer * sY.nNumer, sX.nDenom * sY.nDenom}); // one-liner?
    }

    void printFraction(Fraction sFrac)
    {
    cout << static_cast<double>(sFrac.nNumer) / sFrac.nDenom;
    }

    Second exercise, main.h:
    #ifndef MAIN_H
    #define MAIN_H

    struct Fraction
    {
    int nNumer;
    int nDenom;
    };

    Fraction multFractions(Fraction sX, Fraction sY);
    void printFraction(Fraction sFrac);

    #endif

  • Ollie999

    This is such a great website. Thanks very much for these tutorials. They’re absolutely spot on.

    Just one quick question. What would be the best way to make a struct declaration visible throughout the entire project. ie if you declared the struct type in one file and then wanted to create an instance of it in another.

    Would it be good practice to use a separate file for creating all struct and then use an #include in each file where you need to use it?

  • JMan

    why my code always produce interger output?Can anyone please help me?:(

    // Enumeraor.cpp : Defines the entry point for the console application.
    //

    #include “stdafx.h”
    #include

    struct Fraction
    {
    int nNumerator;
    int nDenomintor;
    };

    double Input()
    {
    using namespace std;
    Fraction sFrac;
    cout<<"Enter the numerator of fraction :"<>sFrac.nNumerator;
    cout<<"Enter the denominator of fraction :"<>sFrac.nDenomintor;
    return static_cast(sFrac.nNumerator/sFrac.nDenomintor);
    }

    int main()
    {
    using namespace std;
    double dFracA=Input();
    double dFracB=Input();
    double dResult;
    static_cast(dResult=dFracA*dFracB);
    cout<<"The result is :"<<dResult<<endl;
    return 0;
    }

    • Alex

      Because you’re doing integer division and casting the result to a float instead of doing floating point division.

      Instead of:

      try:

  • SWEngineer

    Simply well explained tutorial. Thanks.

  • youjay

    For the Example in the post, I tried the following
    ……
    …….
    int main()
    {
    struct Employee
    {
    int nID, nAge;
    float fWage;
    };
    ….
    …..

    As you can see, I defined struct inside main(). This failed. However, defining Struct outside main() worked? Any reason?

    J

    • WCoaster

      The variables inside the structure are treated as local variables inside main(). When you declare it outside main() they become global variables (evil) πŸ˜‰

      It should work if you initialize your variables before calling a function.

    • Alex

      I’m not sure why this wouldn’t work. You should be able to declare a struct inside a function (I tested it with Visual Studio 2010 Express and it works).

      If you do this, the struct declaration has local scope, so you’ll only be able to create variables of this struct type within that function.

  • Would someone mind telling me why this is rounding down my answer? I’m not sure what i’m doing wrong.

    It’s giving me the answer 0, but i want 0.75.

    I take it i accidently read over one of the earlier lessons which is really confusing me now.

    • JD

      I kinda doubt you’re still wondering this 2 months later, but in case you haven’t figured it out:

      Because the / operator is in the parenthesis, the division is performed first. Both inputs are integers, so it is performed as an integer operation (and rounded to 0). Only after the division is complete does it convert the result to float. You want the division to be a float operation, so you should cast one of the inputs *before* the division, as follows:

      This will make it a float operation, and it will not be rounded.

  • Sudheer

    Thanks a lot Alex

    Really very useful tutorial on web. Thanks again for the effort πŸ™‚

  • Elqno

    Hello Alex, I have a qst:

    From your code:

    when you made the declaration of this:

    shouldn’t be like this:

    or making the struct a typedef?? :

    I think it should be like that, so then you can declare sJoe as you did…
    I hope that if you reply I will get to my inbox that you did it πŸ˜‰ so I can check what you wrote πŸ˜‰

    • Alex

      In C++ using “struct Employee sJoe” or typedefs along with structs isn’t necessary.

      This is one of the areas where C++ improves on C.

  • Rafael

    This is my answer for exercise 1 and 2 :

    Exercise 1 :

    And Exercise 2 :

  • Kiasta

    In your “Quiz 1 Answer” You did not divide the percentage, or at least turn it into a decimal, so the output will be wrong.

    For example:

    5000 * 50 * $.15 = $37,500
    5000 * (50 / 100) * $.15 = $375

    The two are very different. The user will not know to put a decimal instead of a whole number (in fact they might even put a modulus after the number which is bad). I guess I am just a little nit-picky. BTW here is my answer if you are interested:

    • Alex

      The way the problem was originally specified asked the user to enter a percentage as a number between 0 and 1, so the division by 100 wasn’t necessary. But I think more people probably think about percentages the way you do (between 0 and 100 instead of between 0 and 1), so I’ve altered the question and answer.

  • drow

    #include “stdafx.h”

    #include

    struct advertising
    {
    int nAds_shown;
    float percentThatclicked;
    float averageEarned;
    };

    void read()
    {
    using namespace std;
    advertising scompany;
    cin >> scompany.nAds_shown;
    cin >> scompany.percentThatclicked;
    cin >> scompany.averageEarned;
    }

    int main(advertising scompany)
    {
    using namespace std;
    read();
    cout << scompany.nAds_shown << endl;
    cout << scompany.percentThatclicked << endl;
    cout << scompany.averageEarned << endl;
    cout << scompany.nAds_shown*scompany.percentThatclicked * scompany.averageEarned << endl;
    return 0;
    }

    This code compiles and links without a problem, but when I actually run it it returns very odd numbers.
    Can anyone help?

    • Kiasta

      Try making the void Read() function return the struct advertisementy like so:

      What happens is that when you run through the function Read() it stores the data in the SCOPE, and than is destroyed after the function ends. So when you return scompany and store it into scompany on int main(), it passes the values to the main() scope. There might be a little more to it but basically the values are destroyed when Read() ends so you have to pass it on to main() which I have done. BTW I am unsure as to why you included stdafx.h when you have nothing in your program that needs that header, whats even more curious is the way you included it (with “” instead of ) and that it compiled at all.

  • BX

    Shouldn’t structure alignment be mentioned in this chapter?

    BTW, great free tutorials. Thanks for your effort.

    • Alex

      I mention it in the article above. Understanding structure alignment isn’t required to use structures or write effective C++ programs. It’s something that curious people can read about if they desire, or come back to later.

  • Carter

    struct Fractions
    {
    int nNumerator;
    int nDenominator;
    };

    int multiply(int x, int y)
    {
    return x * y;
    }

    float produceValue(Fractions sAny)
    {
    using namespace std;
    cout << "Enter a Numerator " <> sAny.nNumerator;
    cout << "Enter a Denominator " <> sAny.nDenominator;
    return (float)sAny.nNumerator / sAny.nDenominator;
    }
    int main()
    {
    using namespace std;
    cout << "Now, we are going to multiply two fractions. First: " << endl;
    float x = produceValue;
    cout << "Now, once again " << endl;
    float y = produceValue;

    cout << "The answer is " << multiply(float x, float y);

    return 0;
    }

    didnt work though, does anyone know why?

    • mirondanro

      Try this:

      #include
      #include

      struct Fractions
      {
      int nNumerator;
      int nDenominator;
      };

      float multiply(float x, float y)
      {
      return x * y;
      }

      float produceValue()
      {
      using namespace std;
      Fractions sAny;
      cout <> sAny.nNumerator;
      cout <> sAny.nDenominator;
      float result = (float)sAny.nNumerator / sAny.nDenominator;
      return (result);
      }

      int main()
      {
      using namespace std;
      cout << "Now, we are going to multiply two fractions. First: " << endl;
      float x = produceValue();
      cout << "Now, once again " << endl;
      float y = produceValue();
      cout << "The answer is " << multiply(x, y);
      cin.clear();
      cin.ignore(255, '\n');
      cin.get();
      return 0;
      }

  • Hello Alex,

    I have trouble understanding the following situation, where a struct consisting of int, float and bool types shows a complete size as 12. Whereas, if I display the size of each data-type, it shows 4(int), 4(float), 1(bool).

    Please explain.

    The code:

    The Output:

    Regards,
    Mayur

    • Ole

      Let me quote Alex and highligth the important bit that you wonder about:

      “Typically, the size of a struct is the sum of the size of all it’s members. In this case, since each integer is 4 bytes and a float is 4 bytes, Employee would be 12 bytes. However, some platforms have specific rules about how variables must be laid out in memory β€” consequently, the compiler may leave gaps between the variables. As a result, we can say the struct will be at minimum 12 bytes.”

      Cheers
      Ole

  • Shawn

    Alex

    for the quiz 2 how-come the same result is not produced?

    the correct answer is printed in the above code

    but it just prints 0, in the next code, i know i am doing something wrong with type casting, but what exactly am i doing wrong

    thanks

    • Alex

      The precedence in the second statement is wrong. You’re doing two integer divisions, multiplying them together (which will produce an integer) and then casting the answer to a float.

      You probably meant:

  • twilight

    When I understand the solution for 1) right, this is showing us a style of coding this tutorial itself said we should not do.

    The method PrintAdvertising() get’s passed a struct named sAd, while the main-function declares a struct with the name sAd.
    So while in the PrintAdvertising() method, the local sAd is hiding the main()-sAd.

    Am I right or did I understand something wrong? Thx!

    • Alex

      No, it’s fine to do it this way.

      While in function PrintAdvertising(), the struct declared in main() is inaccessible. So it’s fine that PrintAdvertising() uses the same name.

  • surua

    why can’t I print the CEO’s data this way. I think I am doing some silly mistake…

    • manju

      U are actually passing the structure sCEO which is not declared . So the compiler gives u an error . Since the sCEO is a member of sMyCompany. U should use sMyCompany.sCEO in order to access the structure of Employee sCEO .

      The following line would fix ur problem
      PrintInformation(sMyCompany.sCEO);
      instead of
      PrintInformation(sCEO);

  • Kinten

    I have see that some people define structs this way

    What I dont understand is, why they write “something” two times?

    • It’s easier to understand when you use different labels:

      In this case, something is the name of the struct itself. It’s like naming a function, it’s just a label to reference it by later. This doesn’t actually create a struct object, it just lays out what a struct looks like.

      somethingelse declares an actual struct object. It is a variable that takes up memory.

      So when you see something like this:

      It basically means you’re defining a struct named something, and then also defining a local variable with the same name. In my opinion, this is bad coding style since you’re using the same name for two different things.

      Usually when I see this, it’s because people want a “one off” struct -- something that’s only used once. In that case, you can just use an anonymous struct:

      This struct has no name and the type can never be referenced directly. However, something is a variable of that type.

  • som shekhar

    Okay i understood what can go wrong in the above problem???correct me if i am wrong!!
    when i declare Score s1; memory has been allocated lets say its address is 123.
    Now when i am passing this s1 to the function ReadScore, this function will create a copy of s1
    which will have different memory,(let say its address is 125) so it does everything right,
    but doesnt touch at all the address 123..which is still junk. SO i m getting junk correct?

    the above problem i overcame ,when i send the reference of the s1!!!
    Or else i can call the PrintScore inside the ReadScore function..this will also eliminates the problem.

    • Chris Walker

      Exactly! This is a pass-parameter-by-value-or-reference issue. As a general rule: if you need to modify the struct/obj and keep changes in the previous scope, pass by reference, not by value. Your description of what was happening in your code is spot on.

  • som shekhar

    When i print the score i got junk , i understood since initially s1 was not initialised, and
    when it is passed to function ReadScore s1 now become a local scope for that function and
    value got destroyed after the function ends and dats y i got junk value.
    But When i am doing like this

    i m still getting junk.Well i have created another structure variable s2, and then i assigned
    s2 to s1, so it will be copying the score, Correct??? then y the value is getting destroyed after the f
    the function ends

    • Quinn

      This is probably old but I’ll answer it anyway, for future reference.

      ReadScore doesn’t return the s1 struct. Since s1 is declared locally in both the main function and the ReadScore function, it’s two separate Score structs. Therefore, when you called ReadScore, it didn’t change the value of main()s s1 Score struct. To change that code to work right, you’d either need to use pointers, or you could return the struct from ReadScore. The following examples exclude your other perfectly fine function and struct declaration.

      Returning struct:

      Or you could use pointers:

      Hope that helps!

  • siku

    Is there such a thing like typedef pointers? I got into trouble to initialize PCOORD. Finally I figured it out but I am not sure that am I doing the right thing? I thought it would be good to put this code example here since the mechanism is used in many places.

    For example (code):

    typedef struct Coordinate
    {
    short X;
    short Y;
    } COORD, *PCOORD;

    int main()
    {
    COORD pos;
    pos.X = 0;
    pos.Y = 0;
    PCOORD pPos = &pos;
    }

    At the beginning my main function looked like the main() below:

    int main()
    {
    //this is the wrong way to do things
    PCOORD pPos;
    pPos->X = 0;
    pPos->Y = 0;
    //if you use pPos now in somewhere compiler gives you warning that pPos is not initialized and my program crashed if I run that.
    // Why is it so? Why I cant initialize?
    }

    • Yes, you can do this -- PCOORD is a typedef’s poniter to a Coordinate struct. You can’t initialize your pPos because pPos is a pointer, and you haven’t set it to point at anything.

  • Astro

    We can pass a structure to a function.

    But can a function return a structure of data?

    Cheers.

  • Mitul Golakiya

    /* Program of Structure Exerice 1 */

    #include <iostream.h>
    #include <conio.h>

    struct web
    {
    int adv;
    float rate;
    float click;
    };

    float money(int adv,float rate,float click)
    {
    float totalearning;
    totalearning = (adv * rate * click);
    return totalearning;
    }

    web mny;

    void main()
    {
    float res;

    clrscr();

    cout << "n How many advertise was shown: ";
    cin >> mny.adv;

    cout << "n What was rate of click : ";
    cin >> mny.rate;

    cout << "n Average of earning per click : ";
    cin >> mny.click;

    res = money(mny.adv , mny.rate , mny.click);

    cout << "n The Total earning is " << res;

    getch();
    }

    /* End of Program */

  • Skylark

    Is there a ‘with’ keyword that allows you to edit variables of a structure without having to re-type the name repeatedly [like in Visual Basic]?


    Employee sFrank //a variable of the Employee type

    with sFrank
    .nEmployeeID = 15
    .nAge = 28
    [end with statement]

    // or something like that?

    Not that I’m lazy or anything…

    PS: I think there are some unwanted borders/shading on this page.

    Bye!

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