The while statement is the simplest of the three loops that C++ provides. It’s definition is very similar to that of an if statement:
while (expression)
statement;
A while statement is declared using the while keyword. When a while statement is executed, the expression is evaluated. If the expression evaluates to true (non-zero), the statement executes.
However, unlike an if statement, once the statement has finished executing, control returns to the top of the while statement and the process is repeated.
Let’s take a look at a simple while loop. The following program prints all the numbers from 0 and 9:
int iii = 0;
while (iii < 10)
{
cout << iii << " ";
iii++;
}
cout << "done!";
This outputs:
0 1 2 3 4 5 6 7 8 9 done!
iii is initialized to 0. 0 < 10 evaluates to true, so the statement block executes. The first statement prints 0, and the second increments iii to 1. Control then returns back to the top of the while statement. 1 < 10 evaluates to true, so the code block is executed again. The code block will repeatedly execute until iii == 10, at which point 10 < 10 will evaluate to false, and the loop will exit.
It is possible that a while statement executes 0 times. Consider the following program:
int iii = 15;
while (iii < 10)
{
cout << iii << " ";
i++;
}
cout << "done!";
The condition 15 < 10 evaluates to false, so the while statement is skipped. The only thing this program prints is done!.
On the other hand, if the expression always evaluates to true, the while loop will execute forever. This is called an infinite loop. Here is an example of an infinite loop:
int iii = 0;
while (iii < 10)
cout << iii << " ";
Because iii is never incremented in this program, iii < 10 will always be true. Consequently, the loop will never terminate, and the program will hang. We can declare an intentional infinite loop like this:
while (1)
{
// this loop will execute forever
}
The only way to exit an infinite loop is through a return statement, a break statement, an exception being thrown, or the user killing the program.
Often, we want a loop to execute a certain number of times. To do this, it is common to use a loop variable. A loop variable is an integer variable that is declared for the sole purpose of counting how many times a loop has executed. Loop variables are often given simple names, such as i, j, or k. Hungarian Notation is often ignored for loop variables (though whether it should be is another question altogether).
However, naming variables i, j, or k has one major problem. If you want to know where in your program a loop variable is used, and you use the search function on i, j, or k, the search function will return half your program! Many words have an i, j, or k in them. Consequently, a better idea is to use iii, jjj, or kkk as your loop variable names. Because these names are more unique, this makes searching for loop variables much easier, and helps them stand out as loop variables. An even better idea is to use "real" variable names, such as nCount, nLoop, or a name that gives more detail about what you're counting.
Each time a loop executes, it is called an iteration. Often, we want to do something every n iterations, such as print a newline. To have something happen every n interations, we can use the modulus operator:
// Loop through every number between 1 and 50
int iii = 1;
while (iii <= 50)
{
// print the number
cout << iii << " ";
// if the loop variable is divisible by 10, print a newline
if (iii % 10 == 0)
cout << endl;
// increment the loop counter
iii++;
}
This program produces the result:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50
It is also possible to nest loops inside of other loops. In the following example, the inner loop and outer loops each have their own counters. However, note that the loop expression for the inner loop makes use of the outer loop's counter as well!
// Loop between 1 and 5
int iii=1;
while (iii<=5)
{
// loop between 1 and iii
int jjj = 1;
while (jjj <= iii)
cout << jjj++;
// print a newline at the end of each row
cout << endl;
iii++;
}
This program prints:
1 1 2 1 2 3 1 2 3 4 1 2 3 4 5
Quiz
1) In the above program, why is jjj declared inside the while block instead of following immediately following the declaration of iii?
2) Write a program that prints out the letters a-z along with their ASCII codes. Hint: to print characters as integers, you have to use a static_cast.
Quiz Answers
5.6 -- Do while statements
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Index
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5.4 -- Goto statements
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5.6 -- Do while statements
Index
5.4 -- Goto statements
In the code above, shouldn’t the if statement read:
[ Yes, it should. Thanks for catching that. -Alex ]
Why does the last program print
Instead of
? Am I missing something? It seems to only print one number each time it repeats by looking at the code.
This happens because there is a loop inside of a loop.
The outer loop iterates iii from 1 to 5. The inner loop iterates jjj from 1 to iii.
Each time the outer loop iterates, all of the inner loop iterations are restarted.
So the first time, iii = 1 and jjj = 1. The inner loop executes once and prints 1.
The second iteration, iii = 2 and jjj = 1. The inner loop executes twice and prints 1 2.
The third iteration, iii=3 and jjj = 1. The inner loop executes three times and prints 1 2 3.
And so on.
int iii=5; while (iii>=0) { // loop between 1 and iii int jjj = 1; while (jjj <= iii) { cout << "*" ; jjj++; } // print a newline at the end of each row cout << endl; iii--; }notice what i changed in the given example…
how do i get the previous code to read this?
12345
1234
123
12
1
pliz reply asap!!!
replace * with jjj
Even though it’ll never be executed, there is a slight bug in the code from the second example.
i++ should be iii++
int iii = 15; while (iii < 10) { cout << iii << " "; i++ } cout << "done!";Actually it wouldn’t even compile, as the variable “i” hasn’t been declared.
I rerote the code from the first example to request an integer from the user and then add each number to the next number on the row and give the combined sum of the row.
#include <iostream> using namespace std; int main() { int iii=1; int holder2; cout << "Enter the number you want to add up to: "; int iterations; cin >> iterations; // loop between iii and iterations while (iii<=iterations) //while iii is less than or equal to iterations do the following: { // loop between 1 and iii-1 int jjj = 1; int holder; int total=0; int finaltotal=0; while (jjj < iii) { holder=jjj; total=total + holder; cout << holder << " + "; jjj++; //increment jjj by 1 } holder2=total; finaltotal=holder2+jjj; cout << jjj; cout << " = " << finaltotal; // print a newline at the end of each row cout << endl; iii++; } return(0); }The resulting output for 5 iterations is:
Enter the number you want to add up to: 5
1 = 1
1 + 2 = 3
1 + 2 + 3 = 6
1 + 2 + 3 + 4 = 10
1 + 2 + 3 + 4 + 5 = 15
Not sure but If I’m right, but I think this is a fibonacci sequence. (please correct me if I’m wrong)
It took a while to figure out the logic (and I sure this is probably not the most efficient solution, but it works)… sorry for the lack of comments… I’m gonna go back and comment more. Great Tutorial… I finally feel like I’m getting somewhere.
fibonacci should be(i think so and lazy to google it) 1 1 2 3 5 …
so thats not fibon…
;)
#include<iostream> int main() { using namespace std; char chValue=97; while (chValue<=122) { cout << static_cast<char>(chValue) << " " << static_cast<int>(chValue) << endl; chValue=chValue++; } }Is it acceptable to write the quiz question in this way?
It’s acceptable and will compile, but for readability it’s a bit ambiguous. It may not be plainly obvious what is meant by 97 and 122.
I also believe static_cast<char>(chValue) is unnecessary, it should just be chValue.
Instead of doing:
I did:
It compiled and worked fine. It seems easier than the static_cast code. Is that bad coding/is there a reason why it shouldn’t be done like that?
Thanks
(int)chValue is a C-style cast, whereas static_cast<int>(chvalue) is a C++ style one, and they are different in the fact that C-style casts are less type safe than C++ ones. Because of this, it is always suggested that you do, and get into the habit of doing, static_cast instead of C-style casts. Alex talked about this in lesson 4.4.
There is error in code number 2 line 5:
i++; it must be iii++;
For conversion from letter to ASCII and back I used this and it worked:
char ch = ‘d’
cout << ch << " " << int(ch) << endl;
int c = 100;
cout << c << " " << char(c) << endl;
Is there anything wrong in this int(char) / char(int) so that we have to use static_cast(char) ?
Fero
In the solution 2 answer:
jjj is declared inside the while block so that it is recreated (and reinitialized to 1) each time the outer loop executes. If jjj were declared before the outer while loop, it’s value would never be reset to 1, or we’d have to do it with an assignment statement. Furthermore, because jjj is only used inside the outer while loop block, it makes sense to declare it there. Remember, declare your variables in the smallest scope possible!
It is okay if we follow – variables inside the scope.
But isnt it more work for the PC to recreate jjj variable each loot (for example, if our other loop was like 10 000 times long) instead of having the jjj once declared outside and jjj = 1 inside?
Don’t forget that infinite loops like
while(true) { }is usually used in web servers!
output should b
1
123
12345
1234567