In order to properly evaluate an expression such as `4 + 2 * 3`

, we must understand both what the operators do, and the correct order to apply them. The order in which operators are evaluated in a compound expression is called **operator precedence**. Using normal mathematical precedence rules (which state that multiplication is resolved before addition), we know that the above expression should evaluate as `4 + (2 * 3)`

to produce the value 10.

In C++, all operators are assigned a level of precedence. Those with the highest precedence are evaluated first. You can see in the table below that multiplication and division (precedence level 5) have a higher precedence than addition and subtraction (precedence level 6). The compiler uses these levels to determine how to evaluate expressions it encounters.

Thus, 4 + 2 * 3 evaluates as 4 + (2 * 3) because multiplication has a higher level of precedence than addition.

If two operators with the same precedence level are adjacent to each other in an expression, the **associativity rules** tell the compiler whether to evaluate the operators from left to right or from right to left. For example, in the expression `3 * 4 / 2`

, the multiplication and division operators are both precedence level 5. Level 5 has an associativity of left to right, so the expression is resolved from left to right: `(3 * 4) / 2 = 6`

.

**Table of operators**

Notes:

- Precedence level 1 is the highest precedence level, and level 17 is the lowest. Operators with a higher precedence level get evaluated first.
- L->R means left to right associativity.
- R->L means right to left associativity.

Prec/Ass | Operator | Description | Pattern |
---|---|---|---|

1 None |
:: :: |
Global scope (unary) Class scope (binary) |
::name class_name::member_name |

2 L->R |
() () () {} type() type{} [] . -> ++ –– typeid const_cast dynamic_cast reinterpret_cast static_cast |
Parentheses Function call Initialization Uniform initialization (C++11) Functional cast Functional cast (C++11) Array subscript Member access from object Member access from object ptr Post-increment Post-decrement Run-time type information Cast away const Run-time type-checked cast Cast one type to another Compile-time type-checked cast |
(expression) function_name(parameters) type name(expression) type name{expression} new_type(expression) new_type{expression} pointer[expression] object.member_name object_pointer->member_name lvalue++ lvalue–– typeid(type) or typeid(expression) const_cast<type>(expression) dynamic_cast<type>(expression) reinterpret_cast<type>(expression) static_cast<type>(expression) |

3 R->L |
+ - ++ –– ! ~ (type) sizeof & * new new[] delete delete[] |
Unary plus Unary minus Pre-increment Pre-decrement Logical NOT Bitwise NOT C-style cast Size in bytes Address of Dereference Dynamic memory allocation Dynamic array allocation Dynamic memory deletion Dynamic array deletion |
+expression -expression ++lvalue ––lvalue !expression ~expression (new_type)expression sizeof(type) or sizeof(expression) &lvalue *expression new type new type[expression] delete pointer delete[] pointer |

4 L->R |
->* .* |
Member pointer selector Member object selector |
object_pointer->*pointer_to_member object.*pointer_to_member |

5 L->R |
* / % |
Multiplication Division Modulus |
expression * expression expression / expression expression % expression |

6 L->R |
+ - |
Addition Subtraction |
expression + expression expression - expression |

7 L->R |
<< >> |
Bitwise shift left Bitwise shift right |
expression << expression expression >> expression |

8 L->R |
< <= > >= |
Comparison less than Comparison less than or equals Comparison greater than Comparison greater than or equals |
expression < expression expression <= expression expression > expression expression >= expression |

9 L->R |
== != |
Equality Inequality |
expression == expression expression != expression |

10 L->R | & | Bitwise AND | expression & expression |

11 L->R | ^ | Bitwise XOR | expression ^ expression |

12 L->R | | | Bitwise OR | expression | expression |

13 L->R | && | Logical AND | expression && expression |

14 L->R | || | Logical OR | expression || expression |

15 R->L |
?: = *= /= %= += -= <<= >>= &= |= ^= |
Conditional (see note below) Assignment Multiplication assignment Division assignment Modulus assignment Addition assignment Subtraction assignment Bitwise shift left assignment Bitwise shift right assignment Bitwise AND assignment Bitwise OR assignment Bitwise XOR assignment |
expression ? expression : expression lvalue = expression lvalue *= expression lvalue /= expression lvalue %= expression lvalue += expression lvalue -= expression lvalue <<= expression lvalue >>= expression lvalue &= expression lvalue |= expression lvalue ^= expression |

16 R->L | throw | Throw expression | throw expression |

17 L->R | , | Comma operator | expression, expression |

Note: The expression in the middle of the conditional operator ?: is evaluated as if it were parenthesized.

A few operators you should already recognize: +, -, *, /, (), =, <, >, <=, and >=. These arithmetic and relational operators have the same meaning in C++ as they do in every-day usage.

However, unless you have experience with another programming language, it’s likely the majority of the operators in this table will be incomprehensible to you right now. That’s expected at this point. We’ll cover many of them in this chapter, and the rest will be introduced as there is a need for them.

The above table is primarily meant to be a reference chart that you can refer back to in the future to resolve any precedence or associativity questions you have.

**How do I do exponents?**

You’ll note that the ^ operator (commonly used to denote exponentiation in standard mathematical nomenclature) is a Bitwise XOR operation in C++. C++ does not include an exponent operator. To do exponents in C++, #include the <cmath> header, and use the pow() function:

1 2 3 |
#include <cmath> double x = pow(3.0, 4.0); // 3 to the 4th power |

Note that the parameters and return value of pow are of type double. Note that due to rounding errors in floating point numbers, the results of pow() may not be precise (slightly smaller or larger than what you’d expect).

If you want to do integer exponents, you’re best off just using your own function to do so, like this one (that uses the “exponentiation by squaring” algorithm for efficiency):

1 2 3 4 5 6 7 8 9 10 11 12 13 14 |
// note: exp must be non-negative int pow(int base, int exp) { int result = 1; while (exp) { if (exp & 1) result *= base; exp >>= 1; base *= base; } return result; } |

Don’t worry if you don’t understand all of the parts of this function yet. Just beware of overflowing your integer result, which can happen very quickly if either argument is large.

**Quiz**

1) You know from everyday mathematics that expressions inside of parentheses get evaluated first. For example, in the expression `(2 + 3) * 4`

, the `(2 + 3)`

part is evaluated first.

For this exercise, you are given a set of expressions that have no parentheses. Using the operator precedence and associativity rules in the table above, add parentheses to each expression to make it clear how the compiler will evaluate the expression.

Hint: Use the pattern column in the table above to determine whether the operator is unary (has one operand) or binary (has two operands). Review section 1.5 -- A first look at operators if you need a refresher on what unary and binary operators are.

Sample problem: x = 2 + 3 % 4
Binary operator % has higher precedence than operator + or operator =, so it gets evaluated first: x = 2 + (3 % 4) Binary operator + has a higher precedence than operator =, so it gets evaluated next: Final answer: x = (2 + (3 % 4)) We now no longer need the table above to understand how this expression will evaluate. |

a) x = 3 + 4 + 5;

b) x = y = z;

c) z *= ++y + 5;

d) a || b && c || d;

**Solutions**

3.2 -- Arithmetic operators |

Index |

2.10 -- Chapter 2 comprehensive quiz |

Hey there!

Seems to be an error:

On answer a), you write that the "+" sign is left to right but on the sheet mention it’s right to left.

EDIT: I was confused by the + for the unary plus, my bad

Also I agree with Ole and Happilicious

Beside these minus stuff, Great work so far

Hi everybody. I have a question, but there is something that is driving me crazy.

How is this evaluated?

while (exp)

{

.

.

.

}

I’ve seen whiles with statements such as x=>10 or x<=50, but not with only a variable. How is this while triggered?

exp is implicitly converted to a boolean value and that is evaluated. In general, a value of 0, 0.0, or nullptr will evaluate to false, and everything else will evaluate to true.

Great tutorials so far: simple & engaging. Currently on my first read-through. However I’d suggest compiling all the reference tables from all the different tutorial pages into a single (& overly long) reference page/glossary. One could use the CRTL + F search function to find the information that they are looking for. It’d be more convenient than searching the information from different pages & then keeping several tabs opened at the same time. It would be a good & supplementary addition to these tutorials and new programmers could keep the reference page open while debugging their code. Well, at least I would.

Not a bad idea. I’ll keep it in mind as a future enhancement. Thanks for the thought.

why doesn’t the single colon have a precedence?

Single colon isn’t an operator in C++, it’s just part of the syntax for doing certain things.

I know that a few sentences in this program are not in English ,but i don’t think it matters that much because the error i keep getting is that sqrt was not declared.Do you know what the problem is?

#include <iostream>

using namespace std;

int main()

{

int a,b,c,D;

float x1,x2;

cout<<"Vnesi gi a, b i c od kvadratnata ravenka: ";

cin>>a>>b>>c;

D=b*b-4*a*c;

if(D<0) cout<<"Ravenkata nema resenie za ovie koeficienti!";

if(D==0) cout<<"Ravenkata ima edno resenie: "<<-b/2*a;

if(D>0) {

x1=(-b+sqrt(D))/(2*a);

x2=(-b-sqrt(D))/(2*a);

cout<<"Kvadratnata ravenka: "<<a<<"x^2+"<<b<<"x+"<<c<<"=0 ima dve resenija\nx1="<<x1<<"\nx2="<<x2;

}

cout<<"n\n\n";

return 0;

You need to include the cmath header.

here is a recursive definition for pow, just for fun:

Hi Alex,

Can you please explain b) question in quiz? ‘=’ is L-> R operator, so shouldn’t it be (x=y)=z?

Operator= is R->L. Maybe you’re confusing operator= with operator==?

thanks!

Is it worth remembering the different operator precedence levels and their associativity rules, or should you just be aware that precedence and associativity exist?

You don’t need to memorize these. For the most part, things work as expected, and the tutorials that cover these operators will generally point out when something might be non-intuitive in this regard.

New programmers tend to have the most issues with the precedence of || vs && (assuming they are the same, when && is higher), and when using * as a dereference operator when mixed with member access operators . or -> (. and -> take precedence over *).

In one of your early examples of operator precedence, the grouping used by the compiler is irrelevant.

3 * 4 / 2 takes the form a * b / c, which = (a * b) / c = a * (b / c).

Your use of brackets shows the precedence used clearly, but a non-trivial example may be more illustrative to some.

Nice tutorials, by the way.

I see what you’re saying, but at this point we haven’t covered what most of these operators actually do yet. So my pool to draw from for an example is somewhat limited.

A good alternate example would be:

3 * 4 % 3, which evaluates as (3 * 4) % 3, which equals 0 (whereas 3 * (4 % 3) equals 3). But I haven’t covered what operator % does yet.

Could you explain in a bit more detail what the exponent function you propose does? I don’t get the `if (exp & 1)` or the `exp >>= 1` and I don’t think these operators have been covered yet?

Thanks

This function implements exponentiation via the exponentiation by squaring method (it’s just more optimized than the version shown on Wikipedia).

“if (exp & 1)” translates to “if exp is odd”, and “exp >>=1” means “divide exp by 2”.

Hi Alex,

should {} be added as part of level 2 in the Prec/Ass column of the table? Initialization type name{expression} (C++11)

P.s. awesome site and look forward to quizes in all tutorial chapters down the road.

Added.

Hi Alex, in sub-chapter 3.4 — Sizeof, comma, and conditional operators, you mention that assignment has even lower precedence than the ?: operator. However I cannot see the assignment operator listed in the "Table of Operators". Could you please list it?

Thanks in advance.

Both the assignment operator and conditional operator are in precedence row 15. I’ve corrected the erroneous statement in lesson 3.4.

Thank you.

Using pow() for computing exponents of integers and straightforward casting is an unsafe method. String like this will probably output some numbers consisting of nines only:

http://codeforces.com/blog/entry/21844

Yes, which is why I noted that it was only okay to do if you know the arguments to pow() are small.

This is not an overflow problem: this string gives output like "1, 10, 99, 1000, 9999,…". The problem is precision of pow() can cause the result to be 10^2=99.9999999 and then floor rounding by static_cast makes the output 10^2=99 which is totally incorrect. This problem should be avoided by using rounding to the nearest integer rather than direct static_cast or by writing your own pow() for integers (this will also work faster).

I see what you’re getting at, and what you say is true. I’ve updated the article to provide an integer implementation for pow(). It’s simpler than trying to deal with all the edge cases of the double version.

Hello… I don’t understand the associativity of special operators, in my case, static_cast…! (I’m talking about when it’s sitting alone, not with similar operators in the same statment) does left to right mean that it takes the type "int" and apply it to the expression rather than getting the expression first and then applying the type int? It’s confusing 🙁

The left to right is only applicable when mixed with operators at the same level of precedence.

But isn’t it applied on operators sitting alone like the assignment operator? It’s right to left, means it takes the rvalue and assigns it to lvalue…?! Or I’m missing something here…

I’m not sure I understand what you’re asking. Assignment is an operator with a lower precedence, so the static cast is resolved first, and then an assignment is made.

The L->R and R->L tell you which directions things resolve if multiple operators have the same precedence. For example, 3 + 4 - 5 resolves as (3 + 4) - 5. With operators of different precedence levels, the ones with more precedence go first.

That’s not what I meant… I mean that the direction (rule) referred to by the associativity of an operator, well most of the times, makes sense if we apply that "rule" to a single operator sitting alone if it’s binary, without having to exist more than once… I noticed this "theory", that my mind picked up, in some operators like the assignment one that I mentioned, the compiler would go from righ to left with respect to every operand as in

after it detects the two assignment operators it’d go for the rightmost assignment operator (that’s what you mean, I guess), but what I mean is that after it aims at the rightmost operator, it also looks for the rightmost (in this case it’s the only right) operand then assigns it to the left operand, like its associativity is how generally the operator operands the operands.

Another example would probably be the bitwise shift left (or right), it has left to right associativity, thus in

after the compiler detects the presence of the operator, it would look at its left operand and then adds the value of the right operand to it

I dunno how would you look at this (personally I found that my point became silly after explaining it xd), but for the sake of knowledge… Maybe D:

Although nothing serious, the notes say «level 18 is the lowest», but the table lists only 17 levels, however.

Thanks, fixed!

Hi Alex, there is something that I don’t understand. Please look at the following code:

As per the above table both pre- and post-increment ++ operators have higher precedence than the << operator. Then why, in the first case in the above code, does the value of y (which is initially 1) get printed before it is incremented? I thought y++ should be evaluated first and then sent to std::cout by the << operator and get printed. This happens in the second case, where the operator is pre-increment. Why is this so?

Okay I found the answer in the Comments section of Module ‘3.3 — Increment/decrement operators, and side effects’. If anyone has the same doubt, refer Alex’s reply to zhiwei’s question among the comments on that module. Hope this might be useful to someone. 🙂

And Alex, thanks for these great tutorials. They are awesome! 🙂

This is due to the way y++ evaluates. Since post-increment has higher precedence than <<, it evaluates first. Variable y initially has the value of 1. y++ evaluates to the current value of y (1), and then y is incremented. That value of 1 is then sent to be printed, even though y has now been changed.

apah bein yee…!!!

hello

thanks for your useful website

the precedence and associativity table has one problem:

ternary conditional is put above assignment in precedence order while it is not. they should both be put at level 15.

please see this address: http://en.cppreference.com/w/cpp/language/operator_precedence

You are correct. This changed between C and C++, and much of the reference material out there is incorrect (including Microsoft’s C++ precedence table for Visual Studio!). I’ve updated the table. Thanks for the tip.

Thanks a lot. Pre- is level 3 btw in table provided, but I got it now. My thinking was if + and ++ are both in level 3 and level 3 is R->L it should be z *= ((++(y + 5))) but lol at me now I see the difference in + in level 3 ( unary ) and + in level 6 ( addition ). All clear now 🙂 Great tutorial!

Hi, Alex, I can’t understand c) in your quiz. ++/- are both in level 2 and level 3. What is the difference? I assumed that if one in level 2 is called post- and one in level 3 is called pre- it matters on which side of operator they are. But in solution is z *= ((++y) + 5) so it seems ++y is level 2. How is it? Can you explain it little more?

Yes, you are correct on all counts. pre is ++y, and post is y++. So in the quiz, we’re looking at the pre version (++y) which is level 2.

Is it considered acceptable to use parenthesis often when in doubt about this issue, or would the purists poo-poo it and say it should be done without parenthesis and according to the rules in the above table? I, for one, doubt i’ll ever memorize all that stuff!

You can never have too many parenthesis in my opinion. Parenthesis make it clear what your intent is, and reduce precedence mistakes. Use them liberally!

Hello,

I encountered a really strange thing with ++ and -- operators!

look at this snipped :

I expected that the result (result of m) of above code should be 10 (even when I compiled it in the php the result was 10) but in C++ the result was 9!

how is that possible!?

-----

an other question >

result(m) of following code in code::blocks : 19

result(m) of following code in borland C++ : 20

How is that possible!?

This kind of inconsistency usually results from compilers evaluating things in a different order. The C++ specification doesn’t specify the order that parameters should be evaluated, for example.

Generally speaking, in C++ you should never use a variable that has a side effect applied to it more than once in a given expression. Doing so is asking for trouble.

Thank Info

what is the difference between :

which one is recommended ?

int(myChar) is a C-style cast, and is not recommended in C++.(myChar) is a C++-style static cast, and should be preferred in C++.

static_cast

This is covered in lesson 4.4a -- Explicit type conversion (casting).

Just to be pedantic but

is called a functional cast. A C-style cast would be

. However both use the same underlying mechanism to re-cast a variable; essentially C++ replaces the C-style or functional cast with a static_cast.

Thanks, I’ve updated the article.

It’s totally acceptable.

My personal rule of thumb is that if the exponent is both an integer and 3 or less, I just do the multiplication by hand (it’s easy enough to understand, and it saves me from having to #include the math header. If the exponent is 4 or more, or a floating point number, then I’ll favor pow().

I found the pow() function interesting so incorporated it into the quiz from the previous lesson. You can see where I changed "* seconds * seconds" to the pow() function. My question is, is this considered acceptable and/or recommended use? Thanks v much.

Thanks a ton.

hi!Alex.

I wanted to know if it is necessary to learn the entire table for the correct sequence of these operators. Or there are some specific operators needed for daily programming.

Thank you.

It’s not really necessary to memorize all of the precedence levels, as most of them are pretty intuitive, and you can always look them up if something isn’t working like it should. If you have any doubts, you can always disambiguate using parenthesis.

If you try to disambiguate with parenthesis, you’ll get a compiler error.

If you do it with parentheses, you’ll have pairs 😉

Great tutorials ! 🙂

I didn’t realize there was a difference. Thanks for enlightening me. 🙂

hi Alex! good job keep it up (y)

i have a question, i’m wondering! is those tutorials are all about console programs?!

Thanks!

Yes.

It’s possible to create other kinds of applications in C++, such as windowed applications or graphical applications, but I don’t cover those here, as doing so requires operating system specific functionality or add-ons.

For someone who wants to learn GUI programming, is it worth it to continue with all those tutorials or should i start over with maybe other tutorials or books …!

Much of the GUI stuff builds on the fundamentals you’ll learn here. If you finish these tutorials, you’ll have a great foundation for anything you want to do later.

Hey I have a question 🙂

What does the % operator do?

Returns the remainder in an integer division. For example: 9 % 4 = 1 (9 / 4 = 2 remainder 1, so 9 % 4 = 1).

This is discussed in section 3.2 -- arithmetic operators.

Hi there, Alex!

Am I getting right that "+" operator (#6 Table) would be evaluated before "*" operator (#5 in Table) due to precedence level?

But mathematically "*" should be done first…

No, operator* is evaluated before operator+. I’ll try to make this more clear in the table.

"In C++, all operators are assigned a level of precedence. Those with the highest precedence are evaluated first."

So, + [Addition 6 L->R] is evaluated before * [Multiplication 5 L-> R] ?

No.

> Precedence level 1 is the highest precedence level, and level 17 is the lowest. Operators with a higher precedence level get evaluated first.

Hello Alex,

Thank you for this very well made tutorial.

I’d like to suggest you to put 3.2 (arithmetic operators) before 3.1 (operator precedence and associativity), I think it will be easier to understand the quizz as unary arithmetic operators (+x and -x) has higher precedence than binary operators (x + y and x - y) which make this a bit confusing.

The quiz is really unclear in this one. I don’t understand even a bit of it.

I’ve updated the quiz question and answer to try and make it more comprehensible. Let me know if it makes more sense now.

Hey Alex,

I cannot seem to find the exponent operator, ^. Am I missing it, or was it not included.

Thanks

Wintur

Strangely enough, C++ does not include an exponent operator. You have to use the pow() function.

pow(base, exponent) will return the results of base.

^{exponent}. To access the pow() function, you need to #includeSorry to correct you, but you left out what header to include here! If I remember right it is #include <cmath> ,but I could be wrong.

Yes, pow() lives in the <cmath> header.

just a suggestion; if you could make the chart in to an image file as well so people could save it and pull it up at any time it might be easier for referencing in the future.

Just a suggestion towards readability and clarity. The R->L and L->R could maybe be changed into R -> L and R <- L ?

How about just make an arrow ->(going right) or <-(going left)? Makes it much easier to read in my opinion 🙂