In mathematics, an **operation** is a mathematical calculation involving zero or more input values (called **operands**) that produces an output value. Common operations (such as addition) use special symbols (such as +) that denote the operation. These symbols are called **operators**. Operators in programming work the same way except the names may not always be a symbol. Operators work analogously to functions that take input parameters and return a value, except they are more concise. For example, 4 + 2 * 3 is much easier to read than add(4, mult(2, 3))!

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++, when the compiler encounters an expression, it must similarly analyze the expression and determine how it should be evaluated. To assist with this, 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 **operator 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.

That said, if you have an expression that uses operators of different types, it is a best practice to use parenthesis to explicitly disambiguate the order of evaluation.

*Rule: If your expression uses different operators, use parenthesis to make it clear how the expression should evaluate, even if they are technically unnecessary.*

**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 |

Hello! In the first table on this page there’s the operator *. The description should be "Refrence" it’s "Dereference"

Thank you : )

No. * is definitely a dereference (indirection). & is reference (also address-of and bitwise AND, depending on context).

Oops! Sorry 😀

Hi Alex, I think writing a thing like this :

has an associativity from right to left because member_name belongs to class_name ans not class_name belongs to member_name; idem for namespaces and global scopes.

Definitely not. The other choices are defensible. Microsoft considers :: to have no associativity, because what comes before the operator and what comes after is always treated as a single unit. Other reference material treats :: as having left to right associativity, since you can’t access an identifier without knowing what namespace it’s in (the namespace has to be known first).

Thanks, I got!

Hey there!

I just wanted to say I’m going through your tutorials and I’m loving them! I also have a question about the Quiz part d, and this might be that I really don’t understand how those operators work yet… could that series of conditions be simplified to just b && c? If I’m following it right, once b && c is true, then the remaining questions are also true, right? This isn’t a question about the answer, I’m just wanting to know if I am following how that works. Thanks!

> could that series of conditions be simplified to just b && c

No.

> If I’m following it right, once b && c is true, then the remaining questions are also true, right?

Yes, but we can’t assume b && c will evaluate to true.

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.