C++ 运算符优先级

http://en.cppreference.com/w/cpp/language/operator_precedence


The following table lists the precedence and associativity of C++ operators. Operators are listed top to bottom, in descending precedence.

Precedence Operator Description Associativity
1::Scope resolution Left-to-right
2 ++   -- Suffix/postfix increment and decrement (a++)
() Function call
[] Array subscripting
. Element selection by reference
−> Element selection through pointer
3 ++   -- Prefix increment and decrement (++a) Right-to-left
+    Unary plus and minus
!   ~ Logical NOT and bitwise NOT
(type) Type cast
* Indirection (dereference)
& Address-of
sizeof Size-of
newnew[] Dynamic memory allocation
deletedelete[] Dynamic memory deallocation
4.*   ->*Pointer to member Left-to-right
5*   /   %Multiplication, division, and remainder
6+   Addition and subtraction
7<<   >>Bitwise left shift and right shift
8 <   <= For relational operators < and ≤ respectively
>   >= For relational operators > and ≥ respectively
9==   !=For relational = and ≠ respectively
10&Bitwise AND
11^Bitwise XOR (exclusive or)
12|Bitwise OR (inclusive or)
13&&Logical AND
14||Logical OR
15 ?: Ternary conditional Right-to-left
= Direct assignment (provided by default for C++ classes)
+=   −= Assignment by sum and difference
*=   /=   %= Assignment by product, quotient, and remainder
<<=   >>= Assignment by bitwise left shift and right shift
&=   ^=   |= Assignment by bitwise AND, XOR, and OR
16throwThrow operator (for exceptions)
17,CommaLeft-to-right

When parsing an expression, an operator which is listed on some row will be bound tighter (as if by parentheses) to its arguments than any operator that is listed on a row further below it. For example, the expressions std::cout<<a&b and *p++ are parsed as (std::cout<<a)&b and *(p++), and not asstd::cout<<(a&b) or (*p)++.

Operators that are in the same cell (there may be several rows of operators listed in a cell) are evaluated with the same precedence, in the given direction. For example, the expression a=b=c is parsed as a=(b=c), and not as (a=b)=c because of right-to-left associativity.

An operator's precedence is unaffected by overloading.

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