C++ 运算符优先级

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
	type()   type{}	Functional cast
	()	Function call
	[]	Subscript
	.   ->	Member access
3	++   --	Prefix increment and decrement	Right-to-left
	+   -	Unary plus and minus
	!   ~	Logical NOT and bitwise NOT
	(type)	C-style cast
	*	Indirection (dereference)
	&	Address-of
	sizeof	Size-of[note 1]
	new   new[]	Dynamic memory allocation
	delete   delete[]	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 operators = and ≠ respectively
10	&	Bitwise AND
11	^	Bitwise XOR (exclusive or)
12	|	Bitwise OR (inclusive or)
13	&&	Logical AND
14	||	Logical OR
15	?:	Ternary conditional[note 2]	Right-to-left
	throw	throw operator
	=	Direct assignment (provided by default for C++ classes)
	+=   -=	Compound assignment by sum and difference
	*=   /=   %=	Compound assignment by product, quotient, and remainder
	<<=   >>=	Compound assignment by bitwise left shift and right shift
	&=   ^=   |=	Compound assignment by bitwise AND, XOR, and OR
16	,	Comma	Left-to-right

1. ↑ The operand of sizeof can't be a C-style type cast: the expression sizeof (int) * p is unambiguously interpreted as(sizeof(int)) * p, but not sizeof((int)*p).
2. ↑ The expression in the middle of the conditional operator (between ? and :) is parsed as if parenthesized: its precedence relative to ?: is ignored.

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

Operators that have the same precedence are bound to their arguments in the direction of their associativity. 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 of assignment, but a + b - c is parsed (a + b) - c and not a + (b - c) because of left-to-right associativity of addition and subtraction.

Associativity specification is redundant for unary operators and is only shown for completeness: unary prefix operators always associate right-to-left (delete ++*p is delete(++(*p))) and unary postfix operators always associate left-to-right (a[1][2]++ is ((a[1])[2])++). Note that the associativity is meaningful for member access operators, even though they are grouped with unary postfix operators: a.b++ is parsed (a.b)++ and not a.(b++))

Operator precedence is unaffected by operator overloading.

Notes

Precedence and associativity are compile-time concepts and are independent from order of evaluation, which is a runtime concept.

The standard itself doesn't specify precedence levels. They are derived from the grammar.

const_cast, static_cast, dynamic_cast, reinterpret_cast, typeid, sizeof..., noexcept and alignof are not included since they are never ambiguous.

Some of the operators have alternate spellings (e.g., and for &&, or for ||, not for !, etc.).

Relative precedence of the ternary conditional and assignment operators differs between C and C++: in C, assignment is not allowed on the right-hand side of a ternary conditional operator, so e = a < d ? a++ : a = dcannot be parsed. Many C compilers use a modified grammar where ?: has higher precedence than =, which parses that as e = ( ((a < d) ? (a++) : a) = d ) (which then fails to compile because ?: is never lvalue in C and =requires lvalue on the left). In C++, ?: and = have equal precedence and group right-to-left, so that e = a < d ? a++ : a = d parses as e = ((a < d) ? (a++) : (a = d)).