输出流中的C ++运算符优先级 [英] C++ operator precedence in output stream

问题描述

int a = 1, b = 2;
int c = a*b + b==0; // c = 0
cout << a*b + b==0; // outputs 4

c 的计算结果为 0 ，因为 * 和 + 运算符的运算符优先级高于 == ，其结果是 c 本质上计算为(a * b + b)== 0 ，这是错误的.

c evaluates to 0 because the operator precedence of the * and + operators is higher than == as a result of which c essentially evaluates to (a*b+b)==0 which is false.

为什么将相同的表达式放在 cout 语句输出4中?

Why does putting the same expression in a cout statement output 4?

推荐答案

因为优先级这些运算符是 operator * > operator + > operator<< > operator == .然后 cout<<a * b + b == 0; 等效于(cout<<((a * b)+ b))== 0; .

Because the precedence of these operators are operator* > operator+ > operator<< > operator==. Then cout << a*b + b==0; is equivalent with (cout << ((a*b) + b)) == 0;.

然后将打印出((a * b)+ b))的结果，即 4 ，然后返回(cout<<(((a * b)+ b))，即将 cout 与 0 进行比较.在C ++ 11之前，可以通过

Then the result of ((a*b) + b)), i.e. 4 will be printed out, then the returned value of (cout << ((a*b) + b)), i.e. cout is compared with 0. Before C++11 cout could be implicitly converted to void* via operator void*, which returns a null pointer when steram has any errors. So here it's compared with 0 (i.e. the null pointer), and does nothing further more with the result.

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