Conditional operator used in cout statement

问题

By trying, I came to know that it is necessary to put parentheses around a conditional operator in a cout statement. Here a small example:

#include <iostream>

int main() {
  int a = 5;
  float b = (a!=0) ? 42.0f : -42.0f;
  // works fine
  std::cout << b << std::endl;
  // works also fine
  std::cout << ( (a != 0) ? 42.0f : -42.0f ) << std::endl;
  // does not work fine
  std::cout << (a != 0) ? 42.0f : -42.0f;

  return 0;
}

The output is:

42
42
1

Why are these parentheses necessary? The resulting type of the conditional operator is known in both cases, isn't it?

回答1:

The ?: operator has lower precedence than the << operator i.e., the compiler interprets your last statement as:

(std::cout << (a != 0)) ? 42.0f : -42.0f;

Which will first stream the boolean value of (a!=0) to cout. Then the result of that expression (i.e., a reference to cout) will be cast to an appropriate type for use in the ?: operator (namely void*: see cplusplus.com), and depending on whether that value is true (i.e., whether cout has no error flags set), it will grab either the value 42 or the value -42. Finally, it will throw that value away (since nothing uses it).

回答2:

Because << has higher precedence than ?.

Fun exercise:

float ftest = std::cout << (a != 0) ? 42.0f : -42.0f;

Take that, Coding Horror!!!

Your code is equivalent to:

if ( std::cout << (a != 0) )
     42.0f;
else
    -42.0f;

It outputs 1 because, well, (a != 0) == true;

来源：https://stackoverflow.com/questions/9619424/conditional-operator-used-in-cout-statement

标签

c++

conditional-operator