cout语句中使用的条件运算符 [英] Conditional operator used in cout statement

问题描述

通过尝试，我了解到在cout语句中有必要在一个条件运算符周围加括号。这里有一个小例子：

By trying, I came to know that it is necessary to put brackets 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;
}

输出为：

42
42
1

为什么需要此支架？在这两种情况下，结果类型的条件运算符是已知的。

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

推荐答案

？：运算符的优先级低于<< 运算符。也就是说，编译器将您最后一个语句解释为：

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;

这将首先传递（a！= 0）的布尔值到cout。然后，该表达式的结果（即对cout的引用）将被转换为适当的类型以供在：：操作符中使用（即 void * ：see http://www.cplusplus.com/reference/iostream/ios/operator_voidpt/ ），并且取决于是否该值为真（即，cout是否没有设置错误标志），它将获取值42或值-42。最后，它会抛出那个值（因为没有什么用）。

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 http://www.cplusplus.com/reference/iostream/ios/operator_voidpt/), 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).

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