为什么没有隐式类型转换的警告？ [英] why no warning for implicit type conversion?

问题描述

我终于找出了我的程序中的一个错误，这是由返回类型上的隐式类型转换造成的。即使使用 g ++ -Wall ，也没有警告。

I finally figured out a bug in my program, which caused by the implicit type conversion on return type. Even with g++ -Wall there is no warning for this.

我想知道是否有快速找出这种无意义的错误？

I wonder if there is someway to quickly find out such mindless errors?

#include <iostream>

// return type should be int, but I wrote bool by mistake
bool foo(int x) {
  return x;
}

int main() {
  for (int i = 0; i < 100; ++i) {
    std::cout << foo(i) << std::endl;
    // 0 1 1 1 1 1  ..
    // should be 0 1 2 3 4 ...
  }

  return 0;
}

推荐答案

if（i） i 的类型 int 也正确。

This is correct code. if (i) where i has type int is correct too.

n3376 4.12 / 1

算术，无范围枚举，指针或指向成员类型的prvalue可以转换为bool类型的
prvalue 。零值，空指针值或空成员指针值将转换为false;
任何其他值将转换为true。

n3376 4.12/1

A prvalue of arithmetic, unscoped enumeration, pointer, or pointer to member type can be converted to a prvalue of type bool. A zero value, null pointer value, or null member pointer value is converted to false; any other value is converted to true.

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