如何防止从char数组隐式转换为bool [英] How to prevent implicit conversion from char array to bool

问题描述

struct Foo {
  void setBar(bool bar_) { bar = bar_; }
  bool bar;
};

int main() {
  Foo f;
  f.setBar("true");
}

上述代码由于类型转换而成功编译，即使char数组通过 bool 。

The above code compiles successfully due to type conversion, even though a char array is passed where a bool is expected.

是否可能导致此代码编译失败？ （C ++ 03解决方案首选，因为我的工作场所的编译器是古老的。）

Is it possible to cause this code to fail compilation? (C++03 solution preferred, since the compiler at my workplace is ancient.)

我看了下面的相关问题在StackOverflow，但他们不是解决这个问题。 防止C ++中的隐式转换，

I have looked at the following related questions on StackOverflow, but they don't quite address this problem. Preventing implicit conversion in C++, Why does the compiler choose bool over string for implicit typecast of L""?

推荐答案

您可以声明一个函数 const char * 并且不提供定义：

You can declare a function that takes const char* and don't provide a definition:

void setBar(const char*);

这将使其在链接时失败。你还剩下所有其他隐式转换，从任何指针到bool，积分到bool，浮动到bool ...

This will make it fail at link time. You're still left will all other implicit conversion, though - from any pointer to bool, integral to bool, floats to bool...

另一个选项：

struct Foo {
  void setBar(bool bar_) {}
private:
  template<typename T>
  void setBar(T bar) {}
};

这样，如果你使用 bool 。

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