如何避免隐式转换非构造函数？ （C ++） [英] How do I avoid implicit casting on non-constructing functions? (C++)

问题描述

如何避免对非构造函数的隐式转换？

我有一个函数，它接受一个整数作为参数，

但是该函数也将采用字符，bools ，和longs。

我相信这是通过隐式转换来实现的。

我如何避免这种情况，使函数只接受匹配类型的参数，并拒绝编译否则？

有一个关键字explicit，但它不适用于非构造函数。 ：\

我该怎么办？

How do I avoid implicit casting on non-constructing functions?
I have a function that takes an integer as a parameter,
but that function will also take characters, bools, and longs.
I believe it does this by implicitly casting them.
How can I avoid this so that the function only accepts parameters of a matching type, and will refuse to compile otherwise?
There is a keyword "explicit" but it does not work on non-constructing functions. :\
what do I do?

以下程序编译，虽然我不想这样做：

The following program compiles, although I'd like it not to:

#include <cstdlib>

//the function signature requires an int
void function(int i);

int main(){

    int i{5};
    function(i); //<- this is acceptable

    char c{'a'};
    function(c); //<- I would NOT like this to compile

    return EXIT_SUCCESS;
}

void function(int i){return;}

*请务必指出任何滥用术语和假设。

*please be sure to point out any misuse of terminology and assumptions

推荐答案

c $ c> char 自动被提升为 int 。

You can't directly, because a char automatically gets promoted to int.

您可以借助一个技巧：创建一个以 char 为参数的函数，它。它会编译，但你会得到一个链接器错误：

You can resort to a trick though: create a function that takes a char as parameter and don't implement it. It will compile, but you'll get a linker error:

void function(int i) 
{
}
void function(char i);

使用 char 参数调用函数打破构建。

Calling the function with a char parameter will break the build.

请参见 http://ideone.com/2SRdM

术语：非建构函数？你的意思是一个不是构造函数的函数？

Terminology: non-construcing functions? Do you mean a function that is not a constructor?

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