如何避免隐式转换非构造函数？ （C ++）

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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 ideone/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?