如何避免对非构造函数进行隐式转换? [英] How do I avoid implicit conversions on non-constructing functions?
问题描述
如何避免在非构造函数上进行隐式转换?
我有一个函数将一个整数作为参数,
但该函数也将转换为字符,bools ,和longs。
我相信这是通过隐式地转换它们。
如何避免这种情况,以便该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?
下面的程序会编译,尽管我不想这样做:
#include< cstdlib>
//函数签名需要一个int
void函数(int i);
int main(){
int i {5};
函数(i); //< - 这是可接受的
char c {'a'};
函数(c); //< - 我不想这样编译
return EXIT_SUCCESS;
void function(int i){return;}
<请务必指出任何滥用术语和假设的情况。 您不能直接,因为 char
自动被提升为 int
。
你可以使用一个技巧:创建一个以 char
作为参数的函数,但不要实现它。它会编译,但你会得到一个链接器错误:
pre $ void函数(int i)
{
}
void function(char i);
//或者,在C ++中11
void function(char i)= delete;
使用 char
参数调用函数将会打破构建。
术语:非构造函数?你的意思是一个不是构造函数的函数吗?
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?
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;}
*please be sure to point out any misuse of terminology and assumptions
You can't directly, because a char
automatically gets promoted to int
.
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);
//or, in C++11
void function(char i) = delete;
Calling the function with a char
parameter will break the build.
Terminology: non-construcing functions? Do you mean a function that is not a constructor?
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