为什么它不给歧义带来错误? [英] Why it is does not give error about ambiguity?

问题描述

请参见此代码

Please see this code

#include "stdafx.h"
#include "iostream"

void fun(int i)
{
    std::cout<<"Int Overload called";
}

void fun(int *i)
{
    std::cout<<"Pointer Overload called";
}

int main ()
{

    fun(0); \\gets called successfully 

  return 0;
}

输出:称为Int重载

我的问题是为什么调用fun不会导致歧义，编译器如何调用"int"实现而不是指针的实现.

Output : Int Overload called

My question is why call to fun does not leads to ambiguity, how compiler calls ''int'' implementation and not pointer''s implementation.

推荐答案

好问题！

在 http://www.learncpp.com/cpp-tutorial/76-function-overloading找到了答案/ [^ ] .
基本上，(编译器)认为int比int * int *更匹配.
您可能还会喜欢: http://xania.org/200711/ambiguous-overloading [

Nice question!

Found an answer at http://www.learncpp.com/cpp-tutorial/76-function-overloading/[^].
Basically, int is considered (by your compiler) a better match than int * for 0.
You may also like this: http://xania.org/200711/ambiguous-overloading[^]

Hope this helps,
Pablo.

编译器调用int实现，因为int *类型是无符号的int不仅仅是int

现在尝试更改

Compiler calls int implementation because int* type is unsigned int not just int

Now try changing

void fun(int i)
{
    std::cout<<"Int Overload called";
}

到

to

void fun(unsigned int i)
{
    std::cout<<"Int Overload called";
}

编译器摇摇晃晃地说
``错误C2668:``乐趣'':对重载函数的模棱两可的调用''


好问题

The compiler cribs and says
''error C2668: ''fun'' : ambiguous call to overloaded function''


Good question

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