为什么不添加sqrt()在C ++中引起冲突? [英] Why doesn't adding sqrt() cause a conflict in C++?

问题描述

如果我编写一个具有C库函数签名的新函数，由于模棱两可，我预计会发生编译错误.但是，我不明白为什么下面的C ++代码没有错误.

If I write a new function that has the signature of a C library function, I expect a compile error due to the ambiguity. But, I can't understand why there is no error in the following C++ code.

#include <iostream>
#include <cmath>
using namespace std;

double sqrt(double number)
{
    return number * 2;  
}

int main( )
{
    cout << sqrt(2.3) << endl;
    cout << ::sqrt(2.3) << endl;
    cout << std::sqrt(2.3) << endl;
    return 0;
}

如果我将sqrt()的返回类型更改为int，则由于在cmath中带有双sqrt()的声明含糊性而发生编译错误.如何覆盖double sqrt()?(实际上，所有cmath函数都可以被覆盖，我不知道为什么.)

If I change the return type of sqrt() to int, then a compile error occurs due to the declaration ambiguity with double sqrt() in cmath. How is it possible to override double sqrt()? (Actually, all the cmath functions can be overridden, and I don't know why.)

推荐答案

程序具有未定义的行为.

The program has undefined behaviour.

[保留名称]
1 C ++标准库保留以下类型的名称:
1.1)—宏
1.2)—全局名称
1.3)—具有外部链接的名称
2如果程序在保留它的上下文中声明或定义了一个名称(本条明确允许的情况除外)，则其行为未定义.

[reserved.names]
1 The C++ standard library reserves the following kinds of names:
1.1) — macros
1.2) — global names
1.3) — names with external linkage
2 If a program declares or defines a name in a context where it is reserved, other than as explicitly allowed by this Clause, its behavior is undefined.

[外部名称]
4使用外部链接声明的C标准库中的每个函数签名均保留给既可以用作带有extern"C"和extern"C ++"链接的函数签名，也可以用作全局名称空间中名称空间作用域的名称.

[extern.names]
4 Each function signature from the C standard library declared with external linkage is reserved to the implementation for use as a function signature with both extern "C" and extern "C++" linkage, or as a name of namespace scope in the global namespace.

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