为什么C ++链接程序对违反ODR保持沉默?

问题描述

让我们考虑一些综合但富有表现力的例子.假设我们有Header.h:

Let's consider some synthetic but expressive example. Suppose we have Header.h:

Header1.h

Header1.h

#include <iostream>

// Define generic version
template<typename T>
inline void Foo()
{
    std::cout << "Generic\n";
}

Header2.h

Header2.h

void Function1();

Header3.h

Header3.h

void Function2();

Source1.cpp

Source1.cpp

#include "Header1.h"
#include "Header3.h"

// Define specialization 1
template<>
inline void Foo<int>()
{
    std::cout << "Specialization 1\n";
}

void Function1()
{
    Foo<int>();
}

后来，我或其他人在另一个源文件中定义了类似的转换.Source2.cpp

Later I or some else defines similar conversion in another source file.Source2.cpp

#include "Header1.h"

// Define specialization 2
template<>
inline void Foo<int>()
{
    std::cout << "Specialization 2\n";
}

void Function2()
{
    Foo<int>();
}

main.cpp

#include "Header2.h"
#include "Header3.h"

int main()
{
    Function1();
    Function2();
}

问题是什么将打印Function1()和Function2()?答案是不确定的行为.

The question is what will print Function1() and Function2()? The answer is undefined behavior.

我希望在输出中看到:专业1专业化2

I expect to see in output: Specialization 1Specialization 2

但是我看到:专业化2专业化2

But I see: Specialization 2Specialization 2

为什么C ++编译器对违反ODR保持沉默?在这种情况下，我希望编译失败.

Why C++ compilers are silent about ODR violation? I would prefer compilation to be failed in this case.

我只找到一种解决方法:在未命名的命名空间中定义模板函数.

I found only one workaround: define template functions in unnamed namespace.

推荐答案

编译器处于静默状态，因为 [basic.def.odr/4] :

The compiler is silent, because it's not required to emit anything by [basic.def.odr/4]:

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