c++ - 为什么 C++ 链接器对 ODR 违规保持沉默？

Question

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:

Header1.h

#include <iostream>

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

Header2.h

void Function1();

Header3.h

void Function2();

源1.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。 源2.cpp

#include "Header1.h"

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

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

主要.cpp

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

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

问题是什么将打印 Function1() 和 Function2()？答案是未定义的行为。

我希望在输出中看到: 特化1 特化2

但我看到: 专精2 特化2

为什么 C++ 编译器对 ODR 违规保持沉默？在这种情况下，我宁愿编译失败。

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。

Answer 1

编译器是静默的，因为它不是要求通过[basic.def.odr/4]发出任何东西:

Every program shall contain exactly one definition of every non-inline function or variable that is odr-used in that program outside of a discarded statement; no diagnostic required. The definition can appear explicitly in the program, it can be found in the standard or a user-defined library, or (when appropriate) it is implicitly defined (see [class.ctor], [class.dtor] and [class.copy]). An inline function or variable shall be defined in every translation unit in which it is odr-used outside of a discarded statement.