为什么获取已声明函数的地址才有效？
Why is Taking the Address of a Function That is Declared Only Working?
我问过 question here about whether taking the address of a function forces the compilation of said function specifically with regard to Substitution-Failure-Is-Not-An-Error. The most direct answer to this can be found here:
Informally, an object is odr-used if its address is taken, or a reference is bound to it, and a function is odr-used if a function call to it is made or its address is taken. If an object or a function is odr-used, its definition must exist somewhere in the program; a violation of that is a link-time error.
但我测试过的所有编译器都表明这是完全可行的：
void foo(int);
auto bar = &foo;
这不合法吧？但如果不是，为什么要建设？
Every program shall contain exactly one definition of every non-inline function or variable that is odr-used in
that program; no diagnostic required.
foo 是 odr-used，但没有定义（大概 - 否则问题是没有意义的）。该程序格式错误，但由于不需要诊断，因此编译有效。
通常，是链接器发现了定义的缺失，而不是编译器，因为定义很容易出现在不同的翻译单元中。典型的例子是试图将缺少定义的 static const int 传递给对 std::max() 或 std::min() 的调用。
您的示例可以正常工作，因为从未使用过该地址，因此链接器从不搜索该符号。
如果您尝试打印 bar，链接将失败。
void foo(int);
auto bar = &foo;
cout << (void*) bar;
我问过 question here about whether taking the address of a function forces the compilation of said function specifically with regard to Substitution-Failure-Is-Not-An-Error. The most direct answer to this can be found here:
Informally, an object is odr-used if its address is taken, or a reference is bound to it, and a function is odr-used if a function call to it is made or its address is taken. If an object or a function is odr-used, its definition must exist somewhere in the program; a violation of that is a link-time error.
但我测试过的所有编译器都表明这是完全可行的：
void foo(int);
auto bar = &foo;
这不合法吧？但如果不是，为什么要建设？
Every program shall contain exactly one definition of every non-inline function or variable that is odr-used in that program; no diagnostic required.
foo 是 odr-used，但没有定义（大概 - 否则问题是没有意义的）。该程序格式错误，但由于不需要诊断，因此编译有效。
通常，是链接器发现了定义的缺失，而不是编译器，因为定义很容易出现在不同的翻译单元中。典型的例子是试图将缺少定义的 static const int 传递给对 std::max() 或 std::min() 的调用。
您的示例可以正常工作，因为从未使用过该地址，因此链接器从不搜索该符号。
如果您尝试打印 bar，链接将失败。
void foo(int);
auto bar = &foo;
cout << (void*) bar;