c++ - 可变参数模板在以下情况下如何工作？

Question

http://coliru.stacked-crooked.com/a/5d16c7e740a31a02

#include <iostream>
template<typename T> void P(T x) { std::cout << x; }
void foo(char a) {
    P(3);
    P(a);
}
template <typename... A>
void foo(int a, A... args) {
    foo(args...);
    P(a);
}
template <typename... A>
void foo(char a, A... args) {
    P(a);
    foo(args...);
}
int main()
{
    foo('1', '2', 48, '4', '5');
}

//1243548        My result (VS2015)
//12355248       correct result (clang, gcc)

如何生成“正确”的结果？

Answer 1

这里的技巧是 foo 的 int 重载不知道 char 重载，因为它还没有被看到，因此该级别的调用对自身递归:

template <typename... A>
void foo(int a, A... args) { /* only ever calls itself and above*/}

因此调用 foo(48, '4', '5')

将通过对 foo(int a, A...) 的额外调用进行递归，其中 chars 将被解释为 int!

因此您将按预期获得初始 12，但随后您将只打印字符的 ASCII 值，因此您将获得 52 的值'4'，'5' 的值为 53。 Clang 和 gcc 是正确的；您的 Visual Studio 版本打印了错误的值。 ( I was able to reproduce with VC 19.00.23506 )

这是调用跟踪(instrumented with some help from C++17):

Called foo(char a, A... args) with 1,2,48,4,5
Called P(T) with 1
Called foo(char a, A... args) with 2,48,4,5
Called P(T) with 2
Called foo(int a, A... args) with 48,4,5
Called foo(int a, A... args) with 52,5
Called foo(char a) with 5
Called P(T) with 3
Called P(T) with 5
Called P(T) with 52
Called P(T) with 48

现在，如果我们移动您的代码以便我们转发声明我们的模板(这始终是一个好习惯)，您将获得您想要的行为:

template<typename T> 
void P(T x);

template <typename... A>
void foo(int a, A... args);

template <typename... A>
void foo(char a, A... args);

// actual definitions below...

然后你会得到你想要的输出。这是该版本的堆栈跟踪 (demo):

Called foo(char a, A... args) with 1,2,48,4,5
Called P(T) with 1
Called foo(char a, A... args) with 2,48,4,5
Called P(T) with 2
Called foo(int a, A... args) with 48,4,5
Called foo(char a, A... args) with 4,5
Called P(T) with 4
Called foo(char a) with 5
Called P(T) with 3
Called P(T) with 5
Called P(T) with 48