c++ - 模板参数推导错误

Question

template <typename T>
void foo(int i)
{
  //nothing inside
}

int main()
{
   foo(5); //fails
   foo<int>(5); //works
}

为什么 foo(5) 失败但 foo< int >(5) 有效？

Answer 1

你可能想写

template <typename T>
void foo(T i)          // note the type of i
{
  //nothing inside
}

---

更新

完整代码如下

#include <iostream>
using std::cout;

template <typename T>
void foo( T i ) {
    cout << __PRETTY_FUNCTION__ << " called with " << i << "\n";
}

int main() {
    foo( 5 );
    foo<int>( 7 );
}

和输出:

void foo(T) [with T = int] called with 5
void foo(T) [with T = int] called with 7