具有指针参数的C ++模板函数

Question

I have this code

template <class N>
inline N swap(N a, N b) {
    int c = *a;
    *a = *b;
    *b = c;
}

with this function I get this error: error: 'N' does not name a type Error compiling.

This is my normal function.

inline void swap(int *a, int *b) {
    int c = *a;
    *a = *b;
    *b = c;
}

My problem is that I need to use this function with unsigned and normal ints. Is it possible to do this with templates.

Answer 1

I think you want something like this:

template<typename T>
inline void swap(T* a, T* b) // Use T* if you're expecting pointers
       ^^^^ // Notice the return type
{
    T c = *a;
    *a = *b;
    *b = c;
}

Answer 2

将指针的声明从int c更改为N* c因为它可能会使用其他数据类型作为参数，此外您不能将指针的值放在普通变量中。