小编Cam*_*zar的帖子

在这个程序中,我创建了两个指针(a,b),指向x和y的内存地址.在我创建的函数中,它应该交换a和b的内存地址(So b = a和a = b).当我编译它时给我一个错误(从'int'到'int*'的无效转换)这是什么意思？我正在传递一个指向该函数的指针,还是将其作为常规int读取？

#include <iostream>
using std::cin;
using std::cout;
using std::endl;
void pointer(int* x,int* y)// Swaps the memory address to a,b
{
    int *c;
    *c = *x;
    *x = *y;
    *y = *c;
}

int main()
{
    int x,y;
    int* a = &x;
    int* b = &y;
    cout<< "Adress of a: "<<a<<" Adress of b: "<<b<<endl; // Display both memory address

    pointer(*a,*b);
    cout<< "Adress of a: "<<a<<" Adress of b: "<<b<<endl; // Displays the swap of memory …