c - C中的指针传递/引用传递

Question

我知道如何在函数中传递指针，我觉得我仍然混淆两个术语，例如按引用传递和按指针传递。

有人问我:

Write a program that uses pointers to pass a variable x by reference to a function where its value is doubled. The return type of the function should be void. The value of the variable should be printed from main() after the function call. The original value of x is 5.

我写了下面的代码。能否根据问题的需求告诉我哪个评论号是对的，解开我的疑惑？

#include <stdio.h>
//double pointer
//void double_ptr(int **x_ptr)
//{
 //   **x_ptr *=2;

//}

void double_ptr(int *x_ptr)
{
    *x_ptr *=2;
}


int main()
{
    int x=5;
    int *x_ptr=&x;      // single pointer

    //double_ptr(*x_ptr);  // this is out of question's demand.
    //int**x_x_ptr=&x_ptr;    // double pointer
    //double_ptr(&x_ptr);     // 1. pass with double pointer
    //double_ptr(x_x_ptr);      //2. pass with double pointer
    //printf("%d",**x_x_ptr); //3.print with double pointer

    double_ptr(x_ptr);     //4.pass by pointer?
    //double_ptr(&x);          //5.pass by reference?
    printf("%d",*x_ptr);
    return 0;
}

Answer 1

指针传递实际上是值传递，C中没有引用传递这种概念。

在 C 中，通过引用传递 == 通过指针传递。

void double_ptr(int *x_ptr)
{
      *x_ptr *=2;
}
int main()
{
      int x=5;
      double_ptr(&x); /* a value is being passed*/
}

按引用传递(在 C++ 中):

void double_ptr(int& x_ptr)
{
   x_ptr *=2;
}
int main()
{
     int x=5;
     int &x_ptr=x;
     double_ptr(x_ptr); /* no temporary is being created i.e. no extra memory allocation is here*?
}