通过动态分配参考了解传递的参数 [英] understand passing parameters by reference with dynamic allocation

问题描述

我正在尝试了解如何使用C语言通过引用传递参数。
因此，我编写了这段代码来测试参数传递的行为：

I'm trying understand how to pass a parameter by reference in C language. So I wrote this code to test the behavior of parameters passing:

#include <stdio.h>
#include <stdlib.h>

void alocar(int* n){
   n = (int*) malloc( sizeof(int));
   if( n == NULL )
      exit(-1);
   *n = 12;
   printf("%d.\n", *n);
}
int main()
{
   int* n;
   alocar( n );
   printf("%d.\n", *n);
   return 0;
}

此处打印：

12.
0.

示例2：

#include <stdio.h>
#include <stdlib.h>

void alocar(int* n){
   *n = 12;
   printf("%d.\n", *n);
}

int main()
{
   int* n;
   n = (int*) malloc(sizeof(int));
   if( n == NULL )
      exit(-1);
   alocar( n );
   printf("%d.\n", *n);
   return 0;
}

打印：

12.
12.

这两个程序有什么区别？

What's the difference of this two programs?

推荐答案

C是传递值，它不提供传递引用。
在您的情况下，指针（而不是它指向的对象）被复制到函数参数中（指针通过值传递-指针的值是地址）

C is pass-by-value, it doesn't provide pass-by-reference. In your case, the pointer (not what it points to) is copied to the function paramer (the pointer is passed by value - the value of a pointer is an address)

void alocar(int* n){
   //n is just a local variable here.
   n = (int*) malloc( sizeof(int));
  //assigning to n just assigns to the local
  //n variable, the caller is not affected.

您想要以下内容：

int *alocar(void){
   int *n = malloc( sizeof(int));
   if( n == NULL )
      exit(-1);
   *n = 12;
   printf("%d.\n", *n);
   return n;
}
int main()
{
   int* n;
   n = alocar();
   printf("%d.\n", *n);
   return 0;
}

或：

void alocar(int** n){
   *n =  malloc( sizeof(int));
   if( *n == NULL )
      exit(-1);
   **n = 12;
   printf("%d.\n", **n);
}
int main()
{
   int* n;
   alocar( &n );
   printf("%d.\n", *n);
   return 0;
}

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