Sane通用清理？

Question

我想学习一种适用于以下场景的通用清理方法.请记住,这只是 - = SAMPLE = - .

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

int main(void)
{
  unsigned char *uchar1, *uchar2, *uchar3;

  if ((uchar1 = malloc(sizeof(*uchar1) * 10)) == NULL) {
    fprintf(stderr, "Error: malloc(uchar1);\n");

    return 1;
  }

  if ((uchar2 = malloc(sizeof(*uchar2) * 10)) == NULL) {
    fprintf(stderr, "Error: malloc(uchar2);\n");

    free(uchar1);
    return 1;
  }

  if ((uchar3 = malloc(sizeof(*uchar3) * 10)) == NULL) {
    fprintf(stderr, "Error: malloc(uchar3);\n");

    free(uchar1);
    free(uchar2);
    return 1;
  }

  /* do something */

  free(uchar1);
  free(uchar2);
  free(uchar3);
  return 0;
}

Answer 1

我想我知道你在做什么.像下面这样的Somehing可以更容易地确保您正确地管理资源.我相信这是一个不需要被视为有害的情况.

编辑反映了Summy0001优秀的观察并修复了错误价值的回报.

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

int main(void) 
{ 
  unsigned char *uchar1 = NULL;
  unsigned char *uchar2 = NULL;
  unsigned char *uchar3 = NULL; 
  int result = 0;

  if ((uchar1 = malloc(sizeof(*uchar1) * 10)) == NULL) { 
    fprintf(stderr, "Error: malloc(uchar1);\n");
    result = 1;
    goto CLEANUP0;
  } 

  if ((uchar2 = malloc(sizeof(*uchar2) * 10)) == NULL) { 
    fprintf(stderr, "Error: malloc(uchar2);\n"); 
    result = 1;
    goto CLEANUP1;
  } 

  if ((uchar3 = malloc(sizeof(*uchar3) * 10)) == NULL) { 
    fprintf(stderr, "Error: malloc(uchar3);\n"); 
    result = 1;
    goto CLEANUP2;
  } 

  /* do something */ 


  free(uchar3);
CLEANUP2:
  free(uchar2);
CLEANUP1:
  free(uchar1);
CLEANUP0:
  return result; 
}