Question

#include<stdio.h>
#include<string.h>
int main()
{
  unsigned char *s;
  unsigned char a[30]="Hello world welcome";
  memcpy(s,&a,15);
  printf("%s",s);
  return 0;
}

这使我陷入分裂的错误。请帮助我确定这一错误

Answer 1

<代码>a已成为指点器,在您通过<代码>a>查询时,即:a ;a> 查询,而不是在座标。此外,你们还需要分配一些记忆,以复制大街的雕像。

Another error is that via memcpy you copy only 15 bytes, this means that your string is not zero terminated ( ) this results in printf() trying to print s until it reaches 0 which could never occur. So you must use strcpy, give the proper length argument or terminate the string to zero yourself.

#include<stdio.h>
#include<string.h>
int main()
{
  unsigned char *s;
  unsigned char a[30]="Hello world welcome";

  s = malloc(strlen(a) + 1); // + 1 for the 0 character
  if (s == NULL)
  {
      perror("malloc");
      exit(1);
  }

  // copy the whole string
  memcpy(s, a, (strlen(a) + 1)); // + 1 for the 0 character
  printf("%s",s);

  // copy the first 15 bytes and 0 terminate
  memcpy(s, a, 15);
  s[15] =   ;
  printf("%s",s);

  // copy via string copy
  strcpy(s, a);
  printf("%s",s);

  free(s)

  return 0;
}

Answer 2

页: 1 从现在起,它只是一个未入选的点,即(很可能)点到:

unsigned char *s = malloc(16);

如同所有记忆分配一样,在你重新使用时,应取消:

free(s);

EDIT:另一个错误(我忽略了)是,你在打电话memcpy后,你必须终止NCL。

memcpy(s,a,15);
s[15] =   ;

或者,你可以使用<代码>strcpy(),将体格划分为15个特性,但你需要分配足够时间储存所有<代码>a/代码>(包括其NUL-决定因素):

unsigned char a[30]="Hello world welcome";
unsigned char *s = malloc(strlen(a) + 1);   //  Allocate
strcpy(s,a);        //  Copy entire string
s[15] =   ;       //  Truncate to 15 characters by inserting NULL.
printf("%s",s);
free(s);            //  Free s

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