memcpy（）似乎不起作用

Question

I'm having trouble with memcpy() and I don't have a clue on where I went wrong.

Code can be seen here: http://pastebin.com/tebksExR

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

typedef struct tmp__ {
   unsigned int num;
   unsigned short id;
   unsigned short type;
} tmp_str;

int
main(int argc, char **argv)
{
   tmp_str hdr;
   char *str = NULL;

   str = calloc(18, sizeof(char));
   memset(&hdr, 0, sizeof(hdr));

   hdr.num = 0;
   hdr.id = 0;
   hdr.type = 21845;
   memcpy((void *) str, (void *) &hdr, sizeof(hdr));
   printf("STR: %s\n", str);

   free(str);
   return 0;
}

On executing it, all I see is just "STR". Nothing is seen on the memory region pointed by str too.

(gdb) b 23
Breakpoint 1 at 0x8048494: file memcpy.c, line 23.
(gdb) run
Starting program: /home/a.out 

Breakpoint 1, main (argc=1, argv=0xbffff234) at memcpy.c:23
23     memcpy((void *) str, (void *) &hdr, sizeof(hdr));
(gdb) n
24     printf("STR: %s\n", str);
(gdb) n
STR: 
26     free(str);
(gdb) info locals 
hdr = {num = 0, id = 0, type = 21845}
str = 0x804b008 ""

Where did I go wrong?

Thanks!

Answer 1

When you specify %s , printf expects a null-terminated string. It can't tell that you allocated 18 bytes and you want the contents of those bytes printed. It looks at *str , sees a null byte, and stops looking.

Answer 2

Your data is getting moved. If you declared:

tmp_str* str2 = (tmp_str*)str;

then it would display properly in the debugger. You're just looking at the data wrong-- your code is executing properly.

Answer 3

As was pointed out, since the first byte of str is '\\0' . a simple printf("%s",str); statement will do nothing (it will stop at the first byte which indicates "end of string"). Instead, you could try

int ii;
printf("STR in hex:\n");
for(ii = 0; ii < sizeof(hdr); ii++) {
  printf("%02x ", str[ii]);
}
printf("\n");

This will now print every byte in str as a hex number; you will see that the copy went OK.

If you would like, you could replace this with

int ii;
printf("STR in hex:\n");
for(ii = 0; ii < sizeof(hdr); ii++) {
  printf("%c", str[ii]);
}
printf("\n");

And you will see the actual characters (but some of them may be "unprintable" and might have unexpected side effects in your terminal).

Answer 4

str is a string ( char * terminated by '\\0' ) and hdr is a structure, which doesn't even have a string field. If you want to convert hdr to a human-readable string, you'll have to use printf or sprintf with the proper conversion specifiers.

Answer 5

Well, you set hdr.num = 0; , so the first byte of copied chunk of memory is equal to 0 , so the first character of your string is NULL, which marks it's ending, so nothing is printed.