How to remove first character from C-string?

Question

Can anyone please help me? I need to remove the first character from a char * in C.

For example, char * contents contains a \'\\n\'

Answer 1

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

int main ()
 {
char src[50] = "123456789123434567678";

char dest[16]={0};
 memcpy(dest, src+1,sizeof(src));
 printf("%s\n",dest);
 return(0);
}

src+1 -> indicate how many char you want to remove

Answer 2

Do not just increment the pointer if you have malloc'd any memory or your program will crash. free needs the original pointer. You can copy the pointer, make a new chunk of memory and memcpy it, access it as ptr+1 or any of a bunch of other ways, but the people who say just increment the pointer are giving you dangerous advice. You can run this sample program and see what happens when you "just increment the pointer".

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
int main(void)
{
    char *str = (char *)malloc(10);
    strcpy(str, "1234567890");
    printf("%s\n", str);
    str++;
    printf("%s\n", str);
    free(str);
}

Hint: Here's the result:

[mfisch@toaster ~]$ ./foo
1234567890
234567890
*** glibc detected *** ./foo: free(): invalid pointer: 0x08c65009 ***
======= Backtrace: =========
/lib/tls/i686/cmov/libc.so.6(+0x6b591)[0x724591]
/lib/tls/i686/cmov/libc.so.6(+0x6cde8)[0x725de8]
/lib/tls/i686/cmov/libc.so.6(cfree+0x6d)[0x728ecd]
./foo[0x80484e3]
/lib/tls/i686/cmov/libc.so.6(__libc_start_main+0xe6)[0x6cfbd6]
./foo[0x80483f1]
======= Memory map: ========
001c9000-001e4000 r-xp 00000000 08:01 2883609    /lib/ld-2.11.1.so
001e4000-001e5000 r--p 0001a000 08:01 2883609    /lib/ld-2.11.1.so
001e5000-001e6000 rw-p 0001b000 08:01 2883609    /lib/ld-2.11.1.so
006b9000-0080c000 r-xp 00000000 08:01 3015690    /lib/tls/i686/cmov/libc-2.11.1.so
0080c000-0080d000 ---p 00153000 08:01 3015690    /lib/tls/i686/cmov/libc-2.11.1.so
0080d000-0080f000 r--p 00153000 08:01 3015690    /lib/tls/i686/cmov/libc-2.11.1.so
0080f000-00810000 rw-p 00155000 08:01 3015690    /lib/tls/i686/cmov/libc-2.11.1.so
00810000-00813000 rw-p 00000000 00:00 0
00e4d000-00e4e000 r-xp 00000000 00:00 0          [vdso]
00fe0000-00ffd000 r-xp 00000000 08:01 2883667    /lib/libgcc_s.so.1
00ffd000-00ffe000 r--p 0001c000 08:01 2883667    /lib/libgcc_s.so.1
00ffe000-00fff000 rw-p 0001d000 08:01 2883667    /lib/libgcc_s.so.1
08048000-08049000 r-xp 00000000 08:01 9700477    /home/mfisch/foo
08049000-0804a000 r--p 00000000 08:01 9700477    /home/mfisch/foo
0804a000-0804b000 rw-p 00001000 08:01 9700477    /home/mfisch/foo
08c65000-08c86000 rw-p 00000000 00:00 0          [heap]
b7600000-b7621000 rw-p 00000000 00:00 0
b7621000-b7700000 ---p 00000000 00:00 0
b776f000-b7770000 rw-p 00000000 00:00 0
b7780000-b7783000 rw-p 00000000 00:00 0
bfc22000-bfc37000 rw-p 00000000 00:00 0          [stack]
Aborted

Answer 3

Here is my code.
So simple

#include <stdio.h>
#include<stdlib.h>
int main()
{
    char *str="hello";
    printf("%s",&str[1]);
    return 0;
}

Answer 4

char* contents_chopped = contents + 1;

This will result in contents_chopped pointing to the same string, except the first char will be the next after \n

Also, this method is faster.

Answer 5

It sounds as if you're under the impression that a char* "contains" characters. It does not. It merely points at a byte. The rest of the string is implied to consist of the subsequent byte in memory up until the next null byte. (You should also be aware that although the 'char' data type is a byte, by definition, it is not really a character - please be aware of Unicode - and nor is a byte necessarily an octet.)

The char* is not an array, either, although there may exist an array of characters such that the pointer is pointing to the beginning of that array.

Answer 6

if (contents[0] == '\n') 
    memmove(contents, contents+1, strlen(contents));

Or, if the pointer can be modified:

if (contents[0] == '\n') contents++;