char ** Problem

[parad0x]

Hello everyone,

I have a problem, I have a char ** called "buffer". I want to have the 'buffer' point to a number of char *'s which store up to "bufsize" characters. I use malloc to allocate the space which gives me the code:

nbufs = 8;
bufsize = 32;
char **buffer;
buffer = malloc(nbufs);
for(int i = 0; i<nbufs; i++){
buffer[i] = malloc(bufsize+1);
}
First off... is this right?

Now when I try to free the buffers I get an error.
If i try to free the buffer like this...

for(int i = 0; i<nbufs; i++){
free(buffer[i]);
}

I get this error at compile time:

*** glibc detected *** free(): invalid next size (fast): 0x0804a008 ***
Aborted

If I just ignore the for statement and try this...

free(buffer);

I get the error at compile time:

*** glibc detected *** free(): invalid next size (fast): 0x0804a008 ***
Aborted

Is my char ** buffer not malloc'd right? What am I missing? Thank you ahead of time for your help.

[ZuK]

it should be

buffer = malloc(nbufs * sizeof( char *) );

Kurt

[vart]

First off... is this right?

no it is not
buffer[i] takes more then 1 byte, so the first malloc allocates not enough memory

buffer = malloc(nbufs * sizeof (*buffer));

read the FAQ about malloc

[parad0x]

Thanks for the help!

[ssharish2005]

use code tags. here is a sample code

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

int main()
{
    char **buf;
    int i;
        
    buf = malloc(5 * sizeof (*buf));
    
    for(i=0;i<5;i++)
        buf[i] = malloc(20);
        
    for(i=0;i<5;i++)
        strcpy(buf[i],"This is a test");
    
    for(i=0;i<5;i++)
        printf("%s\n",buf[i]);
    
    for(i=0;i<5;i++)
        free(buf[i]);
    
    getchar();
    return 0;
}
/* my output
This is a test
This is a test
This is a test
This is a test
This is a test
*/

ssharish2005

[quzah]

for(i=0;i<5;i++)
        free(buf[i]);
free( buf );

Missed something.

Quzah.