Incompatible pointer type

[slushpuppy]

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

int main (int argc, char *argv[]) {
	char **pointer,*p;
	int i = 0;
	pointer = malloc(9);
	while (i <= 8) {
		pointer[i] = malloc(31);
		pointer[i] = "hello";
		i++;
	}
	 printf("%s\n",pointer[0]);
	 pointer = realloc(pointer,30);
	 while (i <= 19) {
		 pointer[i] = malloc(31);
		 pointer[i] = "hellllllooo";
		 i++;
	 }
}

The output message is a memory dump. I am compiling using gcc and the command $ gcc test.c. Thanks!

[Bachstelze]

Replace

        pointer = malloc(9);

with

        pointer = malloc(9*sizeof(char*));

Or better yet:

        pointer = calloc(9, sizeof(char*));

[slushpuppy]

Replace

        pointer = malloc(9);

with

        pointer = malloc(9*sizeof(char*));

Or better yet:

        pointer = calloc(9, sizeof(char*));

Hi, thanks. Could you kindly explain the significant of using the sizeof of a char pointer?

[Bachstelze]

firas@nobue ~ % cat test.c
#include <stdio.h>

int main(void)
{
        printf("%lu\n", sizeof(char*));
        return 0;
}
firas@nobue ~ % cc -o test test.c
firas@nobue ~ % ./test
8

The size of a char* is thus 8 bytes. Therefore, if you do pointer = malloc(9), pointer will not be allocated enough memory to hold all the data you will put in it.

[slushpuppy]

firas@nobue ~ % cat test.c
#include <stdio.h>

int main(void)
{
        printf("%lu\n", sizeof(char*));
        return 0;
}
firas@nobue ~ % cc -o test test.c
firas@nobue ~ % ./test
8

The size of a char* is thus 8 bytes. Therefore, if you do pointer = malloc(9), pointer will not be allocated enough memory to hold all the data you will put in it.

Thanks!