Thread: can't get it to work!!!

char (*p)[8];
char line[8];

Two things:

1) Why are you using parenthesis there? If you want an array of pointers, just do:

char *p[8];

2) That's a mighty short line.

Regarding printing them all, you have to use a loop to print them:

for(x=0;x<8;x++)printf("%s\n",p[x]);


Quzah.

OK, keeping your format the same.... here's a fixed version.

Code:

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

int main(void)
{
    char    *p[8];
    char    line[10];
    int     k;

    for (k = 0; k < 8; k++)
    {
        scanf("%s", line);
        if (feof(stdin)) break;

        p[k] = malloc(sizeof line);
        if (p == NULL)
        {
            printf("storage: no storage available\n");
            return(1);
        }

        strcpy(p[k], line);
    }

    for (k = 0; k < 8; k++)
    {
        printf("%s\n", p[k]);
        free(p[k]);
    }

    return(0);
}

I don't recommend using scanf() to get a "line" as it won't, it will get a word (using %s, it inputs up to the next white space character.

Note that the free() is done with the for loop, this is because you malloc()'d 8 times within the previous loop. You cannot free all 8 allocations in one go.

>malloc(sizeof line)
There's no pointing in mallocing this size, the string might be shorter. Use this instead
>malloc(strlen(line)+1)
Also, malloc() requires stdlib.h

>if (feof(stdin)) break;
if you reach eof, because the user hits CTRL+d/z, you must not do the printf() loop 8 times, as the array won't have 8 valid pointers in it.

Originally posted by Unregistered
I just need a clarification what does the +1 do in
malloc(strlen(line)+1)

The amount of memory needed to hold the string is:
- The number of bytes within the string as returned by strlen()
- Plus one more for the NULL terminator.

If you forget the +1, you'll allocate 1 byte short, and when you do the strcpy(), the NULL will overwrite data that doesn't belong to that variable, maybe not even your program.

here is my suggestion for your program
not sure if this is what you wish but i find ur code a bit too cryptic :P

Code:

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

int main()
{
	char line[25];	// change if u wish larger lines :)
	char *p[8];
	int index;

	printf("type 8 lines of garbage");
	for(index = 0; index < 8; index++) {
		p[index] = malloc(sizeof line);
		scanf("%s", p[index]);
	}

	printf("now printing the 8 lines");
	for(index = 0; index < 8; index++) {
		printf("%s\n", p[index]);
	}

	for(index = 0; index < 8; index++) {
		free(p[index]);
	}

	return 0;
}

p[index] = malloc(sizeof(line[index]));

You realize that you're only allocating one character here, right? sizeof( array[x] ) gives you the size of one array element. Not the size of the entire line.

Quzah.

Originally posted by iNFERNaL
hmm would changing that line to

Code:

p[index] = malloc(sizeof line);

fix it? cause i tryed that way and didnt have any error..

Yes, it would. But I'd still recommend reading into a buffer, then malloc()'ing only the required amount of memory.

Also, you must check the return from malloc. If it fails, it will return NULL, and in your code, the program will crash.