Thread: Dynamic Memory Allocation

In this exercise I can't see what am I doing wrong. Compiler says:

Code:

wk2prog4.c: In function `main':
wk2prog4.c:20: warning: char format, different type arg (arg 2)

I don't understand it, it should be right. And My code is:

Code:

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

int main()
{
    int n,i;
    char **a;
    char buffer[101];
    
	
    printf("How many words do you want to enter?\n");
    scanf("%d", &n);
	
    a = (char **) malloc(n * sizeof(char *));
	
    printf("Enter %d words, each containing no more than 100 characters\n", n);
    for(i = 0; i < n; i++)
    {
	scanf("%s", &buffer);
        strcpy(a[i], buffer);
    }
	
    printf("\nYou entered:\n");
    for(i = 0; i < n; i++)
    {
        printf("%s\n", a[i]);
    }


    for(i = 0; i < n; i++)
    {
        free(a[i]);
    }

    return 0;
}

Where's line 42?

Originally Posted by joeprogrammer

>also if i compile short cut "gcc" the program does not give any error but when the program runs segmentation fault happens.
That's probably because of the 2 problems mentioned already: you're not allocating memory properly, and there's no reason to use the & symbol because it's already a pointer (if you see what I mean).

In the exercise it wants me to allocate the memory by

Code:

a = (char **) malloc(n * sizeof(char *));

AND wants me to use

Code:

char **a

I removed the ampersand , the program compiles fine but I get segmentation error at the middle of the program.

Thank you Joeprogrammer and MacGyver, I fixed the problem.

Code:

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


int main()
{
    int n,i;
    char **a;
    char buffer[101];
    
    printf("How many words do you want to enter?\n");
    scanf("%d", &n);
	
    a = (char **) malloc(n * sizeof(char *));
	
    printf("Enter %d words, each containing no more than 100 characters\n", n);
    
	for(i = 0; i < n; i++)
    {
	scanf("%s", &buffer);
	a[i] = (char *) malloc(strlen(buffer) + 1);
        strcpy(a[i], buffer);
    }
	
    printf("\nYou entered:\n");
    
	for(i = 0; i < n; i++)
    {
        printf("%s\n", a[i]);
    }

    for(i = 0; i < n; i++)
    {
        free(a[i]);
    }

    return 0;
}

> a = (char **) malloc(n * sizeof(char *));
You're still casting malloc - this is C
Also, there is no free(a) at the end.

> scanf("%s", &buffer);
You don't need the & in this case - it's the exception to the rule.
Arrays are turned into a pointer to the first element when you pass them to a function anyway.