Question

Possible Duplicate:
Problem with processing individual strings stored in an array of pointers to multiple strings in C

Ok so I’m trying to change a char of a string to another char in C. The thing is, each string is an element of a 1D array so essentially all together its a 2D array because a string itself is an array of chars. Anyways I have a problem creating code to do this. Is it even possible to do this? Any help is appreciated.

Here’s the code:

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

int main ()
{ 
int i, size;
char **a;

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

printf("Enter the size of the array:");
scanf("%d", &size);

for(i=0;i<size;i++){
 a[i]=(char*)malloc(sizeof(char)*8);
}

a[3]="Read";


while(*(a[3])!='\0'){
 if(*(a[3]) == 'e'){
    *(a[3]) = 'r';
 }
}

 printf("%s\n", a[3]);

 system("pause");
 return 0;

}

Answer 1

You didn’t allocate enough space for a. Instead of

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

you need

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

and obviously this must move to be after size has been read.

Once you sort that rather mundane problem out the fundamental problem is here:

a[3]="Read";

This makes the pointer a[3] point at a literal which cannot be modified. Instead you need to copy the contents of that literal into a[3]. Like this:

strcpy(a[3], "Read");

You must understand that a[3]=... assigns just the pointer a[3] and does not modify the string to which a[3] points.

Now, your code will obviously be in error if size is less than 4 since then a[3] would be out of bounds, but I guess a[3] is just transient while you debug this.

Your while loop is all wrong. Judging from your comments you want something like this:

char *p = a[3];
while (*p != '\0')
{
    if (*p == 'e')
        *p = 'r';
    p++;
}

No need to cast the return value of malloc in C, so remove the casts. sizeof(char) is always equal to 1 so you can remove that too.