C Programming - Changing chars of a string array element [duplicate] - c

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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

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);
}
Nope. You've allocated 1 char*. Then you treat it like it's size elements. You need to allocate size * sizeof(char*) bytes. (Note that this multiplication could also overflow.)
a[3]="Read";
Bad times. You are overwriting a[3] (which previously pointed to an allocation of 8 chars) with the location of a string literal, "Read". This leaks the previous allocation and also puts a non-modifiable string into a[3]. You should look into strncpy et al. for this.

Answer 2

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.

Answer 3

This:
a=(char**)malloc(sizeof(char*));
allocates space for one string. What you probably intend is something like:
char **a = NULL;
size_t number_of_strings = 8; /* for argument's sake */
a = malloc(number_of_strings * sizeof(char*));
if (!a)
return NOT_ENOUGH_MEMORY_ERROR;
At this point, you can dereference elements of a, e.g., a[3]. You'll still want to allocate space for those guys too:
char *staticStr = "Read";
a[3] = malloc(strlen(staticStr) + 1);
strncpy (a[3], staticStr, strlen(staticStr) + 1);
Start with that and see if rethinking how you're allocating memory will help you fix your other bugs.
Some notes:
You don't need to cast the result of malloc in C
You don't need to use sizeof(char) for allocating memory, which is always 1
You should be using a corresponding free() for each a[i] and for a itself, to prevent memory leaks

Answer 4

I see another problem, when you allocate memory for the:
a = (char **) malloc(sizeof(char *));
You are allocating memory for just one position, but you are using size positions. Then your code should be:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main ()
{
int i, size;
char **a;
char *ptr;
printf("Enter the size of the array:");
scanf("%d", &size);
a=(char**)malloc(size * sizeof(char*));
for(i=0;i<size;i++){
a[i]=(char*)malloc(sizeof(char)*8);
}
strcpy(a[3], "Read");
ptr=a[3];
while(*ptr!='\0'){
if(*ptr == 'e'){
*ptr = 'r';
}
ptr++;
}
printf("%s\n", a[3]);
system("pause");
return 0;
}
and of course, you need to free the allocated memory.
In your while when you try to change the 'e' to 'r' you are always point to the same char. You need a new pointer to walk throw the array.

Answer 5

Your while loop doesn't do anything.
while(*(a[3])!='\0'){
if(*(a[3]) == 'e'){
*(a[3]) = 'r';
}
}
It doesn't advance the pointer, it just keeps it at the first position.
A more proper-ish way would be to make a temporary pointer and use it to walk the string
char *temp = a[3];
while (*temp != '\0') {
if (*temp == 'e') *temp = 'r';
temp++;
}