using Array of pointers

Question

My first question here. I am using an array of pointers instead of a 2D array. Now, to display an element I can use *(arr[i]+j), where arr is the array i denotes the row and j denotes the column. However, when I try to assign a value to any element using the same notation the code stops working. I dont get a compilation error but when I run it it just stops working. Could anybody help me?

Here is my code

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

int main(void)
{
    int i,j,k;
    char temp2, temp, *arr[] = {
        "Brinda Roy", 
        "Rakesh Bai", 
        "Neha Saxen", 
        "Ankit Jain"};

    printf("%c",*(arr[3]+8));

    for(i=0;i<4;i++){
        for(j=0, k=9; j<=4, k>=5; j++, k--){
            temp =*(arr[i]+j);
            *(arr[i]+j)=*(arr[i]+k);
            *(arr[i]+k)=temp;
        }

    }

    printf("\nThe array is ");

    for(i=0; i<4; i++){
        printf("\n%s",arr[i]);
    }

    getch();
    return 0;
}

Open your debugger and inspect the values of your counters when the problem occurs. It will probably be very obvious what the issue is. — Ed Swangren
– Ed Swangren, Commented Jun 28, 2012 at 23:01
An array of pointers is just a 2D-array. arr[i] is equivalent to *(arr+i). — Trevor
– Trevor, Commented Jun 28, 2012 at 23:02
Your error might be a segmentation fault error, which means the array is going out of bounds somewhere. It usually occurs when your playing with pointers. Try debugging your code. — ppsreejith
– ppsreejith, Commented Jun 28, 2012 at 23:12
I have a problem with this line for(j=0, k=9; j<=4, k>=5; j++, k--) more specifically these exact characters j<=4, k>=5. What exactly do you expect to happen here? The comma operator will cause the for loop to only check the right most comparison. — Eric des Courtis
– Eric des Courtis, Commented Jun 28, 2012 at 23:15
Your code is like doing this: int *j = &2; *j = 3;. Where is the number you are trying to change stored? If a pointer points to a constant, you can't de-reference it and modify the result! — David Schwartz
– David Schwartz, Commented Jun 28, 2012 at 23:55

AusCBloke · Accepted Answer · 2012-06-28 23:14:31Z

5

The runtime crash is due to you modifying the string literals in arr. The string literals "Brinda Roy", "Rakesh Bai", "Neha Saxen" and "Ankit Jain" should be treated as const char[], as they may be stored in a read-only section of your program. Attempting to change their values causes undefined behaviour, as mentioned in the comments below.

If you want to edit these read-only strings, you're going to need to make copies of them into non-const buffers.

If you do:

char *str = "message"

you shouldn't modify the data pointed to by str because it's pointing to read-only memory.

If you do:

char str[] = "message"

you can modify the data pointed to by str because the read-only string literal "message" is being copied into the array str.

Your code falls into the first scenario above.

answered Jun 28, 2012 at 23:14

AusCBloke

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2 Comments

Michael Over a year ago

Minor nit, the string literals may be stored in read-only section. Writing to them is undefined behavior, and the compiler is free to put them into writable memory.

AusCBloke Over a year ago

@Michael: Yeah that's a better way of putting it.

sarnold · Accepted Answer · 2012-06-28 23:15:42Z

3

C allows strings defined in the source via this style to be stored in a read-only data section. Check this output:

$ cat storage.c
int main(int argc, char* argv[]) {
    char *arr[] = {"hello", "world"};
    arr[1][1]='a';
    return 0;
}
$ make storage
cc     storage.c   -o storage
$ readelf -p.rodata storage

String dump of section '.rodata':
  [     4]  hello
  [     a]  world

My compiler has placed the strings into the .rodata read-only data section. When I try to update the o to an a in the program, the program dies:

$ ./storage 
Segmentation fault (core dumped)

This is general applies to any code that defines strings as the following:

char *foo = "string";

If you want to create a modifiable string in the program source, you'd write it instead like this:

char foo[] = "string";

I'm not entire sure what the syntax would be to create an array of modifiable strings; at least the following ugly approach works, but I'm sure there's something better:

char a1[] = "hello";
char a2[] = "world";
char *arr[] = {a1, a2};

answered Jun 28, 2012 at 23:15

sarnold

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4 Comments

Swa Over a year ago

Ah So, instead of using an array of pointers, if I use a pointer to an array I would be able to modify the strings right?

sarnold Over a year ago

Note that the last line in my answer builds an array of pointers-to-char -- it isn't a pointer to an array.

Swa Over a year ago

Oh yeah. I see your point. Let me try it and get back to you. EDIT-It's working. Thanks. But is there a way to not declare separate arrays for each string?

sarnold Over a year ago

The only way I've found is something even worse: char arr[][4] = {{'h', 'i', 0}, {'m', 'o', 'm', 0}};. You have to know how long the longest string is -- and it allocates that much space for every string.

jpalecek · Accepted Answer · 2012-06-28 23:23:08Z

2

You try to modify string literals. This is forbidden. Note that even though this is not enforced by the type system ("abc" is not const char*), modifying a character literal "abc"[1]='c' leads to undefined behavior.

You have to move your data to a writable location. The easiest, though nonstandard way of doing this would be using the strdup function:

for(i=0; i<4; i++)
  arr[i]=strdup(arr[i]);

That will allocate memory on the heap for the string and copy the string to it. If you don't have the strdup function, you have to implement it using malloc, strlen and memcpy.

Note that as always with dynamically allocated memory, you have to free it; otherwise, you get a memory leak.

BTW the normal way to access both double pointer arrays and multidimensional arrays is arr[i][j].

edited Jun 28, 2012 at 23:23

answered Jun 28, 2012 at 23:14

jpalecek

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1 Comment

Swa Over a year ago

Thanks for your help. I just wanted to practice array of pointers so I tried to modify all my 2D array codes to array of pointers. I do know about arr[i][j] notation.Did not know about string literals. Thanks

Evans · Accepted Answer · 2012-06-28 23:17:17Z

1

You cannot modify the memory reserved this way:

{"Brinda Roy", 
 "Rakesh Bai", 
 "Neha Saxen", 
 "Ankit Jain"};

When you define a char* this way, it is only read memory.

answered Jun 28, 2012 at 23:17

Evans

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