CODEWITHSUNDEEP
cpointersvisual-studio-2013
Dani
Dani

Reputation: 79

In C: Unhandled exception : 0xC0000005: Access violation writing location. program won't run

Visual Studio tells me the problem with my code is in the line where 

*p = *p1;

But I don't see what's wrong there.
The objective in this code is to build a function that will get a string and an int and will sort the string and return the letter in the index of the int.
For example for the string "build" it will sort it to "bdilu" if the int was 2 would return "i". Here is the code I wrote:

#include <stdio.h>
#include <conio.h>
#include <stdlib.h>
#include <string.h>
char letterAtIndex(char *s, int n);

void main(){
    char s, *arr = { "a string" };
    s=letterAtIndex(arr, 7);
    printf("%c", s);
    getch();
}
char letterAtIndex(char *s, int n){
    int t;
    char *p1 = s, *p2 = s, *p;
    while (*(++p2));
    while (p1 < p2){
        p = p1;
        while (p < p2){
            if (*p1>*p){
                t = *p;
                *p = *p1;
                *p1 = t;
            }
            p++;
        }
        p1++;
    }
    return s[n];
}

And this is the error: 

Unhandled exception at 0x00F81436 in justforstudy.exe: 0xC0000005: 
Access violation writing location 0x00F85859.

Upvotes: 1

Views: 4046

Answers (2)

Rabbid76
Rabbid76

Reputation: 210878

You pass a string literal to your function letterAtIndex. you can't write to it. Change char *arr = { "a string" }; to char arr[] = { "a string" }; and it will work.char arr[] is an array of elemts with type char which is read and writeable:

int main(){
    char arr[] = { "a string" };
    char s = letterAtIndex( arr, 7 );
    printf( "%c", s );
    return 0;
}

An other possibility would be to allocate dynamic memory and copy the string into it.

#include <string.h> // strcpy
#include <malloc.h> // malloc, free

int main(){
    const char *str = "a string";
    char *arr = malloc( strlen(str ) + 1 );
    strcpy( arr, str );
    char s = letterAtIndex( arr, 7 );
    printf( "%c", s );
    free( arr );
    return 0;
}

Upvotes: 3

a3f
a3f

Reputation: 8657

It's undefined behaviour to attempt changing a string literal. Use the type const char * for variables pointing to string literals to have the compiler help you spot when you attempt to do so.

You can fix your code by:

  • making a copy of the passed in string and sort that one and return it (You will need to free it afterwards).
  • Pass in a mutable buffer, e.g. by declaring arr as char arr[]

Upvotes: 2

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