Access violation writing location when working with pointers to char

Question

I am writing a very simple program that removes duplicate chars from a string. I ran it visual studio and got the error:

Unhandled exception at 0x00d110d9 in inteviews.exe: 0xC0000005: Access violation writing location 0x00d27830.

I really don't see what the problem is. current cell gets the value of the next cell.

void remove(char *str, char a) {
    while (*str != '\0') {
        if (*(str+1) == a) {
            remove(str + 1, a);
        }

        *str = *(str +1 );//HERE I GET THE ERROR
        ++str;
    }
}


int _tmain(int argc, _TCHAR* argv[])
{
    char *str = "abcad";

    while (*str != '\0') {
        remove(str,*str);
        str++;
    }

    std::cout << str << std::endl;

    return 0;
}

EDIT:

I already tried to change it to char str[] = "abcad" but I still get the same error.

possible duplicate of [Getting Segmentation Fault](http://stackoverflow.com/questions/5834032/getting-segmentation-fault) — MByD, Dec 10 '11 at 19:00
@rob You have asked 50 questions here but voted only once. This is rather poor. Please attempt to participate more fully in future. — David Heffernan, Dec 10 '11 at 19:33
first perhaps come clear with yourself if you are writing C or C++ — Jens Gustedt, Dec 10 '11 at 19:57

score 11 · Accepted Answer · answered Dec 10 '11 at 18:59

11

You're attempting to modify a string literal. You can't do that.

char *str = "abcad";

That's a string literal. It's created in read-only memory therefore attempting to write to it is an access violation.

answered Dec 10 '11 at 18:59

Brian Roach

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what is the reason i can't change it? and what is the way to by pass it? thanks – Asher Saban Dec 10 '11 at 19:00
@rob - initialize as an array or `malloc()` the memory and use `snprintf()` to copy `abcad` into that memory. – Brian Roach Dec 10 '11 at 19:04
This is just the tip of the iceberg. – David Heffernan Dec 10 '11 at 19:15
@David - yeah ... I didn't try to decipher his code any further as that was the the specific issue causing the access violation. – Brian Roach Dec 10 '11 at 19:33

score 6 · Answer 2 · answered Dec 10 '11 at 19:00

6

One problem is that you created a read-only string literal and attempted to modify it:

char *str = "abcad"; // String literals are read-only!

You could use a char array instead:

char str[] = "abcad";

answered Dec 10 '11 at 19:00

Mark Byers

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already tried that and i get the same error at the same line.. any ideas? – Asher Saban Dec 10 '11 at 19:04

David Heffernan · Answer 3 · 2011-12-10T19:48:36.427

There are all sorts of problems with your program. I begun by trying to write them all down but I feel that code is irredeemable. It has indexing errors, parameter passing errors, dubious recursion and so on.

The other answers that point out the error of trying to modify a read-only literal are correct. That is the cause of the error in the code you posted.

The main reason for your troubles, in my view, is that the code is harder to write when you only have a single buffer. You have tied yourself in knots trying to get around this limitation in your design, but with a second buffer to work with, the code is trivial.

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

int main(void)
{
    const char *input = "abcad";
    char *output = malloc(strlen(input)+1);

    char *in = input;
    char *out = output;
    while (*in)
    {
        if (*in != input[0])
        {
            *out = *in;
            out++;
        }
        in++;
    }
    *out = '\0';

    printf("%s\n", output);

    free(output);

    return 0;
}

If you want to get really clever you can in fact manage happily with just a single buffer, so long as you keep two distinct pointers for iteration.

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

int main(void)
{
    char str[] = "abcad";
    char compare = str[0];

    char *in = str;
    char *out = str;
    while (*in)
    {
        if (*in != compare)
        {
            *out = *in;
            out++;
        }
        in++;
    }
    *out = '\0';

    printf("%s\n", str);

    return 0;
}

Note that we had to take a copy of first character in the buffer, the character being removed, since that may be modified by the iteration.

So now you are back where you started, with a single buffer. But now the code works and is easy to understand.

Note that my answer is written in C as per your tag, but note that your code is C++.

score 0 · Answer 4 · answered Jun 29 '22 at 00:10

Since string literal is created in read-only memory, attempting to write to it is an access violation. What you can do is strcpy(dst, src) to a character array.

#include <stdlib.h>
int _tmain(int argc, _TCHAR* argv[])
{
    char *str = "abcad";
    char str2[10];
    strcpy(str2, str);
    while (*str2 != '\0') {
        remove(str2, *str2);
        str2++;
    }
}

Access violation writing location when working with pointers to char

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