What means "write access violation" in C and how do I solve this?

Question

So I am completly new to the C language. I come from a Visual Basic and C# background, languages where I usually don't need to care about low-level stuff. Recently started to learn about pointers and I am doing some of the leetcode challenges to get some practice.

**Here is the challenge text: ** Write a function to find the longest common prefix string amongst an array of strings. If there is no common prefix, return an empty string "".

Here is my attempt:

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

char* longestCommonPrefix(char** strs, int strsSize) {
    if (strsSize == 1) return strs[0];

    if (strs[0] == "") return "";

    char* result = strs[0];
    size_t result_length = strlen(result) - 1;
    char* result_end = strs[0] + result_length;

    for (size_t i = 1; i < strsSize; ++i) {
        
        size_t length = strlen(strs[i]) - 1;
        
        if (length < 1) return "";
        
        if (length < result_length) {
            result_end = strs[0] + length;
            result_length = length;
        }

        char* temp1 = result;
        char* temp2 = strs[i];

        while (*temp1 == *temp2 && temp1 < result_end) {
            ++temp1;
            ++temp2;
        }

        if (temp1 < result_end) {
            result_end = temp1 - 1;
            result_length -= strlen(result_end) - 1;
        }
    }

    *(result_end + 1) = '\0';
    
    return result;
}


int main(int argc, char* argv[]) {
    char* a[3];
    a[0] = "flower";
    a[1] = "flight";
    a[2] = "flow";

    char* b = longestCommonPrefix(a, 3);
    printf(b);
}

Why do I get a write access violation error on the line *(result_end + 1) = '\0'; ? Googling this I found articles saying c strings are readonly, why tho and what would be a work around? Also any general C tips or links to such things are appreciated, thank you.

EDIT: Thanks @Chris for the solution. Here is my new code which works just fine. Also thanks at @@WeatherVane for a general C tip, I also included his knowledge into my new code.

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

char* longestCommonPrefix(char** strs, int strsSize) {
    if (strsSize == 1) return strs[0];

    if (strs[0][0] == 0) return "";

    char* result = _strdup(strs[0]);
    size_t result_length = strlen(result) - 1;
    char* result_end = strs[0] + result_length;

    for (size_t i = 1; i < strsSize; ++i) {

        char* str = _strdup(strs[i]);
        
        size_t length = strlen(str) - 1;
        
        if (length < 1) return "";
        
        if (length < result_length) {
            result_end = str + length;
            result_length = length;
        }

        char* temp = result;

        while (*temp == *str && temp < result_end) {
            ++temp;
            ++str;
        }

        if (temp < result_end) {
            result_end = temp - 1;
            result_length -= strlen(result_end) - 1;
        }
    }

    *(result_end + 1) = '\0';
    
    return result;
}


int main(int argc, char* argv[]) {
    char* a[3];
    a[0] = "flower";
    a[1] = "flight";
    a[2] = "flow";

    char* b = longestCommonPrefix(a, 3);
    printf(b);
}

Answer 1

It means you've written to some memory you shouldn't.

In main you have an array of char pointers. But each points to a string literal which is const char *.

In longestCommonPrefix you then attempt to modify these string literals, which invokes undefined behavior.

If you wish to modify these values in your longestCommonPrefix function, you should copy these string literals into allocated memory. The strdup is a very straightforward way to do this, as long as you ensure you free the dynamically allocated memory afterwards.

char *a[3];
a[0] = strdup("flower");
a[1] = strdup("flight");
a[2] = strdup("flow");

If you're tools don't allow use of this function, you can either declare a to hold 3 fixed length strings, or manually handle the dynamic memory allocation and copying.