Why does the program stop if I'm using strcpy() on two strings with the exact same size in C?

Question

I use CodeBlocks for Windows as my IDE. In this program, I am trying to copy two different strings (string[8][0] with string[9][0]), and I can't, even though they have the exact same length. I don't understand why the program isn't working because, when I used two different strings with the same length in an exercise the program, worked.

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

int main()
{
    char* string[10][10];
    int i;
    string[8][0] = "ooo";
    string[9][0] = "uuu";
    puts(string[8][0]);
    puts(string[9][0]);
    printf("%d %d", strlen(string[8][0]), strlen(string[9][0]));//This line is just to make sure that both strings have the same lenght
    strcpy(string[8][0], string[9][0]);//I want to copy the content of the string "string[9][0]" to the string "string[8][0]" and replace what was on that string
    puts(string[8][0]);
    puts(string[9][0]);
    return 0;
}

Does this answer your question? [What should happen, when we try to modify a string constant?](https://stackoverflow.com/questions/27211884/what-should-happen-when-we-try-to-modify-a-string-constant) — manveti, Apr 16 '20 at 17:59
regarding `string[8][0] = "ooo";` and simolar statements: the code is trying to force a string into a single char, Suggest: `strcpy( string[8], "ooo" ); — user3629249, Apr 17 '20 at 21:12

Adrian Mole · Answer 1 · 2020-04-16T18:15:12.013

What you are seeing is undefined behaviour¹. Why? Because the line string[8][0] = "ooo"; assigns, to the pointer at string[8][0], the address of a constant string literal. Then, when you later call strcpy(string[8][0], string[9][0]);, you are trying to write to that constant data.

A 'quick fix' to make your program work is to copy the string literal into a non-constant char array, like so:

int main()
{
    char* string[10][10];
    int i;
    //    string[8][0] = "ooo";
    char buffer[4] = "ooo"; // This line COPIES the literal into the non-const array
    string[8][0] = buffer;  // ... and here we give the pointer that array's address
    string[9][0] = "uuu";
    puts(string[8][0]);
    puts(string[9][0]); // Note, I've added a newline below to make the output tidier!
    printf("%d %d\n", strlen(string[8][0]), strlen(string[9][0]));//This line is just to make sure that both strings have the same lenght
    strcpy(string[8][0], string[9][0]);//I want to copy the content of the string "string[9][0]" to the string "string[8][0]" and replace what was on that string
    puts(string[8][0]);
    puts(string[9][0]);
    return 0;
}

¹ The undefined behaviour can cause any number of things to happen! On some platforms, the program may silently 'ignore' your attempt to overwrite the constant data; on my platform, the program crashes (as it appears to on yours).

Thanks for the response! This programm was only an example, and in my actual programm the content of the array string[8][0] is random, and that's why I can't just use the char buffer with always the same value. Is there another way to fix this problem? — KazzioBots, Apr 16 '20 at 20:59
The only thing you have to ensure is any pointer passed as the *first* argument to the `strcpy` function (that is, the *destination* string) does not point to a constant (literal) string. How to check for (or prevent) this in your real-world program depends on many factors; maybe you need to re-think your design? — Adrian Mole, Apr 16 '20 at 21:27
How do I acess a specific char of the string[9][0]? I want to acess only the first 'u' for example. — KazzioBots, Apr 17 '20 at 10:25

score 0 · Answer 2 · answered Apr 17 '20 at 21:23

regarding

string[8][0] = "ooo";

and similar statements:

the code is copying a point to read-only memory

And remember that read-only memory cannot be changed by the program.

Suggest modifying the code so the string(s) are actually placed in the array, similar to.

char string[10][10];
...
strcpy( string[0], "ooo" );

regarding:

printf("%d %d\n", strlen(string[8][0]), strlen(string[9][0]));

the function: strlen() returns a size_t, not an int so the output format conversion' specifiers should be %zu not %d

If you compiler did not warn you about the problems in the code, then enable the warnings then fix those warnings.

For gcc, at a minimum use: -Wall -Wextra -Wconversion -pedantic -std=gnu11

Note the other compilers use different options to produce the same results

Why does the program stop if I'm using strcpy() on two strings with the exact same size in C?

2 Answers2