Segmentation error in C when trying to concatenate a string and char

Question

I keep getting a segmentation error when I try to add a character to the string test. I have tried multiple iterations and can't figure out why I get the error. I tried having the test set to \0. I can't figure out how I am accessing outside test.

#include <stdio.h>
#include <cs50.h>
#include <ctype.h>
#include <string.h>

#define _XOPEN_SOURCE
#include <unistd.h>
#include <crypt.h>

//this intializes the argument count and argument vector
int main(int argc, string argv[]) {
    //this makes sure it is a string
    if (argv[1] != NULL) {
        //this makes sure that the user typed 2 arguments, 
        //  the second being the keyword
        if (argc != 2) {
            printf("no hash inputed\n");
            return 1;
        }

        //main program once the keyword has been verified
        int h_len = strlen(argv[1]); 
        string hash = argv[1];
        string test = "A";
        int p_len = 0;
        printf("H len is %i and P len is %i\n", h_len, p_len);
        printf("%s and test: %s\n", hash, test);

        //this will iterate through the 1st space and test the charachters
        for (int j = 0; j < 4; j++) {

            //iterates through the characters that can be used in the  password
            //for (int i = A; i < z; i++)
            //for (char ch = 'A' ; ch <= 'z' ; ch == 'Z' ? ch = 'a' : ++ch )
            for (char i = 'A'; i <= 'z'; i++) {
                //this skips the ASCII inbetween Z and a
                if (i > 'Z' && i < 'a') {
                    i = 'a';
                }
                printf("%c", i); 
                //test[j] = test[j] + (char)i;
                test = strncat(test, &i, 1);
                printf("  test is %s\n", test);
                ...

Answer 1

In this declaration

string test = "A";

there is declared the pointer test to the string literal "A". You may not change string literals.

According to the C Standard (6.4.5 String literals)

7 It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array, the behavior is undefined

However in this statement

test = strncat(test, &i,1);

there is an attempt to modify the string literal pointed to by the pointer test.

You should declare a character array that has enough space to store additional characters.

Moreover if you are using the function strncat to copy n characters from the source string you should reserve memory in the destination array for n + 1 characters because the function always appends the terminating zero.

That is (The C Standard, 7.23.3.2 The strncat function, footnote 302)

302) Thus, the maximum number of characters that can end up in the array pointed to by s1 is strlen(s1)+n+1.

Answer 2

Note that in this horrible cs50, string is defined as char *, so string test = "A"; is equivalent to char *test = "A"; which will make test point to a constant string of length 1 plus a null!

So even if it would be long enough to concatenate anything, you can't because it is read-only, and even if it was not read-only, you can't because it is not long enough.

Answer 3

You have a string with one character in it (test="A") so it has a length of 2 (A + null char) and are trying to append other characters to it, so it WILL access outside of test.

Answer 4

everything is wrong here

you cant strcat char * and the address of the char as you never know if there is zero after that character. Your string is too short to accomodate anything longer than one character string. Next - you try to do something with RO memory as your test points to the string literal.

Buy a good book about C