change value of a char array in a function

Asked Jan 18 '23 at 16:42

Active Jan 18 '23 at 17:00

Viewed 19 times

I try to change a value of a char inside a function, but it doesn't work, I can't change the main function and the function call


typedef char* char_array;

int change(char_array arr, int index, char c){

    int length = 6;
    char *new_array = calloc(length, sizeof(char_array));
    new_array = arr;
    str = calloc(length, sizeof(char_array));

    if(index >= 0 && index <= length){
        for(int i = 0; i < length; i++){
            if(i == index){
                arr[i] = c;
            }else{
                arr[i] = new_array[i];
            }
            
        }
        return 0;
    }else{
        return -1;
    }
}
int main(){
    char_array arr = "Hello!";
    change(arr, 5, ' ');
    printf("%s", arr);

}

correct print: Hello but print: Hello!

edited Jan 18 '23 at 17:00

asked Jan 18 '23 at 16:42

Cebolexx

`arr` points to a literal string. These are not modifiable. Use `char arr[] = "Hello!";` – Barmar Jan 18 '23 at 16:46
Where is str declared? In any case, str is completely local the change function and doesn't exist after the the call. Review how pointers work, what does 'arr' point to? – john elemans Jan 18 '23 at 16:46
@Cebolexx Where is there the identifier astr declared? – Vlad from Moscow Jan 18 '23 at 16:47
@Cebolexx If to update the typo then this code snippet char *new_array = calloc(length, sizeof(astr)); new_array = str; produces a memory leak. – Vlad from Moscow Jan 18 '23 at 16:49

change value of a char array in a function

0 Answers0