Question

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

char * reverse(char *string);
int main(int argc, char *argv[])
{
    char array[10];
    array[0] = 'a';
    array[1] = 'b';
    array[2] = 'c';
    array[3] = 'd';
    array[4] = 'e';
    printf("1%s\n",array);
    char *p = reverse(array);
    printf("4%s\n",p);
    printf("5%s\n",array);
}

char * reverse(char *string)
{
    int size = strlen(string);
    char reversed[size];
    int i;
    int j = 0;
    for(i = size-1; i >= 0; i--)
    {
        reversed[j] = string[i];
        j++;
    }
    printf("2%s\n",reversed);
    string = reversed;
    printf("3%s\n",string);
    return reversed;
}

This code basically just initializes an array of values and passes it into a method that reverses these values.

I am not sure if this is the best way to execute the task, since I am new to pointers and arrays in C.

But the real question is this:

Can anyone figure out why in this line

printf("4%s\n",p);

if you remove the preceding ‘4’, so it looks like so

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

the line won’t print at all?

Answer 1

You are returning a pointer to local variable(reversed) in the function reverse the question should actually be: Why did it work in the first place?.

This code string = reversed; will only copy the pointer, and again the local copy of the pointer so it has no effect outside the function.

To reverse a string you don’t need additional memory – this can be done in-place.