C program for how to reverse a number explanation needed - c

Question

While searching for a C program on how to reverse a string, i came across the below program. I am already familiar with a program where we take the length of the string and then minus each character and find the reverse. But this a different program. So can someone please tell me how this code works? Help will be gratefully accepted.
compiler used is Borland Turbo c.
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main() {
char str[50];
char rev[50];
int i, j, n;
clrscr();
printf("enter the string to be reversed:");
scanf("%s", &str);
for (i = 0; str[i] != 0; i++) {
n = i - 1;
}
for (j = 0; j <= i - 1; j++) {
rev[j] = str[i];
n--;
}
printf("the reverse of the string is:%s", rev);
getch();
}

Answer 1

In C, strings are NUL-terminated meaning that it has '\0' at the end signifying the end of the string. The code you've posted currently has two issues:
This:
scanf("%s", &str);
should be
scanf("%s", str);
as %s expects a char*, not a char(*)[50].
This:
rev[j] = str[i];
should be
rev[j] = str[n];
rev should be NUL-terminated before printing. Add
rev[j] = '\0';
just before
printf("the reverse of the string is:%s", rev);
to avoid Undefined Behavior.

Answer 2

Your code doesn't work as it is supposed to.Have you tried it ?
Consider this approach to reverse a string :
#include <stdio.h>
#include <string.h>
int reverse(char *Str);
void swap(char *x, char *y);
int main(int argc, char *argv[]) {
char Str[255];
printf("enter the string to be reversed : ");
scanf("%s", Str);
reverse(Str);
printf("the reverse of the string is : %s\n", Str);
}
int reverse(char *Str) {
size_t len = strlen(Str);
size_t n = len / 2;
char *begin = Str;
char *end = (Str + len) - 1;
while (n > 0) {
swap(begin, end);
begin++;
end--;
n--;
}
return 0;
}
void swap(char *x, char *y) {
char tmp;
tmp = *x;
*x = *y;
*y = tmp;
}