array return type

Question

#include
#include
#define MAX 30

void push(char );


char stack[MAX];
int tos=0;

int main(){
    char str[]=\"Arijit Saha\";
    char *f         


        

Answer 1

In this code:

char* rev(char s[]) {
    char reverse[strlen(s)];
    ...
    return reverse;
}

reverse is an temporary array with automatic storage duration that is being deallocated once the execution leaves the scope of this function. You return a pointer that becomes a dangling pointer.
Trying access the memory that this pointer points to produces undefined behavior.

Apart from the fact that you should allocate it dynamically by using malloc, note that strlen returns the length of the string, you will also need the space for the terminating character ('\0'). You should create reverse like this:

char* reverse = malloc(strlen(s) + 1);

and don't forget to assign '\0' to the last character of reverse. Also don't forget that the caller of this function becomes responsible for deallocating the memory that has been allocated by malloc, i.e. the caller should call free on the returned pointer.