C pointer mistakes in my understanding of the problem

The output result after execution is______<br/>

#include<stdio.h>
int main()
{  char str[]="xyz",*ps=str;
   while(*ps) 
       ps++;
   for(ps--;ps-str>=0;ps--)
       puts(ps);
   return 0;
}

The right answer is <br/>
z<br/>
yz<br/>
xyz<br/>

My logic is<br/>
After while(*ps) ps++; *psrepresents " " the last element of the array<br/>
When execute codefor(ps--;ps-str>=0;ps--)puts(ps);,*ps go back two positions everytime.
So I think t should be <br/>
z<br/>
xyz<br/>

Probably closer to what you want:

int main()
{  
   char str[]="xyz";
   size_t length = sizeof(str)-1;
   // size_t length = strlen(str); // safer version of above.
 
   for (size_t i = 0; i < length; i++)
   {
       char* ps = str+length-i-1;
       puts(ps);
   }
}

> "After while(*ps) ps++; *ps represents " ""

No, *ps is then '\0'.

> When execute code for(ps--;ps-str>=0;ps--)puts(ps);, *ps go back two positions everytime

No, it does not. It starts by going back one step, then in each iteration it only goes back one step. It's the same as doing this:

{
    ps--;                 // for(ps--; ...; ...)
    while(ps-str >= 0) {  // for(...; ps-str >= 0; ...)
        puts(ps);
        ps--;             // for(...; ...; ps--)
    }
}

ps-str>=0 is not a good test though. If you step ps "behind" the start of str the comparison is invalid.

A safe version could simply be:

#include <stdio.h>

int main() {
    char str[] = "xyz", *ps = str;
    
    while (*ps) ps++;
    
    while (ps != str) {
        --ps;
        puts(ps);
    }
}