Thread: Some question about strings

I've been given a code:

Code:

char *advance(char *ptr)
{
   return ptr++;
}

int main() {
    char *ptr="12";
    advance(ptr);
    ptr=advance(ptr);
    printf("%c",*ptr);
    return 0;
}

And I'm asked what is the output. I found out that the output is 1 by compiling the code, but I don't really know why. Seems like advance() doesn't change the string, although it gets a pointer to it.

I'll appreciate any help.

use

Code:

return ++ptr;

instead.

ptr++ retuns the original pointer and then increments.

Kurt

Maybe you have a link to some material that explains the difference between prefix and postfix operators a little more clearly?

And about this question, again, what about the line advance(ptr)? It doesn't change anything?

Thanks again.

That's what your program actually does

1) a copy of ptr in main is passed to advance.
2) the value of that pointer is copied to a temporary.
3) the argument ptr of advance is incremented
4) the temporary ( the same value that was passed ) is returned
effectively the call to advance does not change anything in main.
Kurt

And if I would want to change advance() so that it actually will advance the pointer, what I need to do? Send advance(&ptr) and change to advance(char **ptr)?

Just changing the parameter to a reference wouldn't change anything.
You would also have to change advance to not return the pointer, or remove the assignment in main.
The easiest solution is in my 1st reply.
But why don't you just modify your code and see what happens ?
Kurt