Hi there, Could someone tell me why this wont work? #include <stdio.h> int main() { char p[10] = "Success"; char *t = p; while (*t++ != '\0'){ printf("%c ", *t); *t = 'i'; } return 0; } If I move the increment part from while to the assignment statement, it works fine. ie., *t++ = 'i'; It prints Success. The code above doesn't print anything. Let me know if I'm missing something here. Thanks a lot. Tina
Seems like this is your first post. So, welcome to the forum :clap::clap: Now, about your code ... hmmm First of all, using while(*t++ != '\0'){ is bad because it will NEVER print the first letter as t is incremented immediately. Next, I don't think your code will print NOTHING. Which compiler do you use ? I think it will print "u c c e s s" and then some garbage. On my machine it shows : u c c e s s " ╥ ‼ @ l ‼ @ ≡ ² ⌂ ░ * " K ↕ @ ☺ H ? > └ , > @ @ ñ * " * * * * ¿ * " æ î O Ç └ , > ≡ ² ⌂ └ * " ÿ ↕ @ ☺ ≡ * " g p ü | Ö x ┤ ≈ " ≡ ² ⌂ ╕ ╢ T Ç ╚ * " ¿ ═ ╥ ü * * * * └ Ü â | p p ü | Ç ↕ @ A Now why does this happen ? Let's analyze step-by-step : Loop 1 : Step 1 : *t is checked for '\0' and returns false as *t = 'S'. .............Step 2 : t is incremented and points to 2nd char = 'u'. .............Step 3 : *t is printed with a space after it. .............Step 3 : *t is changed to 'i'. Loop 2 : Step 1 : *t is checked for '\0' and returns false as *t = 'i' (look at last step of Loop1). .............Step 2 : t is incremented and points to 3nd char = 'c'. .............Step 3 : *t is printed with a space after it. .............Step 3 : *t is changed to 'i'. Loop 3 : Step 1 : *t is checked for '\0' and returns false as *t = 'i' (look at last step of Loop1). .............Step 2 : t is incremented and points to 4th char = 'c'. .............Step 3 : *t is printed with a space after it. .............Step 3 : *t is changed to 'i'. Do you see what is going on ? Whenever *t is checked for '\0', it contains 'i'. So, the program falls into an infinite loop and terminates when an exception occurs. t should be incremented after changing *t to 'i'. But you, increment t first and then change it to 'i'. So, it's useless ! Hope this will help. Best of luck.
Hi, Thanks for responding! I see the error now. I got confused with the strcpy version : void strcpy(char *s, char *t) { while(*s++ = *t++); } So, the point is when you've increment in while loop, then you shouldn't use that operand in the function body. Btw, I used gcc compiler and I dint see any print on the screen. Thanks again, Tina
Anyway, *s++ and *t++ are interpreted as *(s++) and *(t++) because of higher precedence of ++ than *. See here : http://www.cppreference.com/wiki/operator_precedence
Ummm....Doesn't this fetch the character first and then increment? Like (*s)++. If it increments the pointer first, this version of strcpy wont work. Is that right? Tina
I don't see any reason why that version should not work ! :surprised: It works fine on my machine. Adding brackets, as suggested by it_career is unnecessary as the compiler automatically interprets *s++ as *(s++). Look, I think you don't get it properly. Observe at the loop : while(*s++ = *t++); What this does is : (1) Assign *t to *s. (2) Increase t and increase s. (3) As nothing to do inside loop body, so go to step (1). It does not increment the pointer first ! It's post-fix ++. In the case you mentioned in post #1, first the while condition was executed, then t was increased and then the body was executed. So, the first char was missed out. But here, nothing is done in the loop-body. And one more thing, never do anything as crazy as (*s)++ because that will change your string by increasing each character by 1. Hope I made it clear to you. Best of luck.
Umm..My mistake again...That's what I meant but I got confused with the brackets. Its crystal clear now Thanks! Tina
