looping question

[ranjit89]

#include<stdio.h>
int main()
{
  int i;
  for (i = 0; i < 5; i++) {
    int i = 10;
    printf(" %d", i);
    i++;
  }
  return 0;
}

Output:10 10 10 10 10
Here after i++ (after printf) value of i should be 11 and it should break out of loop after printing only once?? Why is the output like this then?

[laserlight]

Notice that you declared a variable named i that is local to the loop body. This variable hides the variable named i that you declared prior to the loop.

[itsme86]

You're using 2 different variables called i. The loop condition is using the i that's outside the loop, and the printf() and everything else inside the loop is using the i that you declare inside the loop.

[ranjit89]

Thanks.. that sums it up.

[CommonTater]

Thanks.. that sums it up.

Ok, you have 4 concurrent threads with questions about garbage code...

Exactly what are you trying to accomplish?

If you are trying to learn C by quesswork, it ain't gonna happen... Get yourself a decent book or tutorial and work it page by page, example by example, beginning to end... At least then you'll have some clue how the language works.

And as you were told in a different thread... you will not discover how the compiler works by continuously trying to go around it. There's an old story about teaching a pig to sing... in the end all you do is annoy the pig and waste your time.

[spaghetticode]

There's an old story about teaching a pig to sing... in the end all you do is annoy the pig and waste your time.

I've never heard of that before... it made me laugh. Still I'm glad living in times where wasting your time is even possible, plus an annoyed pig is still better than a slaughtered one.