How the last printf in this code work?

Question

This code makes an ASCII representation of a Sierpinski triangle of order 4, and I have no idea how the last printf works. If anyone can explain it to me, I would be very grateful.

#include <stdio.h>

#define SIZE (1 << 4)
int main()
{
    int x, y, i;
    for (y = SIZE - 1; y >= 0; y--, putchar('\n')) {
        for (i = 0; i < y; i++) putchar(' ');
        for (x = 0; x + y < SIZE; x++)
            printf((x & y) ? "  " : "* ");
    }
    return 0;
}

Which part of it don't you understand? Do you know what the & operator does? Do you know the ? : conditional operator? Break it down into pieces and it should become clear. — Barmar
– Barmar, Commented Apr 19, 2018 at 23:46
If (x & y) evaluates to true, it prints a " ", otherwise a "* ". — Andy J
– Andy J, Commented Apr 19, 2018 at 23:54

Mark Ransom · Accepted Answer · 2018-04-19 23:54:12Z

1

The ? is a ternary operator. It evaluates the expression on the left, and if it's non-zero (true) it selects the first value before the colon : and if it's zero (false) it selects the second.

answered Apr 19, 2018 at 23:54

Mark Ransom

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Michael Burr · Accepted Answer · 2018-04-19 23:56:39Z

1

printf((x & y) ? " " : "* "); is more or less equivalent to:

if ((x & y) != 0) {
    printf("  ");
}
else {
    printf("* ");
}

answered Apr 19, 2018 at 23:56

Michael Burr

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2 Comments

Barmar Over a year ago

"more or less"? I'd say it's exactly equivalent. I resisted posting an almost identical answer.

Michael Burr Over a year ago

I've gotten into the habit of saying "more or less" in answers like this because there's often some detail that's not important to the question/answer but is technically different. It's a CYA that I just use without too much thought.

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