output of c code

Question

Why output is giving 3 , expecting -3. How to handle such preprocessing in c?

#include<stdio.h>
#include<math.h>
#define sq(x) ((x<0)?sqrt(-x):sqrt(x))

int main()
{
    int x;
    x=sq(-9);
    printf("%d\n",x);
    return 0;
}

Why would you expect the result of any square root to be a negative number??? — Mithrandir, Jan 27 '12 at 18:38
ALWAYS put macro arguments in their own parentheses! As in `#define sq(x) (((x)<0)?sqrt(-(x)):sqrt(x))`. Imagine if you had `sq(a+b)` in an expression ... `-(a+b)` is not the same as `-a+b`. — pmg, Jan 27 '12 at 18:42
Presumably you really intended something like `(((x)<0)?-sqrt(-(x)):...`. — Jerry Coffin, Jan 27 '12 at 18:46

Keith Nicholas · Accepted Answer · 2012-01-27T19:44:48.363

4

because your # define "sq" checks if its a negative number and turns it into a positive number before calculating a square root

its doing sqrt(-x) which is sqrt(-(-9)) ( taking the negative of a negative is the positive)

so its doing sqrt(9)

edited Jan 27 '12 at 19:44

answered Jan 27 '12 at 18:34

Keith Nicholas

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if we write sqrt(--9) directly than it give error...and in c (--9) doesnt have any mean...so how can u tell here sqrt(--9) ( taking the negative of a negative is the positive) – anil Jan 27 '12 at 18:54
@anil write `sqrt(-(-9))` c thought you meant to use the decrement operator. – Dave Jan 27 '12 at 19:00
@anil: He did not say `--9`. He said `-(-9)`. – Lightness Races in Orbit Jan 27 '12 at 19:46

score 1 · Answer 2 · answered Jan 27 '12 at 18:35

1

You have this define:

define sq(x)  ((x<0)?sqrt(-x):sqrt(x))

Since you're passing -9, x<0 is true, so it's doing sqrt(9), which is 3.

You then print the 3.

answered Jan 27 '12 at 18:35

Reed Copsey

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score 1 · Answer 3 · answered Jan 27 '12 at 18:35

1

The code is doing exactly what it's being told, I think.

sq(x) tests first for x < 0 - in your case, this is true, so sq(x) calls sqrt(-(-9)) - which is 3.

An attempted solution is (x < 0) ? -sqrt(-x) : sqrt(x), which will return negative roots for negative x (my best interpretation of your intent).

answered Jan 27 '12 at 18:35

mzhang

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but actually preprocessor just replace x with passed value..so sq(-9) should be seem like this ((-9<0)?sqrt(--9):sqrt(9))...and it should be give an error...but u tell it replace like this sqrt(-(-9))...how it is possible – anil Jan 27 '12 at 19:00
I believe `-1 * x` will work. (`--9` being `-1 * -9` being passed into `sqrt`) – mzhang Jan 27 '12 at 23:03

score 1 · Answer 4 · answered Jan 27 '12 at 18:38

1

Wait, wait, wait.

Are we trying to break the basic rules of Math here?

The square root of -9 is 3 i.

That's because (-3)^2 is 9. Negative numbers have 'imaginary' square roots. sqrt(-1) is i. Not -1. (-1)^2 is 1.

answered Jan 27 '12 at 18:38

ThatOtherPerson

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score 0 · Answer 5 · answered Jan 27 '12 at 18:37

0

If you expect -3, then you probably meant

#define sq(x)  ((x<0)?-sqrt(-x):sqrt(x))

but why would you expect negative square root of negative number is beyond me.

answered Jan 27 '12 at 18:37

Michael Krelin - hacker

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He probably does not want to do `sqrt(-9)`... maybe `-sqrt(-x)`? – Reed Copsey Jan 27 '12 at 18:39
So, basically, due to absurdity of expectation, it might have been `#define sq(x) -3` :-) – Michael Krelin - hacker Jan 27 '12 at 18:42

output of c code

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