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divide by zero - c programming

I have a question about the next code:

int main { 
double x = 0;
double y = 0/x;

if(y==1) {.....}
....
....
return 0;
}

When I run the code on my computer, I get no runtime error and I see that y = -nan(0x8000000000000). Why it is not a runtime error to divide by zero?

Additionally, when I change the first line to int x = 0; now there is a runtime error. What is the difference?

like image 536
wantToLearn Avatar asked Oct 28 '12 16:10

wantToLearn


2 Answers

You can't rely on this "working" (i.e. doing the same thing all the time, portably) at all, it's undefined behavior in C for the second case, and also for the first if your implementation doesn't define __STDC_IEC_559__ (this is, I believe, rare these days).

C99, §6.5.5/5

The result of the / operator is the quotient from the division of the first operand by the second; the result of the % operator is the remainder. In both operations, if the value of the second operand is zero, the behavior is undefined.

The fact you're getting a "Not a Number" in one case and and not in the other is that one is done in floating-point arithmetic, where, on your implementation (conforming to IEEE 754 division by zero semantics), 0/0 gives a NaN.

In the second case, you're using integer arithmetic – undefined behavior, there's no predicting what will happen.

like image 154
Mat Avatar answered Sep 22 '22 10:09

Mat


The reason you don't get an exception or error is because for a double, infinity and NaN are defined (see IEEE floating point) but when you try the same for integer, you'll get an error because NaN/Infinity aren't defined

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Suhail Patel Avatar answered Sep 22 '22 10:09

Suhail Patel