Why is this returning zero in c++?

Question

When I run the result is zero, except for -5. I checked, it has nothing to do with the last line.

#include <iostream>
using namespace std;

int main ()
{
    int x;
    float E;
    cin >> x;
    if((float)x!=-5)
    {
        E=(x+3)/(x+5);
        cout << "Valoarea expresiesi este: " << E;
    }
    else 
        cout << "You cannot divide by zero";
    cin.get();
    return 0;
}

What did I miss? Thank you.

Look up *integer division*. You are dividint two ints. If the magnitude of the denominator is larger than that of the numerator you get `0`. There are SO duplicates around. — juanchopanza, Jan 10 '14 at 08:53
[Another duplicate on integer division](http://stackoverflow.com/questions/3602827/what-is-the-behavior-of-integer-division-in-c) — Mike, Jan 10 '14 at 12:20

score 2 · Answer 1 · answered Jan 10 '14 at 08:54

2

(x+3)/(x+5) is evaluated as an integer division, which rounds toward zero. It doesn't matter that E is a float, because neither x+3 nor x+5 is a float. As you've written it, the conversion to floating-point will happen after the rounding has already occurred. Convert an operand to floating-point before performing the operation:

E = (1.0*x+3) / (x+5);

answered Jan 10 '14 at 08:54

user2357112

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And how can I specify the number of decimal places? f2.22E doesn't work. – Tamás Jan 10 '14 at 09:02
@Tamás: I have no idea what `f2.22E` is supposed to be. In C, we'd use `printf` with a format specifier indicating 2 decimal places. I dunno about C++. – user2357112 Jan 10 '14 at 09:03
OK. I'm in college and they are supposed to be teaching me c++ and we used the above format specifier along with printf – Tamás Jan 10 '14 at 09:18

score 2 · Answer 2 · edited Jan 10 '14 at 08:56

2

You are dividing two integers and the result will be zero since the denominator is larger. To simply solve this, use 3.0 and 5.0 as your literals.

E = (x + 3.0) / (x + 5.0);

edited Jan 10 '14 at 08:56

Zeta

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answered Jan 10 '14 at 08:54

It'sPete

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score 1 · Answer 3 · answered Jan 10 '14 at 08:57

1

For any x > -5, (x+3)/(x+5), the denominator will always be greater than numerator and the integer division will always return 0. Only if you provide x < -5 the answer will never be 0.

answered Jan 10 '14 at 08:57

kvivek

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score 0 · Answer 4 · edited May 23 '17 at 10:33

0

Don't ever compare floats/doubles with strict comparisson == nor !=. What you should do is

if ( fabs(x-5) < 0.00001 ) { // logically x == 5
}
else {
}

EDIT

Check also my post about infinite loop

edited May 23 '17 at 10:33

Community

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answered Jan 10 '14 at 09:02

SOReader

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This doesn't answer the question. Also, `x` is an integer, so using a tolerance for the comparison is pointless. – user2357112 Jan 10 '14 at 09:20
I know but it looks like he's beginning his journey with C++ so I wanted him to know about floating point numbers comparison. – SOReader Jan 10 '14 at 09:21

Why is this returning zero in c++?

4 Answers4