Why is the volume always calculated by zero?

Question

I'm beginner in using functions, and I wrote this simple code. But I don't know why the volume is always calculated as zero.

 #include <iostream>
 using namespace std;

 double area (double) ;
 double volume (double) ;
 int main () {
     double radious ;
     cout << "please enter the Radious \n" ;
     cin >> radious ;
     cout << "The area = " << area(radious) << "\n" ;
     cin >> radious ;
     cout << "The volume = " << volume(radious) << "\n" ;
     cout << radious << "\n" ;
 }

 // defintion function of the area 
 double area (double R) {
     return ( (4) * (3.14) * (R * R) ) ;
 }

 // defintion function of the volume 
 double volume (double R) {
     return ( (3/4) * (3.14) * (R * R * R) ) ;
 }

[Please do not post images of code because they are hard to use.](https://meta.stackoverflow.com/questions/285551/why-not-upload-images-of-code-errors-when-asking-a-question) Code should be posted directly **as text** in your question. — MikeCAT, Nov 22 '20 at 14:22
Include the code in the question, don't post pictures. `3/4` is `0`, and `0` multiplied by anything is `0`. — Eljay, Nov 22 '20 at 14:22
Does this answer your question? [Why does this calculation (division) return a wrong result?](https://stackoverflow.com/questions/41827246/why-does-this-calculation-division-return-a-wrong-result) — Raymond Chen, Nov 22 '20 at 14:57

NRUB · Answer 1 · 2020-11-22T14:59:40.007

0

Your function volume will always return zero because 3 divided by 4 is 0 when interpreted as an integer. This is because casting any real value to integer will simply result in discarding decimal part. For example, 2.7 as int will be 2 not 3, there is no rounding, in a mathematical sense.

You can fix this in 2 ways:

A) reorder your equation so division will be the last operation you do, e.g. ((3.14*R*R*R*3)/4). Note that this is often necessary, when you want your result to be int, which is not the case here.

B) explicitly say that either (or both) 3 or 4 have to be treated as a real number (float/double) by adding .0, e.g. 3.0/4 or 3/4.0 or 3.0/4.0. This approach is better in your case since you expect double anyway.

For more information refer to Numeric conversions and this FAQ

edited Nov 22 '20 at 14:59

answered Nov 22 '20 at 14:52

NRUB

404
4
17

"This is because casting any real value to integer will simply result in discarding decimal part." is unclear as code does no casting. Recommend re-wording. – chux - Reinstate Monica Nov 22 '20 at 14:56
One could also do `static_cast(3)/4`, for example. – NotAProgrammer Nov 22 '20 at 15:35
@NotAProgrammer `3.0/4` would be much simpler, as stated in the answer – Remy Lebeau Nov 22 '20 at 20:39
@RemyLebeau I think it's a matter of taste regardless, but as the OP is a beginner, it may help them to know that this way is also possible, for example when they work with variables and can't modify the literal. – NotAProgrammer Nov 22 '20 at 20:58

score 0 · Answer 2 · answered Nov 25 '20 at 07:47

0

The part 3/4 in your code performs an integer division. Integers cannot have floating points so usually the last part of integer is truncated, which leaves 0 in your case.

You can replace 3/4 with 3.0/4.0to make it work.

Good Luck!

answered Nov 25 '20 at 07:47

saadurr

57
10

Why is the volume always calculated by zero?

2 Answers2