Why Is My Double or Int Value Is Always 0 After Division

Why is my double or int value is always 0 after division?

Your current calculation only involves integers and so will be affected by integer division (which truncates the result to the nearest integer value).

(currPixel / modifier) * newValue
     |           |
      ---------------integer division e.g. 10/3 = 3, not 3.333

The result is then cast to double, but the accuracy is lost before this point.

Consider the following:

#include <iostream>
using namespace std;

int main() {
    int val1 = 10;
    int val2 = 7;
    int val3 = 9;

    double outval1 = (val1 / val2) * val3;
    double outval2 = ((double)val1 / val2) * val3;
    cout << "without cast: " << outval1 << "\nwith    cast: "<< outval2 << std::endl;

    return 0;
}

The output of this is:

without cast: 9
with    cast: 12.8571

See it here

Note that the cast has to be applied in the right place:

(double)(val1 / val2) * val3 == 9.0      //casts result of (val1/val2) after integer division
(val1 / val2) * (double)val3 == 9.0      //promotes result of (val1/val2) after integer division
((double)val1 / val2) * val3 == 12.8571  //promotes val2 before division
(val1 / (double)val2) * val3 == 12.8571  //promotes val1 before division

Due to promotion of the other operands, if in doubt you can just cast everything and the resulting code will be the same:

((double)val1 / (double)val2) * (double)val3 == 12.8571  

It is a little more verbose though.

Division in double variable returning always zero

The result of 80/100 (both integers) is always 0.

Change it to 80.0/100.0

c# double - why is the answer always 0?

You are actually performing integer division here because the 2 numbers are whole. That forces the result of the calculation to be int which truncates the precision so you end up with 0.

You need to tell the compiler you are actually working with floating-point numbers and not int's, just adding a floating point to one of the numbers should fix this e.g.

double test = (27096140.0 / 27216140);

Alternatively, you could actually declare the numbers as doubles

double a = 27096140;
double b = 27216140;
double test = (a / b);

Or even cast the number in the calculation

double test = ((double)27096140 / 27216140)

Why does value / 100 return 0?

It's due to integer division. By using a floating value for 100.0 instead of 100, the division is performed using floating point.

Try

int asteriskCount = (int)((aGrade/100.0)*50);
                                  ^^^^^

Integer division always zero

You are doing integer division.

Try the following and it will work as expected:

int x = 17;
double result = 1.0 / x;

The type of the 1 in the expression you have above is int, and the type of x is int. When you do int / int, you get an int back. You need at least one of the types involved to be floating point (float or double) in order for floating point division to occur.

Unlike in Mathematics, division in C++ can either refer to truncated integer division (what you did) or floating point division (what I did in my example). Be careful of this!

In my example, explicitly what we have is double / int -> double.

Java: Why is this double variable coming out 0?

Make that:

double check = 3.0 / 4;

and it'll work. You got 0 because 3 / 4 is an integer division, whose value is 0.

Double value returns 0

That's because 1 and 3 are treated as integers when you don't specify otherwise, so 1/3 evaluates to the integer 0 which is then cast to the double 0. To fix it, try (1.0/3), or maybe 1D/3 to explicitly state that you're dealing with double values.

Why does 1 / 2 == 0 using double?

It's because of the data type.

When you do 1/2 that is integer division because two operands are integers, hence it resolves to zero (0.5 rounded down to zero).

If you convert any one of them to double, you'll get a double result.

double d = 1d/2;

or

double d = 1/2.0;

Why does dividing two int not yield the right value when assigned to double?

This is because you are using the integer division version of operator/, which takes 2 ints and returns an int. In order to use the double version, which returns a double, at least one of the ints must be explicitly casted to a double.

c = a/(double)b;

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