How do you divide integers and get a double in C#?

Question

int x = 73;  
int y = 100;  
double pct = x/y;  

Why do I see 0 instead of .73?

Answer 1

Because the division is done with integers then converted to a double. Try this instead:

double pct = (double)x / (double)y;

Answer 2

It does the same in all C-like languages. If you divide two integers, the result is an integer. 0.73 is not an integer.

The common work-around is to multiply one of the two numbers by 1.0 to make it a floating point type, or just cast it.

Answer 3

because the operation is still on int type. Try double pct = (double)x / (double)y;

Answer 4

Integer division drops the fractional portion of the result. See: http://mathworld.wolfram.com/IntegerDivision.html

Answer 5

It's important to understand the flow of execution in a line of code. You're correct to assume that setting the right side of the equation equal to double (on the left side) will implicitly convert the solution as a double. However, the flow execution dicates that x/y is evaluated by itself before you even get to the double pct = portion of the code. Thus, since two int s are divided by each other, they will evaluate to an int solution (in this case, rounding towards zero) before being implicitly converted to a double.

As other have noted, you'll need to cast the int variables as double s so the solution comes out as a double and not as an int .

Answer 6

That's because the type of the left hand operand of the division ( x ) is of type int , so the return type of x / y is still int . The fact that the destination variable is of type double doesn't affect the operation. To get the intended result, you first have to cast (convert) x to double , as in:

double pct = (double)x / y;