I have this:
double result = 60 / 23;
In my program, the result is 2, but correct is 2,608695652173913. Where is problem?
You can use any of the following all will give 2.60869565217391:
double result = 60 / 23d;
double result = 60d / 23;
double result = 60d/ 23d;
double result = 60.0 / 23.0;
But
double result = 60 / 23; //give 2
Explanation:
if any of the number is double it will give a double
EDIT:
Documentation
The evaluation of the expression is performed according to the following rules:
If one of the floating-point types is double, the expression evaluates to double (or bool in the case of relational or Boolean expressions).
If there is no double type in the expression, it evaluates to float (or bool in the case of relational or Boolean expressions).
It will work
double result = (double)60 / (double) 23;
Or equivalently
double result = (double)60 / 23;
To add to what has been said so far... 60/23 is an operation on two constants. The compiler recognizes the result as a constant and pre-computes the answer. Since the operation is on two integers, the compiler uses an integer result The integer operation of 60/23 has a result of 2; so the compiler effective creates the following code:
double result = 2;
As has been pointed out already, you need to tell the compiler not to use integers, changing one or both of the operands to non-integer will get the compiler to use a floating-point constant.