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

How come that in the following snippet

int a = 7;
int b = 3;
double c = 0;
c = a / b;

c ends up having the value 2, rather than 2.3333, as one would expect. If a and b are doubles, the answer does turn to 2.333. But surely because c already is a double it should have worked with integers?

So how come int/int=double doesn't work?

Solution 1:

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;

Solution 2:

Here it is:

a) Dividing two ints performs integer division always. So the result of a/b in your case can only be an int.

If you want to keep a and b as ints, yet divide them fully, you must cast at least one of them to double: (double)a/b or a/(double)b or (double)a/(double)b.

b) c is a double, so it can accept an int value on assignement: the int is automatically converted to double and assigned to c.

c) Remember that on assignement, the expression to the right of = is computed first (according to rule (a) above, and without regard of the variable to the left of =) and then assigned to the variable to the left of = (according to (b) above). I believe this completes the picture.

Solution 3:

With very few exceptions (I can only think of one), C++ determines the entire meaning of an expression (or sub-expression) from the expression itself. What you do with the results of the expression doesn't matter. In your case, in the expression a / b, there's not a double in sight; everything is int. So the compiler uses integer division. Only once it has the result does it consider what to do with it, and convert it to double.