tags:

views:

93

answers:

5

In C#, I am doing something like this:

float a = 4.0f;
float b = 84.5f;
int ans = a * b;

However, the compiler states that a cast is required to go from float -> int in assignment. Of course I could probably do this:

int ans = (int)a * (int)b;

But this is ugly and redundant. Is there a better way? I know in C++ I could do this:

int ans = int(a * b);

At least that looks a little better on the eyes. But I can't do this in C# it seems.

+1  A: 

Try int ans = (int)(a * b);

jdv
+1  A: 
int ans = (int)(a * b);
RedFilter
+3  A: 

Put the int in parentheses as well.

int ans = (int)(a * b);
Brandon
They're called parentheses.
SLaks
@SLaks, my mistake.
Brandon
@SLaks what you call them depends on where you learned English.
ScottS
This is a larger question than what looks good with the code. Depending on when the cast will be done it will change the result.
David Basarab
@David, you are correct but his question specifically asked for the proper way to cast results. (I did upvote your answer for your thorough explanation though).
Brandon
+1  A: 
 int ans = (int)a * (int)b; 

 int ans = (int)(a * b); 

These two statements are not equivalent and will produce different results. In once case, you give up precision before the multiplication, in the other after the multiplication.

cdonner
+7  A: 

You should consider the need of your application before the look of the code. Doing float math to int, is not something to be taken lightly. The real question what are you looking for out of your final answer.

a is cast to 4, and b is cast to 84, which is the result of 336. However if you cast it to an int after you do the math, the result is 338.

Is being off by 2 good enough for you? Then you have to do

int ans = (int)a * (int)b;

// ans = 336

If you want 338 then you have to do

int ans = (int)(a * b);

// ans = 338

I would really consider the side effects about what you are doing. Ideally you should have a policy for rounding the 2 floats before doing the math. Remember casting to an int is just going to cut a decimal off. So 84.9 becomes 84. That could greatly change your final result. You need to consider what is required in your application.

David Basarab
Thank you (and to everyone else) for the answers. What I really want is to truncate it *after* the multiplication, so the latter solution is what I need.
Bob