Multiplying an integer by a float literal - is the "F" required?

Question

Is this:

int i = 100 * 0.6;

less correct than this?

int i = 100 * (0.6F);

I apologize for such a simple question, but I haven't memorized all the rules for data type promotions and I'm not really sure how to even verify this.

Answer 1

It can make a difference. For example:

int i = (1 << 24) + 3;

printf("%d\n", (int)(i * 0.6));    // 10066331
printf("%d\n", (int)(i * 0.6f));   // 10066332

The reason is that the calculation for the first is done in double-precision, the latter in single-precision. Single-precision cannot represent all integers greater than 1 << 24.

Another example (courtesy of @EricPostpischil in the comments below):

int i = 100;

printf("%d\n", (int)(i * 0.29));    // 28
printf("%d\n", (int)(i * 0.29f));   // 29

Here the reason is that the intermediate result in the double-precision case falls slightly below 29, and so is truncated down to 28.

So I would suggest allowing double-precision (i.e. omitting the f/F) (and then using round() rather than relying on implicit truncation) unless you have a good reason to do otherwise.

Answer 2

The F is not required in this particular case. All it does is specify the constant as being of type float instead of type double.