Floating pt. subtraction error depends on abs. values

[jorgen]

I thought I understood subtraction errors (subtracting two almost equal numbers effectively removes one digit of precision per equal digit) but apparently not; in the following snippet:

int main() {

	cout << setprecision(15) << 1-0.9999999999999 << endl;
        cout << setprecision(15) << 100000000000000.0-9999999999999.0 << endl;
	
	return 0;
}

I don't see why the first calculation has an error and the second is correct - I can't see where the dependence on the absolute value comes from.

EDIT: correction

[Bourgond Aries]

You are subtracting an from an integer on line 3.
You need to write 1. instead of 1 to signify that it's a double. Currently you use doubles instead of floats. Put 1.f to tell the compiler to interpret the number like a float.

[Cubbi]

1, 100000000000000.0 and 9999999999999.0 are valid values for the type double.

The decimal number 0.9999999999999 is not a valid double it happens to be between two valid doubles: 0.99999999999989996890548127339570783078670501708984375 and 0.999999999999900079927783735911361873149871826171875. The first one is closer to 0.9999999999999, and is what is actually compiled.