C变量函数va_arg返回错误的双精度

Question

So I'm nearly done reimplementing printf(3) (I can't use any function that would the conversion for me) in C .

Now that I have implemented all the conversions I'm seeing something weird, when I pass my function a double to use with say %g like 100.10 then va_arg gives me 100.09999999999999 so of course my conversion then gives me a string with 100.099999 .

I fetch the double like this:

double d = (double)(va_arg(args, double));

And I am calling both functions like this:

double testd = 100.10;
my_printf("%g", testd); #=> 100.099999
printf("%g", testd); #=> 100.1

I think that I know that the value is wrong because while stepping through the program I'm seeing that double d = 100.09999999999999 .

Should I do it differently ? Because the real printf seems to be getting the correct value.

EDIT:

printf("%#.10g\n", testd); #=> 100.1000000000

Is this due to rounding or that maybe the real printf is getting the good value ?

Answer 1

Just add format specifier, and printf will perhaps show the same issue:

printf("%0.20g\n", d); // 100.09999999999999 on my system

The reason is that in our case the number 100.1 (as most of all numbers) does not have exact binary representation.