Converting double to integer

[theicfire]

This code:

cout.setf(ios_base::fixed, ios_base::floatfield);
double a=5.67;
double b = a*10000;
int c = a*10000;
int d = (double)(b);
printf("a: %lf b: %lf c: %d d: %d", a, b, c, d);

Is giving me this output

a: 5.670000 b: 56700.000000 c: 56699 d: 56700

Why is c 56699 and not 56700?

Thanks

[Bazzy]

Because double aren't precise http://www.parashift.com/c++-faq-lite/newbie.html#faq-29.16

[chrisname]

Not to mention that floating point arithmetic sucks.
A quick google returned this result:
http://www.atalasoft.com/cs/blogs/stevehawley/archive/2008/01/25/never-forget-floating-point-sucks.aspx
Which basically says the same thing. Floating point arithmetic is imprecise and should be avoided (IMO). In fact, Linus Torvalds has virtually banned it from Linux kernel code.

[jsmith]

The obligatory link on the subject of floating point arithmetic is:
http://docs.sun.com/source/806-3568/ncg_goldberg.html

While it is true that Linux does not use floating point arithmetic in the kernel, I think
the larger reason for this is to avoid the need to save/restore state of the FPU
every time a context switch occurs.

[blackcoder41]

maybe it should be:
int c = ( (int)a ) * 10000;

[Disch]

That would be 50000 =P

[blackcoder41]

@Disch
oh sorry,

[blackcoder41]

1 2 3 4 5 6 7 8 9 10 11 12 13 14

#include <iostream> #include <sstream> using namespace std; int main() { cout.setf(ios_base::fixed, ios_base::floatfield); double a=5.67; double b = a*10000; int c = (int)(a*10000); int d = (double)(b); c = (int) b; printf("a: %lf b: %lf c: %d d: %d", a, b, c, d); return 0; }

a: 5.670000 b: 56700.000000 c: 56700 d: 56700