printf rounding behavior for doubles

问题

Can someone explain this behavior? I am well aware of machine-level representation of floating point numbers. This seems to be related to printf and its formats. Both numbers are represented exactly by floating-point notation (check: multiplying by 64 gives an integer).

#include <stdio.h>
#include <iostream>
using namespace std;

int main() {
  double x1=108.765625;
  printf("%34.30f\n", x1);
  printf("%9.5f\n", x1);
  printf("%34.30f\n", x1*64);

  double x2=108.046875;
  printf("%34.30lf\n", x2);
  printf("%9.5f\n", x2);
  printf("%34.30f\n", x2*64);
}

Output:

> 108.765625000000000000000000000000
> 108.76562
> 6961.000000000000000000000000000000
> 108.046875000000000000000000000000
> 108.04688
> 6915.000000000000000000000000000000

Note, the first number gets rounded down, and the second one gets rounded up.

回答1:

It's "round half to even" or "Banker's rounding". The last digit of the rounded representation is chosen to be even if the number is exactly half way between the two.

http://linuxgazette.net/144/misc/lg/a_question_of_rounding_in_issue_143.html:
"For the GNU C library, the rounding rule used by printf() is "bankers rounding" or "round to even". This is more correct than some other C libraries, as the C99 specification says that conversion to decimal should use the currently selected IEEE rounding mode (default bankers rounding)."

回答2:

The %9.5f output gives the number with 5 digits after the decimal dot which is the nearest of the source number.

来源：https://stackoverflow.com/questions/10357192/printf-rounding-behavior-for-doubles

标签

c++

double

printf

rounding