CODEWITHSUNDEEP
c++printf
queen3
queen3

Reputation: 15511

printf zero padding difference between %g and %f

I want to understand why zero padding works differently for %g and %f formats in printf. Here's a sample program:

#include <stdio.h>
int main()
{
  double d = 0.10000000000001;
  printf("%04.2f\n", d);
  printf("%04.2g\n", d);
  return 0;
}

It outputs

0.10
00.1

This is reproducible in both VC++ on Windows and gcc on Linux.

Why it works like this? It is a correct behavior?

UPDATE: I found the answer but I think I'll left the question open...

Upvotes: 3

Views: 9069

Answers (3)

Mike Seymour
Mike Seymour

Reputation: 254531

This is correct. According to the specification of style g in C99 7.19.6.1/8:

unless the # flag is used, any trailing zeros are removed from the fractional portion of the result and the decimal-point character is removed if there is no fractional portion remaining

while style f does not specify that trailing zeros are removed.

Upvotes: 3

Edouard Thiel
Edouard Thiel

Reputation: 6228

When using %g, printf chooses the format which gives the best precision in minimum place. You can guess how it evaluates the precision, knowing that for d = 0.10500000 it prints "00.1" while for d = 0.10500001 it prints "0.11".

Upvotes: 1

Arcturus
Arcturus

Reputation: 548

You can control width and precision by namesake member functions on the std::cout object.

From doc at http://en.cppreference.com/w/cpp/io/basic_ostream precision manages decimal precision of floating point operations (public member function of std::ios_base) width manages field width (public member function of std::ios_base)

Upvotes: 0

Related Questions