printf format float with padding

Question

The following test code produces an undesired output, even though I used a width parameter:

int main(int , char* [])
{
    float test = 1234.5f;
    float test2 = 14.5f;

    printf("ABC %5.1f DEF\n", test);
    printf("ABC %5.1f DEF\n", test2);

    return 0;
}

Output

ABC 1234.5 DEF   
ABC  14.5 DEF

How to achieve an output like this, which format string to use?

ABC 1234.5 DEF   
ABC   14.5 DEF

@hmjd: My real code is C++, but this example is pure C. So you are right, I try to retag it. — nabulke, Mar 07 '13 at 10:05

score 73 · Accepted Answer · answered Mar 07 '13 at 10:01

73

The following should line everything up correctly:

printf("ABC %6.1f DEF\n", test);
printf("ABC %6.1f DEF\n", test2);

When I run this, I get:

ABC 1234.5 DEF
ABC   14.5 DEF

The issue is that, in %5.1f, the 5 is the number of characters allocated for the entire number, and 1234.5 takes more than five characters. This results in misalignment with 14.5, which does fit in five characters.

answered Mar 07 '13 at 10:01

NPE

486,780
108
951
1,012

4

@nabulke: Yes, every character counts. – NPE Mar 07 '13 at 10:03

score 17 · Answer 2 · answered Mar 07 '13 at 10:01

17

You're trying to print something wider than 5 characters, so make your length specifier larger:

printf("ABC %6.1f DEF\n", test);
printf("ABC %6.1f DEF\n", test2);

The first value is not "digits before the point", but "total length".

answered Mar 07 '13 at 10:01

JasonD

16,464
2
29
44

Note that if you have a chance of printing infinity or Nan, you may want to provide more space after the decimal (at least 4 characters). Then it becomes something like: `printf("ABC %10.4f DEF\n", test);` – phyatt Oct 26 '18 at 16:17

printf format float with padding

2 Answers2