[SOLVED] sprintf: how force leading 0 on floating point result

porphyry5 · 09-27-2012, 10:41 AM

I need the result of a FP calculation to always show as xx.xxx, e.g. 00.123, 01.468, 23.739 (the result is always < 60)

As I read man printf

Quote:

The flag characters
The character % is followed by zero or more of the following flags:
...
0 The value should be zero padded. For d, i, o, u, x, X, a, A, e, E, f, F, g, and
G conversions, the converted value is padded on the left with zeros rather than
blanks. If the 0 and - flags both appear, the 0 flag is ignored. If a precision
is given with a numeric conversion (d, i, o, u, x, and X), the 0 flag is ignored.
For other conversions, the behavior is undefined.

the following code should produce the desired result, but as you see, it does not. What am I missing?

Code:

~ $ awk '{ss=$1*60; sss=sprintf("%02.3f", ss); print "x" sss}'
.01
x0.600

firstfire · 09-27-2012, 11:12 AM

Hi.

Try this

Code:

$ awk '{ss=$1*60; sss=sprintf("%06.3f", ss); print "x" sss}'
0.01
x00.600

or this

Code:

$ printf "%06.3f\n" 0.6
00.600

From man 3 printf:

Quote:

The field width
An optional decimal digit string (with nonzero first digit) specifying a minimum field width. If the
converted value has fewer characters than the field width, it will be padded with spaces on the left
(or right, if the left-adjustment flag has been given). Instead of a decimal digit string one may
write "*" or "*m$" (for some decimal integer m) to specify that the field width is given in the next
argument, or in the m-th argument, respectively, which must be of type int. A negative field width is
taken as a '-' flag followed by a positive field width. In no case does a nonexistent or small field
width cause truncation of a field; if the result of a conversion is wider than the field width, the
field is expanded to contain the conversion result.

So 6 here is the minimum number of characters occupied by your number including padding, dot etc.

porphyry5 · 09-27-2012, 11:34 AM

Many thanks!