How to get number of digits in int variable?

[guidosh]

Hello,

I'm trying to write a printf statement that sets the field width when
printing a number. I'm using this:

printf("%*", fieldwidth, num_to_print);

However, I can't figure out how to get the number of digits in the
number (which I would then assign to the 'fieldwidth' variable above).

I know I don't have to explicitly set the field width with printf, but
I would like very much to know how to discover it.

Can anyone suggest how I might get the number of digits for a numeric
variable?

Thanks...

[Michael Mair]

gu*****@gmail.c om wrote:

Hello,

I'm trying to write a printf statement that sets the field width when
printing a number. I'm using this:

printf("%*", fieldwidth, num_to_print);
Which will not do anything useful.
ITYM "%*d" or whatever format specifier is appropriate for
num_to_print.
However, I can't figure out how to get the number of digits in the
number (which I would then assign to the 'fieldwidth' variable above).
If num_to_print is of type int, you could use
int tmp = num_print;

for (fieldwidth = 0; tmp != 0; tmp /= 10)
fieldwidth += 1;
if (num_print < 0)
fieldwidth += 1;
(untested)
Otherwise you want
fieldwidth = (int)ceil(log(f abs(num_print))/log(10)) + (num_print < 0);
which is quite a lot of work for this information.
Alternatively, I would just use 1 + (CHAR_BIT*sizeo f num_print)/3
as an overestimation to get an always wide enough field.

I know I don't have to explicitly set the field width with printf, but
I would like very much to know how to discover it.

Can anyone suggest how I might get the number of digits for a numeric
variable?

Note that the above is about integer types.
Floating point types
- tell you how many (valid) decimal digits they have
- and usually should not printed to their full precision
if not necessary

Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Arafangion]

Michael Mair wrote:

gu*****@gmail.c om wrote:

Hello,

I'm trying to write a printf statement that sets the field width when
printing a number. I'm using this:

printf("%*", fieldwidth, num_to_print);

Which will not do anything useful.
ITYM "%*d" or whatever format specifier is appropriate for
num_to_print.

However, I can't figure out how to get the number of digits in the
number (which I would then assign to the 'fieldwidth' variable above).

If num_to_print is of type int, you could use
int tmp = num_print;

for (fieldwidth = 0; tmp != 0; tmp /= 10)
fieldwidth += 1;
if (num_print < 0)
fieldwidth += 1;
(untested)
Otherwise you want
fieldwidth = (int)ceil(log(f abs(num_print))/log(10)) + (num_print < 0);
which is quite a lot of work for this information.
Alternatively, I would just use 1 + (CHAR_BIT*sizeo f num_print)/3
as an overestimation to get an always wide enough field.

I know I don't have to explicitly set the field width with printf, but
I would like very much to know how to discover it.

Can anyone suggest how I might get the number of digits for a numeric
variable?

Note that the above is about integer types.
Floating point types
- tell you how many (valid) decimal digits they have
- and usually should not printed to their full precision
if not necessary

Cheers
Michael

Wouldn't that be quite slow? Why not bitshift an int of value 1?

[SM Ryan]

gu*****@gmail.c om wrote:
# Hello,
#
# I'm trying to write a printf statement that sets the field width when
# printing a number. I'm using this:
#
# printf("%*", fieldwidth, num_to_print);
#
# However, I can't figure out how to get the number of digits in the
# number (which I would then assign to the 'fieldwidth' variable above).

L = entier(log10(|n |))+1, |n|>=1, is the number of digits left of the decimal point.
For -1<n<1, it is customary to present L=1 digit ('0') left of the decimal point.
If n<0, it is customary to include a negative mark such as '-'.

All reals have an infinite repeating or nonrepeating decimal representation,
so that right of the decimal point there's no specific limit. For integers, you
can use 0. For floats, if you can get the approximate number of significant
digits supported by the hardware, S, you can use R = S-L as the number of digits
to the right for |n|>=1. For |n|<1, n!=0, you can use entier(-log10(|n|))
zero digits followed by S significant digits.

--
SM Ryan http://www.rawbw.com/~wyrmwif/
If you plan to shoplift, let us know.
Thanks

[Paul Mesken]

On Thu, 19 May 2005 06:46:07 +0200, Michael Mair
<Mi**********@i nvalid.invalid> wrote:

If num_to_print is of type int, you could use
int tmp = num_print;

for (fieldwidth = 0; tmp != 0; tmp /= 10)
fieldwidth += 1;
if (num_print < 0)
fieldwidth += 1;
(untested)

Then test it with tmp = 0 :-)

A multiplication would be faster than a division and doesn't really
change the readability of the code (shifts would and multiplications
are so fast nowadays that it doesn't really pay to use them instead of
mulitplications when more than 1 shift is needed).

int tmp = num_print, fieldwidth = 1, testvalue = 10;

if (tmp < 0)
{
++ fieldwidth;
tmp *= -1;
}
if (tmp > SomeOverflowVal ue)
fieldwidth += TheNumberOfDigi tsOfThatOverflo wValueMinus1
else
while(tmp >= testvalue)
{
++ fieldwidth;
testvalue *= 10;
}

For a 32 bit integer "SomeOverflowVa lue" would be 1 billion (we don't
want undefined behaviour which would probably result in an endless
loop).

Of course, a far quicker solution would involve a lot of if's.
Something like this (you'll need to know the maximum value an int can
have as defined in limits.h) :

if (tmp < 10000)
if (tmp < 100)
if (tmp < 10)
fieldwidth += 1;
else
fieldwidth += 2;
else if (tmp < 1000)
fieldwidth += 3;
else
fieldwidth += 4;
else if (tmp < 10000000)

etc.

The basic idea is that the number of executed if's will be minimized
by using some binary tree like structure (checking first for the
situation with half the maximum number of digits, etc.).

I'm sure there must be better ways but, unfortunately, I'm not
mathematically endowed :-)

[CBFalconer]

gu*****@gmail.c om wrote:

I'm trying to write a printf statement that sets the field width
when printing a number. I'm using this:

printf("%*", fieldwidth, num_to_print);

However, I can't figure out how to get the number of digits in
the number (which I would then assign to the 'fieldwidth' variable
above).

I know I don't have to explicitly set the field width with
printf, but I would like very much to know how to discover it.

Can anyone suggest how I might get the number of digits for a
numeric variable?

The following routine may be useful. They have been designed to
avoid the heavy overhead of a complete printf system.
/* Mask and convert digit to hex representation */
/* Output range is 0..9 and a..f only */
static int hexify(unsigned int value)
{
static char hexchars[] = "0123456789abcd ef";

return (hexchars[value & 0xf]);
} /* hexify */

/* ----------------------- */

/* Convert value to stream of digits
A NULL value for f returns a char count with no output
Returns count of chars. output
return is negative for any output error */
int unum2txt(unsign ed long n, int base, FILE *f)
{
int count, err;

if ((base < 2) || (base > 16)) base = 10;
count = 1;
if (n / base) {
if ((err = unum2txt(n / base, base, f)) < 0) return err;
else count += err;
}
if (f && (putc(hexify(n % base), f) < 0)) return -count;
return count;
} /* unum2txt */

--
"If you want to post a followup via groups.google.c om, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson

[Mark]

<gu*****@gmail. com> wrote in message
news:11******** **************@ o13g2000cwo.goo glegroups.com.. .

Hello,
I'm trying to write a printf statement that sets the field width when
printing a number. I'm using this:
printf("%*", fieldwidth, num_to_print); format specification is missing type...
However, I can't figure out how to get the number of digits in the
number (which I would then assign to the 'fieldwidth' variable above).
I know I don't have to explicitly set the field width with printf, but
I would like very much to know how to discover it.
Can anyone suggest how I might get the number of digits for a numeric
variable?

sprintf();
....
char szbuf[40];
fieldwidth = sprintf(szbuf, "%d", num_to_print);

Mark

[Old Wolf]

Mark wrote:

<gu*****@gmail. com> wrote:

I can't figure out how to get the number of digits in the number
Can anyone suggest how I might get the number of digits for a numeric variable?

sprintf();
...
char szbuf[40];
fieldwidth = sprintf(szbuf, "%d", num_to_print);

I'm getting slightly pedantic here, but on a system with 128-bit int
(or larger), this may cause a buffer overflow.
You could either use the preprocessor to determine the maximum
buffer size required, or if using C99, snprintf(0, 0, "%d", num).

[guidosh]

Hi folks,

Thanks muchly for your ideas. Mark's idea of using sprintf seems a good
one, so I'll try that. Old Wolf, thanks for the note about buffer
overflows too...

G.

[Joe Wright]

gu*****@gmail.c om wrote:

Hello,

I'm trying to write a printf statement that sets the field width when
printing a number. I'm using this:

printf("%*", fieldwidth, num_to_print);

However, I can't figure out how to get the number of digits in the
number (which I would then assign to the 'fieldwidth' variable above).

I know I don't have to explicitly set the field width with printf, but
I would like very much to know how to discover it.

Can anyone suggest how I might get the number of digits for a numeric
variable?

Thanks...

I realize I'm late responding here but first, I don't see your problem.

printf("%d\n", num_to_print);

...will print the 'right' number of digits left-justified.

12
3456
7
876

I propose that fieldwidth is not calculated, but defined by the format.
The 'f' in printf is for format.

This allows right-justifying a column of numbers..

int fieldwidth = 5;
printf("%*d\n", fieldwidth, num_to_print);

12
3456
7
876

Of course, I might have missed the whole point.
--
Joe Wright mailto:jo****** **@comcast.net
"Everything should be made as simple as possible, but not simpler."
--- Albert Einstein ---