Getting number of int digits

Question

I'm reading an integer from a file using fscanf(fp, "%d", &n) function.

Is there a way to know how many digits that number has without loops?

There is no reason why you would want to do that without loops. All present answers are inferior to the simple `unsigned int digits = 1; while(n > base) { n /= base; digits++; }`. Using strings, float numbers and other such nonsense is just bad programming. You've started a competition over who can write the worst program, so I think this question is harmful to future readers. — Lundin, Feb 18 '16 at 12:15
`unsigned int digits = 1; not_a_loop: if(n > base) { n /= base; digits++; goto not_a_loop; }` There you go. 100 times more effective than any of the answers posted. — Lundin, Feb 18 '16 at 12:20
@Lundin, certainly you meant `while(n >= base)`. (>= vs >). As for me I would have used: `unsigned int digits = 0; do { digits++; n /= base; } while(n); }`. — chux - Reinstate Monica, Feb 18 '16 at 16:30
similar to http://stackoverflow.com/q/1068849/2410359 (possible dupe) — chux - Reinstate Monica, Feb 18 '16 at 18:21
@Lundin - you may like [`int n = 1; while (n /= 10) n++;`](http://stackoverflow.com/a/1069045/2410359) — chux - Reinstate Monica, Feb 18 '16 at 18:25

score 4 · Accepted Answer · edited May 23 '17 at 11:59

4

You can try:

int count = 0;
fscanf(fp, "%d%n", &n, &count);

The %n specifier puts in count the number of characters read so far.

However, this number can be larger that the number of digits of n because the %d specifier allows skipping whitespace characters before it reads the number.

Check @pmg's answer to see how to count those whitespaces.

edited May 23 '17 at 11:59

Community

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answered Feb 18 '16 at 11:39

axiac

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See my answer. We had the same idea, but I counted the whitespace and subtracted from the result. – pmg Feb 18 '16 at 11:53

pmg · Answer 2 · 2016-02-18T11:42:26.433

3

int optionalwhitespace, characters;
if (fscanf(fp, " %n%d%n", &optionalwhitespace, &n, &characters) != 1) /* error */;
characters -= optionalwhitespace;
// characters now has the number of characters read to interpret the value in n

edited Feb 18 '16 at 11:42

answered Feb 18 '16 at 11:37

pmg

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score 2 · Answer 3 · answered Feb 18 '16 at 11:34

2

yes, by reading it as a string:

char buffer[256];
fscanf(fp, "%s", buffer);
int length = strlen(buffer);

NOTE: if it is a negative number, you might want to discount the - sign...

Suppose you still want to have the integer value:

int n = atoi(buffer);

answered Feb 18 '16 at 11:34

Chris Maes

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Wouldn't that `strlen()` give 256? – Mattia Righetti Feb 18 '16 at 11:37
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@matt95 `strlen` will stop counting at `\0` character. – niyasc Feb 18 '16 at 11:38
Please note that this is _far_ slower than reading it as an int and using a loop. – Lundin Feb 18 '16 at 12:04

score 1 · Answer 4 · answered Feb 18 '16 at 11:39

1

By using the log function:

int number_of_digits(unsigned int i)
{
    return (int)(log(i+1) / log(10)) + 1;
}

int main (void)
{
    int i = 52;
    printf("%d has %d digits\n", number_of_digits(i));
}

answered Feb 18 '16 at 11:39

Mathieu

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chux - Reinstate Monica · Answer 5 · 2016-02-18T18:08:28.740

1

For code golf approach, just return the value of snprintf()

width = snprintf(0, 0, "%u", (unsigned) n);

edited Feb 18 '16 at 18:08

answered Feb 18 '16 at 17:49

chux - Reinstate Monica

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score 0 · Answer 6 · answered Feb 18 '16 at 11:40

0

You can use snprinf, which is a better option than itoa:

int x = 0;
scanf( "%d", &x );
char buffer[20] = "";
snprintf(buffer, sizeof(buffer), "%d", x);
printf( "num of digits = %d", strlen(buffer) );

answered Feb 18 '16 at 11:40

artm

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Why 20? Could use 1 or `printf( "num of digits = %d", snprintf(0, 0, "%d", x) );` – chux - Reinstate Monica Feb 18 '16 at 17:47
@chux - the syntax looks very strange to me, but indeed it works. `snprintf` manpage never mentions the first param can be NULL - so I suppose that exploit must come from inspecting the code of `snprintf` itself?? – artm Feb 19 '16 at 11:32
Rather than derived references like a manapge, consider using the C spec. Draft versions readily google-able on-line. "`snprintf(char * restrict s, size_t n, const char * restrict format, ...);` ... If n is zero, nothing is written, and s may be a null pointer. ... The snprintf function returns the number of characters that would have been written had n been sufficiently large ..." No need for exploit or inspecting the code. – chux - Reinstate Monica Feb 19 '16 at 15:45

Marinos K · Answer 7 · 2016-02-18T11:48:07.627

0

try this:

#include <math.h>
number > 0 ? (int) log10 ((double) number) + 1 : 1;

edited Feb 18 '16 at 11:48

answered Feb 18 '16 at 11:40

Marinos K

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score 0 · Answer 8 · answered Feb 18 '16 at 11:41

0

You can just count log10 of that number and round it up like so:

int length = ceil(log10((double)n)) + 0.5;

+ 0.5 Is for casting floating point number to integer, because number like 3 cant be written explicitly in floating point representation doing (int)3.0 may yield 2. I am assuming that ceil doesn't return answer off by 0.5 which is never going to happen.

answered Feb 18 '16 at 11:41

8bra1nz

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Casting `(double)` can lose precision in `n`. – chux - Reinstate Monica Feb 18 '16 at 17:38
@chux I assumed it was `int` not 64 bit integer. – 8bra1nz Feb 18 '16 at 17:40
Adding the 0.5 serves not purpose here. `ceil()` returns a whole number value, why add 0.5 to it only to truncate when converted to `int` – chux - Reinstate Monica Feb 18 '16 at 17:44
@chux `ceil` doesn't return integer value. http://www.cplusplus.com/reference/cmath/ceil/ – 8bra1nz Feb 18 '16 at 17:46
The C spec says "The `ceil` functions compute the smallest integer value not less than `x`." Are they wrong too? – chux - Reinstate Monica Feb 18 '16 at 17:51
You haven't read the link, it says "Rounds x upward, returning the smallest integral value that is not less than x" but in `double`. Send me reference to that C spec then. – 8bra1nz Feb 18 '16 at 17:53
Ref: `C11 7.12.9.1 The ceil functions`. Easy enough to google and find an on-line version. I prefer the C spec to a c++ site. – chux - Reinstate Monica Feb 18 '16 at 18:02
IAC, so the `ceil()` returns a whole/integer/integral value. Why add 0.5 to that result? I have seen code where folks have used the inferior `ceil(x + 0.5)` vs instead of `round(x)`. This is the first that suggests adding 0.5 _after_ computing `ceil()` and then convert to `int` – chux - Reinstate Monica Feb 18 '16 at 18:06

chux - Reinstate Monica · Answer 9 · 2016-02-18T17:23:03.780

0

Is there a way to know how many digits that number has without loops?

A recursion solution

unsigned digit_count(unsigned x) {
  if (x >= 10) return digit_count(x /10) + 1;
  return 1;
}

edited Feb 18 '16 at 17:23

answered Feb 18 '16 at 16:37

chux - Reinstate Monica

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e.upton · Answer 10 · 2016-02-18T12:20:27.037

-1

#include<math.h>

//rest of program
fscanf(fp, "%d", &n)
int length = (int)(log10(n)) +1;
//rest of program

edited Feb 18 '16 at 12:20

answered Feb 18 '16 at 11:39

e.upton

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Errr... `log10(*n)`? I don't think you meant to dereference that integer as a pointer. – David Hoelzer Feb 18 '16 at 11:52
Oops! Edited. Thanks. – e.upton Feb 18 '16 at 12:20

Getting number of int digits

10 Answers10