Specifying maximum printf field width for numbers (truncating if necessary)?

Question

You can truncate strings with a printf field-width specifier:

printf("%.5s", "abcdefgh");

> abcde

Unfortunately it does not work for numbers (replacing d with x is the same):

printf("%2d",   1234);  // for 34
printf("%.2d",  1234);  // for 34
printf("%-2d",  1234);  // for 12
printf("%-.2d", 1234);  // for 12

> 1234

Is there an easy/trivial way to specify the number of digits to be printed even if it means truncating a number?

MSDN specifically says that it will not happen which seems unnecessarily limiting. (Yes, it can be done by creating strings and such, but I’m hoping for a “printf trick” or clever kludge.)

I'm not able to comment yet so I'm adding a new answer, it's mathematically impossible to do a truncate faster than log, if you ever found one, then you can make a faster log from it. — JimmyZ, Mar 21 '19 at 07:08

score 30 · Accepted Answer · answered Mar 31 '12 at 16:27

30

Like many of my best ideas, the answer came to me while lying in bed, waiting to fall asleep (there’s not much else to do at that time than think).

Use modulus!

printf("%2d\n", 1234%10);   // for 4
printf("%2d\n", 1234%100);  // for 34

printf("%2x\n", 1234%16);   // for 2
printf("%2x\n", 1234%256);  // for d2

It’s not ideal because it can’t truncate from the left (e.g., 12 instead of 34), but it works for the main use-cases. For example:

// print a decimal ruler
for (int i=0; i<36; i++)
  printf("%d", i%10);

answered Mar 31 '12 at 16:27

Synetech

9,643
9
64
96

2

If you want to truncate from the left, assuming the values are integer, just divide. eg.: `1234/100 // for 12` – Ortwin Angermeier Mar 20 '14 at 12:34
@ortang, yes, I know, but it’s [not the same as truncating from the right](http://stackoverflow.com/questions/9953125/9957649?noredirect=1#comment34294951_22533677). – Synetech Mar 20 '14 at 14:25
3

Please note, that using modulus is not safe for negative numbers. -9999 % 1000 gives either 1 or -999, depending on your compiler, the CPU and the compiler flags used. Most modern combinations tend to give -999 though, as they round -9999 / 1000 towards zero (giving -9, not -10, as the result). So there is still a risk of buffer and field overruns! – Kai Petzke Feb 09 '19 at 12:25

Johannes Luong · Answer 2 · 2013-05-12T14:56:53.177

7

If you want to truncate from the right you can convert your number to a string and then use the string field width specifier.

"%.3s".format(1234567.toString)

edited May 12 '13 at 14:56

answered May 12 '13 at 12:08

Johannes Luong

574
5
19

1

I already acknowledged that you convert them to strings; this question is about raw numbers. Presumably (someone would have to check to be sure, but it seems reasonable), the conversion in `printf` is more optimal than doing a manual conversion. – Synetech May 12 '13 at 13:54

score 4 · Answer 3 · answered Apr 20 '17 at 13:09

4

Example from Bash command line:

localhost ~$ printf "%.3s\n" $(printf "%03d"  1234)
123
localhost ~$

answered Apr 20 '17 at 13:09

adrigiga

41
1

score 1 · Answer 4 · answered Feb 09 '14 at 00:57

1

You could use snprintf to truncate from the right

char buf[10];
static const int WIDTH_INCL_NULL = 3;

snprintf(buf, WIDTH_INCL_NULL, "%d", 1234); // buf will contain 12

answered Feb 09 '14 at 00:57

KarlR

11
1

That uses a buffer and hard-coded values, so it’s even less ideal—using dynamically allocated memory would be even worse. – Synetech Feb 10 '14 at 20:30

score 0 · Answer 5 · answered Mar 20 '14 at 12:54

0

Why not from the left? the only difference is to use simple division:

printf("%2d", 1234/100); // you get 12

answered Mar 20 '14 at 12:54

troy

704
8
16

3

When I said that my solution can’t truncate from the left, I didn’t mean it’s not possible; as you demonstrated, you can use division. The reason I didn’t use it is because unlike modulus which can easily give you the right-most *n* digits of *any* number, to truncate from the left, you have to first determine the number of digits in the number before you can divide (in your example, you have to use `1000` instead of `100` if the number were `12345`). This creates a lot more work to first determine the number of digits, then multiply 10 to get a mask, then divide, all just to truncate. `:-/` – Synetech Mar 20 '14 at 14:24
There may be (probably are) edge cases I'm not thinking of, but something like `num / pow(10, (int)log10(num) - digits + 1);` should work for simple cases (assuming base 10). – Chad Schouggins Jun 02 '16 at 20:21

Specifying maximum printf field width for numbers (truncating if necessary)?

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