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.)

Answer 1

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);

Answer 2

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)

Answer 3

Example from Bash command line:

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

Answer 4

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

Answer 5

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

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