You can truncate strings with a printf
field-width specifier:
printf(\"%.5s\", \"abcdefgh\");
> abcde
Unfortunately it does no
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
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)
Example from Bash command line:
localhost ~$ printf "%.3s\n" $(printf "%03d" 1234)
123
localhost ~$
Why not from the left? the only difference is to use simple division:
printf("%2d", 1234/100); // you get 12
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);