#include<stdio.h>
int main()
{
for(int i=0;i<=31;i++)
printf("%c",i);
}
when we try to run this code then nothing prints what is the reason for it ?
C is printing them, but perhaps your terminal is not displaying them. This distinction is important because the terminal is responsible for interpreting the output of your program, printing letters, moving the cursor around, changing colors and such.
By historical convention the first 32 characters of the ASCII table are considered "control characters", some of which are printable, some like backspace which move the cursor, others like BEL which can make your terminal beep.
Different terminals may display these differently, or not at all.
It's worth noting that ASCII pre-dates modern "glass" terminals and that these codes were used to move the print-head around on the page. Early machines used teletypes to communicate with them and a line-feed would crank down the paper one line, a carriage return move the cursor back to the start of the line, much like the physical carriage return on a typewriter which would move the "carriage" back to the first column.
These were pretty elaborate elecromechanical contraptions that didn't have any modern circuitry in them, yet they could still process ASCII data, at least for those using ASCII, as there are other character sets like EBCDIC that co-existed with ASCII.
As these characters were never intended to be printed, so they don't have a standard visual representation in ASCII.
With "extended ASCII", as used in DOS, there are symbols defined for them because it seemed like a waste otherwise. These don't have control-code meanings, typically you write them directly to the console character buffer in order to see them.