This question already has an answer here:
I am curious as to how printf
works internally within Linux. I don't understand how it writes data to STDOUT
.
After a bit of searching for the internals, I downloaded glibc
and took a look at the source code:
__printf (const char *format, ...)
{
va_list arg;
int done;
va_start (arg, format);
done = vfprintf (stdout, format, arg);
va_end (arg);
return done;
}
After finding this, I went deeper into the vfprintf
function - but the file is about 2500 lines of unfamiliar C code. I'm looking for an explanation from 10,000 feet of how printf works with a computer's memory and output to display characters on screen.
If I were a piece of assembly code what would I have to do to accomplish the same task? Is it operating system dependent?
I think you're looking at the wrong layer. The logic in vfprintf
is responsible for formatting its arguments and writing them through the underlying stdio functions, usually into a buffer in the FILE
object it's targetting. The actual logic for getting this output to the file descriptor (or on other non-POSIX-like systems, the underlying device/file representation) is probably in fwrite
, fputc
, and/or some __
-prefixed internal functions (perhaps __overflow
).
来源:https://stackoverflow.com/questions/18279908/how-does-printf-work-internally