why /dev/stdout is "1

Question

SEGMENT .data ; nothing here
SEGMENT .text ; sauce
global _start
_start:
            pop ECX ; get ARGC value
            mov EAX, 4 ; sys_write()
            mov EBX, 1 ; /dev/stdout
           ;^^^^^^^^^^^
            mov EDX, 1 ; a single byte
            int 0x80
            mov EAX, 1 ; sys_exit()
                    mov EBX, 0 ; return 0
            int 0x80
SEGMENT .bss ; nothing here

why mov EBX, ""1"" ; /dev/stdout ? where document I can find "1" ?

("In C, stdin, **stdout**, and stderr are FILE*, which in UNIX respectively map to file descriptors 0, **1** and 2." - [from this answer](https://stackoverflow.com/a/5257718)) — user202729, Mar 04 '18 at 14:28
Here is a good survey of the history: [Standard streams](https://en.wikipedia.org/wiki/Standard_streams) from wikipedia. The article goes back to the 1950's and discusses Unix design, and explains why the streams were standardized and their values. — jww, Mar 04 '18 at 14:44

score -1 · Answer 1 · answered Mar 04 '18 at 14:25

On unix systems, a process normally have 3 common I/O channels connected to it, called stdin/stdout/stderr. These have the corresponding file descriptor values 0, 1 and 2.

This is documented at: http://pubs.opengroup.org/onlinepubs/9699919799/functions/stdin.html (See also the wikipedia page )

why /dev/stdout is "1

1 Answers1