Test platform is 32 bit x64 Linux, coreutils 8.5.
In the source code of base64, fwrite will use stdout to output the base64 encoded string
When I use ltrace to print all the libc call, we can see that stdout is equal to 0xb772da20
__libc_start_main(0x8048eb0, 2, 0xbf892f74, 0x804cb50, 0x804cbc0 <unfinished ...>
strrchr("base64", '/') = NULL
setlocale(6, "") = "en_US.UTF-8"
bindtextdomain("coreutils", "/usr/share/locale") = "/usr/share/locale"
textdomain("coreutils") = "coreutils"
__cxa_atexit(0x804a3b0, 0, 0, 0xbf892f74, 2) = 0
getopt_long(2, 0xbf892f74, "diw:", 0x0804d1a0, NULL) = -1
fopen64("testbase64", "rb") = 0x8591878
fileno(0x8591878) = 3
posix_fadvise64(3, 0, 0, 0, 0) = 0
fread_unlocked(0xbf89225c, 1, 3072, 0x8591878) = 900
fwrite_unlocked("Ly8gcXVpY2tTb3J0LmMKI2luY2x1ZGUg"..., 1, 76, 0xb772da20) = 76
When I modify the code of base64 like this:
int main (int argc, char **argv)
{
printf("%p \n", stdout);
int opt;
FILE *input_fh;
const char *infile;
.....
The output is still 0xb772da20 , it is strange to me as this is the first line of base64.c.
I grep in the lib folder of coreutils
grep stdout *.h
and I don't see any predefine of stdout.
Could anyone give me some help about why stdout will be defined as "0xb772da20" , not 1, not 0?
According to stdout(3), the stdout is a pointer (to some FILE opaque structure), because it is a file stream. It is not a file descriptor. Its file descriptor is STDOUT_FILENO which indeed is 1.
On my Gnu libc Linux system, I have near line 169 of /usr/include/stdio.h :
/* Standard streams. */
extern struct _IO_FILE *stdin; /* Standard input stream. */
extern struct _IO_FILE *stdout; /* Standard output stream. */
extern struct _IO_FILE *stderr; /* Standard error output stream. */
/* C89/C99 say they're macros. Make them happy. */
#define stdin stdin
#define stdout stdout
#define stderr stderr
Before that (line 48) there is
typedef struct _IO_FILE FILE;
See also this answer to a related question.
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