The write system call function is at a lower level than printf. The job
of printf is to convert C data types to character strings for display.
That is, the decimal integer 123 is converted to the characters '1',
'2', '3', and '\0' (the null character).
This character string is sent to write, which writes one byte
(character) at a time to standard out (usually the screen).
printf must take endianness into account when doing its type
conversions.
write is at a much lower level. It knows nothing about data types or
endianess. I simply writes one byte at a time to the specified file
descriptor.
Bob
On Sat, 2005-07-09 at 17:41 +0300, Nanakos Chrysostomos wrote:
> Thanks very much for your response.
> I know that x86 CPU is little endian,but how the printf prints out the
> that number which starts from the first lowest byte in mem /stack which is
> 0x44??
> Maybe internally knows that this is a cast???
> Lest assume the following example:
> #include <stdio.h>
>
> int main(void) {
> unsigned int buf = {0x44,0x43,0x42,0x41};
> unsigned char *x = (unsigned char*)&buf;
> printf("%#x\n", *(int *)x);
> return 0;
> }
>
> prints out the buf array in reverse order ,little-ednian, and treats it as
> a number,
>
> 0x41424344
>
> But check out the following example:
>
> endian.txt
> -----------
> DBCA
> .section .data
> .filename: .string "endian.txt"
>
>
> .text
> .globl _start
>
> _start:
> pushl %ebp
> movl %esp,%ebp
>
> movl $5,%eax
> movl $.filename,%ebx
> movl $0x00,%ecx
> int $0x80
>
> movl %eax,%ebx
> movl $3,%eax
> leal -8(%ebp),%ecx
> movl $4,%edx
> int $0x80
>
> movl $4,%eax
> movl $1,%ebx
> movl $0x0a,-4(%ebp)
> leal -8(%ebp),%ecx
> movl $5,%edx
> int $0x80
>
> movl $1,%eax
> movl $0,%ebx
> int $0x80
>
>
> #as -o example.o example.s
> #ld -o example example.o
> #./example
> DBCA
>
> We print out the memory from the lowest byte-order.
> How can we print out by using the system call 'write' this byte-order and
> treat it like a number,as printf does.????????????????????????????????
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