linux - What is global _start in assembly language?

Question

This is my assembly level code ...

section .text
global _start
_start: mov eax, 4
        mov ebx, 1
        mov ecx, mesg
        mov edx, size
        int 0x80
exit:   mov eax, 1
        int 0x80
section .data
mesg    db      'KingKong',0xa
size    equ     $-mesg

Output:

root@bt:~/Arena# nasm -f elf a.asm -o a.o
root@bt:~/Arena# ld -o out a.o
root@bt:~/Arena# ./out 
KingKong

My question is What is the global _start used for? I tried my luck with Mr.Google and I found that it is used to tell the starting point of my program. Why cant we just have the _start to tell where the program starts like the one given below which produces a kinda warning on the screen

section .text
_start: mov eax, 4
        mov ebx, 1
        mov ecx, mesg
        mov edx, size
        int 0x80
exit:   mov eax, 1
        int 0x80
section .data
mesg    db      'KingKong',0xa
size    equ     $-mesg

root@bt:~/Arena# nasm -f elf a.asm
root@bt:~/Arena# ld -e _start -o out a.o
ld: warning: cannot find entry symbol _start; defaulting to 0000000008048080
root@bt:~/Arena# ld -o out a.o
ld: warning: cannot find entry symbol _start; defaulting to 0000000008048080

Answer 1

global directive is NASM specific. It is for exporting symbols in your code to where it points in the object code generated. Here you mark _start symbol global so its name is added in the object code (a.o). The linker (ld) can read that symbol in the object code and its value so it knows where to mark as an entry point in the output executable. When you run the executable it starts at where marked as _start in the code.

If a global directive missing for a symbol, that symbol will not be placed in the object code's export table so linker has no way of knowing about the symbol.

If you want to use a different entry point name other than _start (which is the default), you can specify -e parameter to ld like:

ld -e my_entry_point -o out a.o