x86 : Puzzles about reset code

[王翔]

source code like this:
-----------------------------------------------------------------------------------------------------
    .section ".reset", "ax", %progbits
    .code16
.globl	_start
_start:
    .byte  0xe9
    .int   _start16bit - ( . + 2 )
    /* Note: The above jump is hand coded to work around bugs in binutils.
     * 5 byte are used for a 3 byte instruction.  This works because x86
     * is little endian and allows us to use supported 32bit relocations
     * instead of the weird 16 bit relocations that binutils does not
     * handle consistently between versions because they are used so rarely.
     */
    .previous

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I tried to modify the code and passed test (use qemu test).
My code like this:
-----------------------------------------------------------------------------------------------------
    .section ".reset", "ax", %progbits
    .code16
.globl	_start
_start:
    jmp   _start16bit
    .previous

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What is the meaning of hand coding?  In 16-bit mode, the last two bytes are ignored.

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王翔

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[Patrick Georgi]

2017-08-16 5:03 GMT+02:00 王翔 :
> What is the meaning of hand coding?  In 16-bit mode, the last two bytes

are ignored.
This is _very_ old code. Back in the day, before we started to strongly
encourage people to use our compiler, we had to deal with tons of different
versions of the toolchain.
As the comment to the code indicates, 16bit relocations worked for some and
failed for others. Therefore we went for the safe route and manually
created a 32bit relocation.
We could probably clean up this part of the code now, but since so few
people are ever concerned with it (because it does exactly what it's
supposed to), there was no pressing need yet.

Regards,
Patrick