fix libproc for big-endian systems

[br]

Summary

use uint32_t for instruction storage. This fixes operation on big endian 64bit MIPS

[rpaulo]

I don't think this is right. Not all big endian systems are LP64.

[br]

In D8250#171624, @rpaulo wrote:

I don't think this is right. Not all big endian systems are LP64.

why? I bet it should work with any sizeof(unsigned long), e.g. 4 or 8

[rpaulo]

In D8250#172166, @br wrote:

In D8250#171624, @rpaulo wrote:

I don't think this is right. Not all big endian systems are LP64.

why? I bet it should work with any sizeof(unsigned long), e.g. 4 or 8

What does this have to do with big endianess? You're basically moving 8 bytes forward and then 4 bytes backwards. What system is this on?

[br]

What does this have to do with big endianess? You're basically moving 8 bytes forward and then 4 bytes backwards. What system is this on?

it is MIPS64EB. Unsigned long is 8 bytes, instruction is 4 bytes, so it copies wrong data (it copies zeroes)

[rpaulo]

In D8250#172233, @br wrote:

What does this have to do with big endianess? You're basically moving 8 bytes forward and then 4 bytes backwards. What system is this on?

it is MIPS64EB. Unsigned long is 8 bytes, instruction is 4 bytes, so it copies wrong data (it copies zeroes)

Then how is it reading the original instruction correctly?

[br]

In D8250#172304, @rpaulo wrote:

In D8250#172233, @br wrote:

What does this have to do with big endianess? You're basically moving 8 bytes forward and then 4 bytes backwards. What system is this on?

it is MIPS64EB. Unsigned long is 8 bytes, instruction is 4 bytes, so it copies wrong data (it copies zeroes)

Then how is it reading the original instruction correctly?

it reads the correct value, e.g. 0x67bdfff000000000. Casting it to uint32_t gives us 0x67bdfff0.

[br]

here is another one possible solution

[rpaulo]

In D8250#172346, @br wrote:

In D8250#172304, @rpaulo wrote:

In D8250#172233, @br wrote:

What does this have to do with big endianess? You're basically moving 8 bytes forward and then 4 bytes backwards. What system is this on?

it is MIPS64EB. Unsigned long is 8 bytes, instruction is 4 bytes, so it copies wrong data (it copies zeroes)

Then how is it reading the original instruction correctly?

it reads the correct value, e.g. 0x67bdfff000000000. Casting it to uint32_t gives us 0x67bdfff0.

Well, this is odd. I wouldn't expect casting would produce such result.

[rpaulo]

In D8250#172348, @br wrote:

here is another one possible solution

I like this solution better.

[br]

use instr_t for instruction storage.

[rpaulo]

lib/libproc/proc_bkpt.c

71 ↗

(On Diff #21651)

Maybe add a comment here explaining why this works for x86.

[br]

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