Re: toolchain's output depends on toolchain used to build the compiler?

[Shachar Shemesh]

Omer Zak wrote:
On Tue, 2010-02-09 at 11:26 +0200, Shachar Shemesh wrote:
  
Orna Agmon Ben-Yehuda wrote:
    

  
If the absolute path is included (I think -g does that). The mount
point in the two environments may be called differently.

      
That's why I'm using objdump -d, which does not print the source
files, and is path location agnostic. In essence, I'm comparing just
the actual assembly produced.
    

On Tue, 2010-02-09 at 11:29 +0200, Shachar Shemesh wrote:
  
Then again, the same source code gets compiled by two supposedly
identical compilers using the same compiler flags. Why should the
result by different?
    

1. Compile the project twice on the same platform, saving the object
files from each compilation - to make sure that there are no timestamps.
  
Tested (and, anyways, if that were the case then the newlib compilation 
would not be the same).
2. From man objdump, I see that there are -d (--disassemble) and -D
(--disassemble-all) flags.  What happens if you use objdump -D instead
of objdump -d?
  
If -d is not the same, -D is guaranteed to be different.
This could catch blocks created from uninitialized (or
differently-initialized) memory areas.
  
In my software or in the compiler's?

Shachar

--
Shachar Shemesh
Lingnu Open Source Consulting Ltd.
http://www.lingnu.com

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