comment on

Not enough information.

Are the values token4 = 0 & token5 = 32 the values passed into func(), or the values returned by func()?

Because if those are the values coming out of func(), your C program is broken:

0 & 95 = 0. (0 & anything == 0)
32 & 32 = 32. (X & X) always == X
32 & 128 = 0.
0 | 32 | 0 = 32.

But if the values you've given are the values going into func(), then as you've only supplied a declaration (was that form ever legal C?), we cannot tell what might come out for those inputs.

Further more, the only way I can interpret that declaration of func(), is that it is expecting a pointer to an unsigned char as its input. And the likelyhood of 0 and 32 being valid pointers is minimal.

Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.

"Science is about questioning the status quo. Questioning authority".

In the absence of evidence, opinion is indistinguishable from prejudice.

"Too many [] have been sedated by an oppressive environment of political correctness and risk aversion."

In reply to Re: help porting a line of bitwise C code to perl by BrowserUk
in thread help porting a line of bitwise C code to perl by blahblahblah

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