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return value for _lread

Hi,

I'd like to know what is the expected behavior for the _lread functions (kerlnel.dll # 82 and kernel32.dll # 596) when uBytes == 0. This is not described in the API spec.

I.e., if we ask to read 0 bytes, what will happen:

an error? which one?

what is returned by the function? 0? -1?

does that depend on wether the EOF has been reached?

does that depend on the windows' version?

Thanks.
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zorzella
Asked:
zorzella
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1 Solution
 
nietodCommented:
There is no problem reading 0 bytes.  It just does nothing and returns 0 (the number of bytes read).
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zorzellaAuthor Commented:
Is this valid if the file reader has just reached the EOF?

For example, imagine that EOF has already been reached, and one asks for a _lread of 0 bytes.

In the specs, it is said:

"If the number of bytes read is less than uBytes, the function has reached the end of file (EOF) before reading the specified number of bytes. "

So, would it return 0 even though EOF was reached?

Thanks
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nietodCommented:
assuming no error, _lread() returns the number of bytes that were read.  This is the lesser of the number requested or the number remaining after the current file position.  If you specify 0 bytes to be read, then the procedure will read 0 bytes (regardless of the current file position) and will return 0.  Thus when you read 0 bytes it doesn't matter what the current file position is.
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zorzellaAuthor Commented:
Thanks. BTW, did you find this documented somewhere or you just know it by use?

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nietodCommented:
both documentation and use---but not of _lread().  'm assuming (and it is a safe assumption) that it works like the 100s of other functions that perform the same task (C and C++ have several that work like this, DOS has a read service that works like this, Win32 has the ReadFile() function.)  Thus there is a sort-of convention to how a read() function is supposed to work and _read() will almost certainly follow it.  If it didn't it would have to have a good reason and would have to make it clear in the documentation because programmers will assume it does.
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