Why do I get warning C4081 on this #pragma?

Question

I am in the habit of removing all warning reported in my code. I just like a clean build if possible. I used

#pragma comment(lib,"some.lib");

I get this warning:

warning c4081: expected 'newline'; found ';'

I am uncertain why that would create a warning. Could I get help on removing it?

Answer 1

#pragma warning(disable: 4081)

will disable the warning. but I can't repro the warning unless the syntax of the comment is wrong.

#pragma comment(lib "some.lib") 

gives me warning 4081 but

#pragma comment(lib, "some.lib") 

does not.

What is the text of your warning message?

Edit: I see now, forget adding a #pragma warning, just remove the ; from the end of the comment line. it's a syntax error.

Answer 2

You can selectively, and temporarily disable all warnings like this:

#pragma warning(push, 0)        

#include <boost/filesystem/operations.hpp>
#include <boost/filesystem/exception.hpp>
#include <boost/filesystem/convenience.hpp>
#include <boost/program_options.hpp>
#include <boost/foreach.hpp>
#include <boost/algorithm/string/replace.hpp>
#include <boost/filesystem/operations.hpp>
#include <boost/filesystem/exception.hpp>
#include <boost/bind.hpp>
#include <boost/iostreams/copy.hpp>
#include <boost/iostreams/filter/zlib.hpp>
#include <boost/iostreams/filtering_stream.hpp>
#include <boost/iostreams/device/back_inserter.hpp>

#pragma warning(pop)

Instead of 0 you can optionally pass in something like:

#pragma warning( push )
#pragma warning( disable : 4081)
#pragma warning( disable : 4706 )
#pragma warning( disable : 4707 )
// Some code
#pragma warning( pop ) 

Answer 3

You have not mentioned the compiler type and version, but I think you need to put the name without the ".lib" in second parameter (see here)

Answer 4

Its the semi-colon at the end of the line. Its not needed for #pragma.

edit: The warning says it all: Expected a newline at the end of the pragma, but found a semi-colon instead.

Tested with VS2008