Oliver Kullmann <O.Kullmann@swansea.ac.uk> writes:
On Tue, Jun 27, 2006 at 02:37:13PM +0100, Reece Dunn wrote:
Oliver Kullmann wrote:
NOTE: Instead of disabling the warnings, you can remove the warning by
updating the code. For example:
void foo( int bar ){} // unreferenced parameter
void foo( int bar ){ bar; } // no warning!
this is FALSE:
No - it is one way to prevent the bug.
bug ? there is no bug here.
I guess you meant "warning" --- but your "repair" is meaningless,
since it introduces a meaningless expression.
And, for good reasons, g++ warns about this(!) (at least versions 4.0.2 and above):
Test.cpp: In function ‘void foo(int)’:
Test.cpp:1: warning: statement has no effect
One should not "repair" code just because of the stupidity of one compiler.
There's nothing wrong with that in principle, We do it routinely. And we have a standard way to suppress unused variable warnings template <class T> inline void suppress_unused_variable_warning(T const&) {} void foo( int bar ){ suppress_unused_variable_warning(bar); } // no warning! you can also just leave out the variable name void foo( int ){} -- Dave Abrahams Boost Consulting www.boost-consulting.com