Board index » cppbuilder » Re: bcb6 incorrectly warns 'declared but never used'

2004-01-15 08:01:33 AM
cppbuilder77
seriously, there's no other legitimate way to supress this warning but
to turn the 8080 compiler warning off. No matter how many times you try
to read that const value, the compiler won't shut up. Unfortunately the
ACE_UNUSED_ARG(a) didn't work. SHUT_UP_ALMOST_ANY_COMPILER(x) neither:
const int nTimes = 3;
(void) nTimes; // still there's a warning!
for (int k=0; k<nTimes; k++){
MessageBox(0,"MESSAGE","TITLE",0);
}
It's a bug, and you can not silence the compiler in any other way but to
turn off the warning itself.
Tom
 

zero-overhead.
Tom
 

OK, finally I got it. It was really tough. Here's a solution that's not
going to crash, has 0 overhead in my opinion, and really works:
#pragma warn +8080
void f()
{
const int nTimes = 3;
for (int k=0; k<nTimes; k++){
MessageBox(0,"MESSAGE","TITLE",0);
}
const int& rnTimes = nTimes;
(void) rnTimes;
}
So you can write a simple macro like this:
#define SHUT_UP_BORLAND(x) const int& temp__##x = x; (void) temp__##x;
You should only use this once within a scope.
Tom
Chris Uzdavinis (TeamB) wrote:

{smallsort}

Tamas Demjen < XXXX@XXXXX.COM >writes:
making it local to the function?
--
Chris (TeamB);
 

or reference to the variable, the compiler believes the variable is
used. If you just try to use its value, the warning will always be issued.
#pragma warn +8080
namespace { const int nTimes = 3; }
void f()
{
for(int k=0; k < nTimes; k++) ;
}
Good idea! It works too. No warning.
Tom