Paul Pluzhnikov wrote in message...
> "BobR" writes:
>
> > Alexandru Mosoi wrote in message...
> >> supposing that I have the following lines:
> >>
> >> long long a = 12345678901234567890LL;
> >> int b = a;
> >>
> >> how can I make g++ (4.1.2) issue an warning because of loss of
> >> precision?
> >>
> >
> > Use -ansi -pedantic.
>
> int main()
> {
> unsigned long long a = 12345678901234567890ULL;
> int b = a;
> return 0;
> }
>
> $ /usr/local/gcc-4.2.1/bin/g++ -ansi -pedantic t.c
> t.c:3:28: warning: use of C99 long long integer constant
> t.c: In function 'int main()':
> t.c:3: error: ISO C++ does not support 'long long'
>
> $ /usr/local/gcc-4.2.1/bin/gcc -ansi -pedantic t.c
> t.c: In function 'main':
> t.c:3: warning: ISO C90 does not support 'long long'
> t.c:3:28: warning: use of C99 long long integer constant
>
> Perhaps you should try your suggestions before suggesting them?
Using only -Wall (no -ansi or -pedantic):
double c( 123.054);
int d = c;
// [Warning] initialization to `int' from `double'
// [Warning] argument to `int' from `double'
But, type 'unsigned long long' is not a 'built-in' type (maybe next ISO),
so, the 'warning: use of C99 long long integer constant' is the only way I
know.
I always do the checks at runtime.
You could do:
unsigned long long a = 12345678901234567890ULL;
int b(0);
if( b == a ){}
// [Warning] comparison between signed and unsigned integer expressions
Kind of silly. If you knew to put the 'if()' there, you'd know to do it
right.
if( sizeof( b ) == 4 ){
b = a & 0x7FFFFFFF;
if( (unsigned long long)(b) != a ){ b = 0; }
}
....or whatever.
You got anything?
--
Bob R
POVrookie
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