Difference between #define and declearing

AnshulMaheshwari · 09-27-2012, 12:42 AM

HI Guys

This is my program source code that give me warning

Code:

#define a 0x8000

main()
{
   short int var = ~a;
}

gcc 4.2 gives me warning

Quote:

overflow in implicit constant conversion [-Woverflow]

There is another code that doesn't give me warning

Code:

int a = 0x8000;
main()
{
   short int var = ~a; 
}

This Compile successsfully

NevemTeve · 09-27-2012, 02:26 AM

And what is your question?

In the first case ~a is a constant value (0xffff7fff, type int), in the second it is an expression. The warning you got in the first case refers to conversion of a constant, so it doesn't apply to the second case.

AnshulMaheshwari · 09-27-2012, 04:16 AM

My Question was Why compiler doesn't give warning in second case ??

Code:

const int a = 0x8000;
main()
{
  short int var = ~a; 
}

According to Nevem Teve this code should have given me warning but even this is not giving me warning..

NevemTeve · 09-27-2012, 05:14 AM

Let me repeat: the warning is about constant-overflow. There is no such thing in the second case.

Edit: in this case 'constant' doesn't mean "variable with const modifier", it means literal value, or an expression that can be evaluated compile time, eg 1+2, 'Z'-'A', "string", ~0x8000 etc

eSelix · 10-01-2012, 05:06 AM

If you want to be warned about converting "int" to "short int", you can use "-Wconversion" flag for gcc.