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List:       gcc
Subject:    Re: Volatile constants?
From:       Richard Earnshaw <rearnsha () arm ! com>
Date:       2000-03-31 11:16:09
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> inline int abs1(int x) {
>     return x<0 ? -x : x;
> }
> #define abs2(x) ((x)<0 ? -(x) : x)
>  
> int r = abs1(-7);
> int s = abs2(-7);
> 
> >An inline function is NOT a macro, it's still a function.  You can't use a 
> >function as an initializer for a static extent variable in C.
> 
> OK, this is not valid in ANSI C
> In C++, however, this is valid.
> Now, take a look at the assembly output (x86,regparm)
> 
[...]

> Why is 'r' not handled as 's'?
> This is slower and occupies more memory.
> 

Because even in C++ you have still specified it as a constructor function, 
not a constant expression.  All this means is that the your inline 
function will get inlined into your constructors rather than left as a 
stand-alone function that is called by them (try taking the "inline" out 
and see how it changes your assembly output).

R.

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