[Bug c/15437] int vs const int computation: different answers

["bangerth at dealii dot org" <gcc-bugzilla at gcc dot gnu dot org>]

------- Additional Comments From bangerth at dealii dot org  2004-05-14 22:13 -------
I see your point, but I don't think you are entirely right. If you 
store something in a 'const int' you tell the compiler that the value 
of the variable won't change. The compiler can then do some simple 
transformations based on this knowledge, without having to resort 
to doing these things at run-time. Note that within the margin of 
errors compiler-internal transformations and run-time behavior might 
differ. 
 
If you just store it in an 'int', the compiler has to do some serious 
investigations to figure out whether a variable is set only once or 
not. Only if it is sure that it can't change can it do these transformation. 
For gcc, these investigations are only performed if you ask for optimization. 
Thus, you _do_ get the same result if you ask for them: 
 
g/x> c++ x.cc ; ./a.out  
f(x) = 299 
f(y) = 300 
 
g/x> c++ x.cc -O2 ; ./a.out  
f(x) = 300 
f(y) = 300 
 
In the first case, one computation is performed at compile- the other at 
run-time. I understand that this seems weird here and maybe frustrating, 
but there are good reasons for this. 
 
W. 

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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=15437