On 21/09/13 21:33, wempwer@xxxxxxxxx wrote: > On Sat, Sep 21, 2013 at 07:27:20PM +0100, Jonathan Wakely wrote: >> On 21 September 2013 19:09, <wempwer@xxxxxxxxx> wrote: >>> >>> It tried -O2 and it also worked. However on my system it only said: >>> >>> warning: integer overflow in expression >>> >>> How did you get gcc to produce this part: >>> >>> [-Woverflow] >>> >>> I use gcc 4.5.2. But when I use gcc like this I also do not get any >>> warnings: >>> >>> gcc -Woverflow main.c -o main >> >> 4.5.2 is quite old now. More recent versions show the warning option >> that triggered the warning, that's what the [-Woverflow] part is. >> >>> I wonder: >>> >>> 1) why -Woverflow appears in your gcc output? >> >> Because it's a recent version. >> >>> 2) why `gcc -Woverflow main.c -o main' does not produce a warning? >> >> Because without optimisation the compiler doesn't do the necessary >> analysis to realise that in the multiplication expression it knows the >> values. >> >>> 3) why does `gcc -O main.c -o main' produce warning in the first place >>> and why does it do this only when two operands are const? >> >> Because the optimiser can track the values of constants (because they >> don't change) and see that the values in the multiplication are known. > > But compiler also knows the operand values even if they are not > constant? These values must be known to produce the multiplication > result. Is this warning produced only when two operands are constant > and optimization is turned on because this is just the way gcc does it > and the warnings could also be produced for non-const operands if a > compiler worked in a different way? > The compiler can only produce warnings here if it knows the values you are using for the calculations. If you write "r = ab * bc;" then compiler assumes that you know what you are doing, and using sensible values for "ab" and "bc". But if the compiler can calculate the values involved (it requires -O1 or above for many of these calculations and analysis passes), and it can see there is definitely an error, then it can give a warning. So it needs the values to be "compile-time constants". That does not necessarily mean values declared with "const", such as "const int ab = 50000;". It also includes values set with #define, enumeration constants, literals (i.e., "50000"), as well as statements, expressions and function calls using these if the compiler knows all the details at compile-time. Note that the compiler /can/ give a warning in such cases, but it does not have to do so. gcc is a fantastic compiler, and gives better compile-time warnings and static analysis than any other compiler I have used - but it has limits. If it can easily do the calculations at compile-time, /and/ you have optimisation flags to enable such passes, /and/ you have enabled the appropriate warnings - /then/ gcc will give you the warnings.
