c++ inconsistency...really, look inside.

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12 comments, last by CTar 17 years, 10 months ago
Quote:Original post by Trienco
Quote:Original post by SiCrane
Another example is a function call: all arguments to the function call must be evaluated before the function is actually called. So if you have foo(++a, ++b), you know that both ++a and ++b were both evaluated before foo() was called.


Just to add one more: while you can be sure that they will be evaluated before the function call, you can't be sure about the order. So doing something like foo(a,++a) and relying on a left to right order isn't a good idea (and real fun to debug when differently optimized builds or builds from different compilers behave differently).


I'm assuming the order its evaluated is based on the calling convention of that function. I'm sure this isn't guarenteed in the standard but it makes sense from a compiler point of view.
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Quote:Original post by SiCrane
C++ has the concept of sequence points. A sequence point is basically a point in the code where everything before it has to have finished evaluating before execution proceeds any further. For example: the end of a statement is a sequence point. So this is well defined:
++a;
++a;
Even though this is not:
++a + ++a;

Another example is a function call: all arguments to the function call must be evaluated before the function is actually called. So if you have foo(++a, ++b), you know that both ++a and ++b were both evaluated before foo() was called.

Logical and and or, the ternary operator and the comma operator also have a sequence point after the first expression. So in A && B, A || B, A , B and A ? B : C the A has to finish evaluating, with all its side effects before B (or C) is evaluated. So ++a && ++a has well defined behaviour even though ++a + ++a</t> doesn't.


So, sequence points consist of
- ;
- ()
- &&
- ||
- ? :
- function call

All other operators are not a sequence point and do not guarentee the order of evaluations.

Anything I'm missing?
Quote:Original post by ph33r
Anything I'm missing?


Yes, the comma operator: ,

++a, ++a; is perfectly well defined.

Note that overloading &&, || or , turns them into a function call, changing the position of the sequence point ang producing undefined behaviour in the above examples.
"Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it." — Brian W. Kernighan
Quote:Original post by ph33r
Quote:Original post by Trienco
Quote:Original post by SiCrane
Another example is a function call: all arguments to the function call must be evaluated before the function is actually called. So if you have foo(++a, ++b), you know that both ++a and ++b were both evaluated before foo() was called.


Just to add one more: while you can be sure that they will be evaluated before the function call, you can't be sure about the order. So doing something like foo(a,++a) and relying on a left to right order isn't a good idea (and real fun to debug when differently optimized builds or builds from different compilers behave differently).


I'm assuming the order its evaluated is based on the calling convention of that function. I'm sure this isn't guarenteed in the standard but it makes sense from a compiler point of view.


No, it might. If we have a function:
void func(int,int,int);
And we call it like this:
func(++a, ++b, ++c)
And the compiler pass parameters from left to right, it could do something like this:
inc ainc binc cpush apush bpush c

But there is no reason it couldn't call inc in the opposite order, especially if it determines that it might give a speedup to change the order (c might be in the cache so we should start incrementing c).

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