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SubjectRe: typedefs and structs

On Wed, 9 Nov 2005, Vadim Lobanov wrote:

> On Wed, 9 Nov 2005, linux-os \(Dick Johnson\) wrote:
>> On Wed, 9 Nov 2005, linas wrote:
>>> On Wed, Nov 09, 2005 at 08:22:15AM -0800, Vadim Lobanov was heard to remark:
>>>> On Wed, 9 Nov 2005, J.A. Magallon wrote:
>>>>> void do_some_stuff(T& arg1,T& arg2)
>>>> A diligent C programmer would write this as follows:
>>>> void do_some_stuff (struct T * a, struct T * b);
>>>> So I don't see C++ winning at all here.
>>> I guess the real point that I'd wanted to make, and seems
>>> to have gotten lost, was that by avoiding using pointers,
>>> you end up designing code in a very different way, and you
>>> can find out that often/usually, you don't need structs
>>> filled with a zoo of pointers.
>> But you can't avoid pointers unless you make your entire
>> program have global scope. That may be great for performance,
>> but a killer if for have any bugs.
> Just to extract some useful technical knowledge from the current ongoing
> "flamewar"...
> I'm not entirely sure if the above statement regarding performance is
> correct. Some enlightenment would be appreciated.
> Suppose you have the following code:
> int myvar;
> void foo (void) {
> printf("%d\n", myvar);
> bar();
> printf("%d\n", myvar);
> }
> If bar is declared in _another_ file as
> void bar (void);
> then I believe the compiler has to reread the global 'myvar' from memory
> for the second printf().

Correct because bar() could have modified (it's global).

> However, if the code is as follows:
> void foo (void) {
> int myvar = 0;
> printf("%d\n", myvar);
> bar(&myvar);
> printf("%d\n", myvar);
> }
> If bar is declared in _another_ file as
> void bar (const int * var);
> then I think the compiler can validly cache the value of 'myvar' for the
> second printf without re-reading it. Correct/incorrect?

Maybe you tried to trick me by showing the variable was not going
to be changed (const *). In that case, the compiler may not re-read
the variable. However, it can re-read the variable.

A "smart" compiler might just do: write(1, "0\n", 2);
... for the first printf() as well. Such compilers make
debugging difficult.

> -Vadim Lobanov

Dick Johnson
Penguin : Linux version on an i686 machine (5589.55 BogoMips).
Warning : 98.36% of all statistics are fiction.

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