about arrays initialization

[uulinux]

Dear Bob plantz:
I am sorry for reply your mail so late. Yes, I see why it is done twice 
too. Thank you for your reminder.

    I think I have got the answer. May god bless you and your family.


于 2011/8/28 下午 11:40, Bob Plantz 写道:
> Ah, yes, I missed the i=0 statement. And now I see why it is done twice.
> Your code sets i=0 twice:
>
> int i = 0;
> and
> for (i=0;....
>
> The order of assigning values to the elements of the array makes sense,
> but it's not part of the C language standard. I would not be surprised
> if some other C compiler does as you suggest with this code. But to
> ensure that your code is portable, you need to do (notice my comment to
> help the maintenance programmer):
> ===============================================
> #include <stdio.h>
> int main()
> {
> int a[5]={1, 2, 3};
> int i;
> // Finish initializing array
> a[3] = a[2];
> a[4] = a[0] + a[1];
>
> for (i=0; i<5; i++) {
> printf("\t%d\n", a[i]);
> }
> return 0;
> }
> ===============================================
>
> By the way, it is my personal programming philosophy that one should NOT
> do arithmetic operations during variable initialization.
>
> Also, I like to use symbolic names and #defines. Again, it helps the
> maintenance programmer. (I have found that this is often me, a few weeks
> later.) So the above code might become something like (depending on what
> the numbers actually represent):
> ===============================================
> #define nCats 1
> #define nDogs 2
> #define nBirds 3
> #define nWinged nBirds
> #define nFourLegged nCats+nDogs
>
> #include <stdio.h>
> int main()
> {
> int myCreatures[5]={nCats, nDogs, nBirds, nWinged, nFourLegged};
> int i;
>
> for (i=0; i<5; i++) {
> printf("\t%d\n", myCreatures[i]);
> }
> return 0;
> }
> ===============================================
>
>
>
> On 08/28/2011 12:53 AM, uulinux wrote:
>> Dear Bob plantz:
>>
>> Thank you very mach. You are very enthusiastic and friendly, and your
>> analysis is very good. The C code is as following:
>> ===============================================
>> #include <stdio.h>
>> int main()
>> {
>> int a[5]={1, 2, 3, a[2], a[0]+a[1]};
>> int i=0;
>>
>> for (i=0; i<5; i++) {
>> printf("\t%d", a[i]);
>> }
>> return 0;
>> }
>> ===============================================
>> So, please read the following again.
>> >> movl %eax, 36(%esp) # a[3] = a[2] (previously computed above]
>> >> movl %edx, 40(%esp) # a[4] = a[0] + a[1] (previously computed above]
>> >> movl $0, 44(%esp) # a[5] = 0
>> >> movl $0, 44(%esp) # a[5] = 0
>> > # I have no idea why 0 is stored beyond the end of the array or why
>> it is done twice. But the assignment to the elements of the array
>> occur as I have always expected in C.
>> It is not a[5] = 0. It is i=0; But I also have no idea about why it is
>> done twice.
>>
>> I have known the reason of this issue is "a[3], a[4]" has not
>> evaluated before the assignment occurred. So the result of this action
>> is not fixed? And the result of this action is determined by the
>> compiler. But there is still a question in my head. Why gcc put
>> "a[3]=a[2]; a[4]=a[0]+a[1];" operation in the front of "a[0]=1;
>> a[1]=2; a[2]=3;"? Is not the array has initialized from a[0] to a[n]?
>> I just wonder why not put "a[0]=1; a[1]=2; a[2]=3;" operation in the
>> front of "a[3]=a[2]; a[4]=a[0]+a[1];".
>>
>> Is there a method to achieve “a[3] = a[2]; a[4]=a[0]+a[1]” for global
>> variable. So, I just need to update the value of a[0]…a[1] when the
>> value should be changed. I think it is useful when the arithmetic
>> expression is very complex. The value of some elements in a array will
>> be auto updated by compiler, just like this:
>> ===============================================
>> static int a[6]={1, 2, 3, a[2], a[0]+a[1], sin(a[4])}; // gcc-4.4.5
>> report error !
>>
>> #include <stdio.h>
>> int main()
>> {
>> int i=0;
>>
>> for (i=0; i<6; i++) {
>> printf("\t%d", a[i]);
>> }
>> return 0;
>> }
>> ===============================================
>