Re: Some confuse about the pass of the arguments by gcc

[åè <yxy dot 716 at gmail dot com>]

2012/2/22 James Courtier-Dutton <james.dutton@gmail.com>:
> The order that function parameters are evaluated is undefined. Therefore it
> is wise to ensure that no matter what order they are evaluated, the result
> should be the same. It is the ++ that breaks it in this case. Didn't you get
> a compiler warning?

Yes you are right. gcc -Wall indeed get the warning.
operation on 'a' may be undefined [-Wsequence-point]

Thanks  for you reminder.
Let me know ,If we want the result we want ,we should do the precision
logic design .

> On Feb 21, 2012 3:19 PM, "åè" <yxy.716@gmail.com> wrote:
>>
>> I do a experiments to check how gcc pass the arguments.
>> here is the code
>>
>> #include <stdio.h>
>> int main(int argc , char *argv[]){
>>        int a=3;
>>        int b=3;
>>        int c=3;
>>        printf("%d %d\n",++a+c,a+c);
>>        printf("%d %d\n",++b,b);
>>        return 0;
>> }
>>
>> the anwer is
>>
>> 8 7
>> 4 4
>>
>> the piece of assembly language:  gcc 4.6.2
>>
>>        movl    $3, 28(%esp)
>>        movl    $3, 24(%esp)
>>        movl    $3, 20(%esp)
>>        movl    20(%esp), %eax
>>        movl    28(%esp), %edx
>>        leal    (%edx,%eax), %ecx
>>        addl    $1, 28(%esp)
>>        movl    20(%esp), %eax
>>        movl    28(%esp), %edx
>>        addl    %eax, %edx
>>        movl    $.LC0, %eax
>>        movl    %ecx, 8(%esp)
>>        movl    %edx, 4(%esp)
>>        movl    %eax, (%esp)
>>        call    printf
>>        addl    $1, 24(%esp)
>>        movl    $.LC0, %eax
>>        movl    24(%esp), %edx
>>        movl    %edx, 8(%esp)
>>        movl    24(%esp), %edx
>>        movl    %edx, 4(%esp)
>>        movl    %eax, (%esp)
>>        call    printf
>>
>> In the first case , gcc first compute the a+c to %ecx ,and pass it
>> stack , the compute ++a+c to %edx ,so the answer is 8 7
>>
>> In the second case , why it didn't do the same thing like
>> compute b=3  and pass it to stack ,then compute ++b and pass it to
>> stack .to the result 4 3. However it  first
>> addl    $1, 24(%esp)  ==>  b+1  I think it compute the expression
>> b+1. the pass it to stack . the b which now is 4 and was passed to
>> stack.
>>
>> I was wondering why gcc handle  the same mode in two ways.
>>
>> Is there some errors I had made ?
>> In my opinion ,for the reason  of the concept of consistent ,it  may be
>> 8  8
>> 4  4
>>
>> thank you advance