Some confuse about the pass of the arguments by gcc

[åè <yxy dot 716 at gmail dot com>]

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