What happened when using -Os with gcc?

["Tech Win" <tech_win at shaw dot ca>]

Hi there, 

I ran across a strange problem in my project, and eventually I tracked 
down to the "-Os" option with gcc. Please take a look at the following 
code: 

================================================== 
static float test_cell() 
{ 
    int i; 
    float sum =0; 
    for ( i= 0; i < 100; i ++ ) 
        sum += i; 
    return sum; 
} 
void strange_test(const char *name) 
{ 
        int i; 
        float t; 
        printf("================= %s ===============\n", name); 
    for ( i = 0; i< 3; i++ ) { 
        int head = !i; 
        float x = test_cell(); 
        float width = 100.0; 
        if ( head ) { 
            t = (width + (int)x) / 2; 
            printf("1 --- %d\n", head); 
        } 
        else { 
            t = (int)x * 10; 
            printf("0 --- %d\n", head); 
        } 
    } 
    printf("t=%.2f\n", t); 
}
 
The normal output should be: 
==============ZZZZZZZZZZZZZZZZZ================ 
1 --- 1 
0 --- 0 
0 --- 0 
t=49500.00
 
But if I use -Os when compiling, the result is: 
================= LAST =============== 
1 --- 1 
1 --- 0 
1 --- 0 
t=2525.00
 
Seems it's not branching! 

When I use -O0, -O1, -O2, or -O3 the output is no problem. But if I use -Os
then get the wrong result. 

And if I change the line 
      t = (width + (int)x) / 2;
to
       t = (width+x)/2

And
            t = (int)x * 10;  
to
       t = x * 10;

It'll be no problem with -Os options in gcc.

Can anybody please explain this? 

My GCC version is 3.4.6.

Thanks a lot, 


Win