[Bug middle-end/22072] bizarre code for int*int/2

["pluto at agmk dot net" <gcc-bugzilla at gcc dot gnu dot org>]

------- Additional Comments From pluto at agmk dot net  2005-06-17 00:53 -------
(In reply to comment #7)  
> (In reply to comment #6)  
> >  SAR r/m32, 1    Signed divide* r/m32 by 2, once  
>   
> Huh, I think that is wrong,  
 
/* 
 * specifically, the division algorithm states that given 
 * two integers a and d, with d != 0, there exists unique 
 * integers q and r such that a = qd + r and 0 <= r < |d|, 
 * where |d| denotes the absolute value of d. 
 * the integer q is the quotient, r is the remainder, 
 * d is the divisor, and a is the dividend. 
 * 
 * examples: 
 * 
 * if a =  7 and d =  3, then q =  2 and r = 1. 
 * if a =  7 and d = -3, then q = -2 and r = 1. 
 * if a = -7 and d =  3, then q = -3 and r = 2. 
 * if a = -7 and d = -3, then q =  3 and r = 2. 
 * 
 */ 
 
#include <stdio.h> 
 
const int a[4] = { 5, -5 }; 
 
void div2_wrong(int* q, int*r, const int a) 
{ 
    *q = a / 2; 
    *r = a % 2; 
} 
 
void div2_correct(int* q, int*r, const int a) 
{ 
    *q = a >> 1; 
    *r = a - (*q << 1); 
} 
 
int main() 
{ 
    int i; 
    for (i = 0; i < 2; i++) 
    { 
        int q, r; 
        div2_wrong(&q, &r, a[i]); 
        printf("[w] a=%d, q=%d, r=%d\n", a[i], q, r); 
        div2_correct(&q, &r, a[i]); 
        printf("[c] a=%d, q=%d, r=%d\n", a[i], q, r); 
    } 
    return 0; 
} 
 
$ ./a.out (in this case d = 2) 
[w] a= 5, q= 2, r= 1 
[c] a= 5, q= 2, r= 1 
[w] a=-5, q=-2, r=-1   <=  wrong, vide 0 <= r < |d| 
[c] a=-5, q=-3, r= 1 
 
gcc div algorithm produces wrong result. 
 
> (...) witness:  
> #include <stdio.h>  
> int f(int a)  
> {  
>   return a >> 1;  
> }  
> int main(void)  
> {  
>   int g = f(-5);  
>   printf("%d\n", g);  
> }  
>   
> prints -3.  
  
-3 looks fine. 
 
a = -5, d = 2  ->  q = -3, r = 1  ->  qd+r = -3*2+1 = -5 = a. 
 
did i miss something in my math? 

-- 


http://gcc.gnu.org/bugzilla/show_bug.cgi?id=22072