| /* |
| * This test stresses masking and sign-extension after int operations |
| * that cause overflow, producing bogus high-order bits. |
| * The basic overflow situation (x * x + y * y, for x = y = 1 << 21) |
| * actually happens in Olden-perimeter, in the function CheckOutside. |
| * |
| * Several things have to happen correctly: |
| * -- correct constant folding if it is done at compile-time |
| * -- correct sign-extensions during native code generation for -, * and /. |
| * -- correct handling of high bits during native code generation for |
| * a sequence of operations involving -, * and /. |
| */ |
| #include <stdio.h> |
| |
| void compareOvf(int x, int y) |
| { |
| int sum = x * x + y * y; |
| if (sum < (1 << 22)) |
| printf("compare after overflow is TRUE\n"); |
| else |
| printf("compare after overflow is FALSE\n"); |
| } |
| |
| void divideOvf(int x, int y) |
| { |
| int sum = x * x + y * y; |
| int rem = (1 << 31) / sum; |
| printf("divide after overflow = %d (0x%x)\n", rem, rem); |
| } |
| |
| void divideNeg(int x, int y) |
| { |
| int sum = x * x - y * y; |
| int rem = sum / (1 << 18); |
| printf("divide negative value by power-of-2 = %d (0x%x)\n", rem, rem); |
| } |
| |
| void subtractOvf(int x, int y) |
| { |
| int sum = x * x + y * y; |
| int rem = (1u << 31) - sum; |
| printf("subtract after overflow = %d (0x%x)\n", rem, rem); |
| } |
| |
| int main() |
| { |
| int b21 = 1 << 21; |
| compareOvf(b21, b21); |
| divideOvf(b21 + 1, b21 + 2); |
| divideNeg(b21 + 1, b21 + 2); /* arg1 must be < arg2 */ |
| subtractOvf(b21 + 1, b21 + 2); |
| return 0; |
| } |