4 ms·
int f(int x, int y) { return x << y; } This is undefined for SOME values of y, namely those bigger than 31 (on platforms where int is 32 bit; adjust
by jabot 10y ago
int f(int x, int y) {
return x << y;
}
This is undefined for SOME values of y, namely those bigger than 31 (on platforms where int is 32 bit; adjust where different).
This is caused by two facts:
a) The arithmetic operation should be compiled to a hardware shift instruction where possible
b) Different architectures do different things when shifting by an amount bigger than the word size:
- on ARM, the "y" operand is evaluated mod 32
- on x86, if y is greater than 31, the result is 0
- wott 10y ago> - on ARM, the "y" operand is evaluated mod 32 > - on x86, if y is greater than 31, the result is 0 You mixed them up, didn't you ? AFAIK, ARM does set all bits to 0 if the shift if greater than 31, but it only uses the lowest byte of the shift amount register, so it feels like a mod 256. x86[^1] does mask the lower 5 bits of the shift amount, so it feels like a mod 32. And the amount can only be 8 bits, so a 32-bit C variable must be truncated somehow. [1] except on 8086, there is no mask there, so no modulo.