i32_popcnt 算術指令BUG

[yanjiuntw]

WasmVM/src/lib/core/runtime/i32_popcnt.c

Lines 10 to 17 in f702cac

	int32_t value = value1->value.i32;
	uint8_t count = 0;
	while (value) {
	if (value & 0x1) {
	count++;
	}
	value = value >> 1;
	}

#L10 這裡 value 如果使用 int32_t 型態，會使後面 #L16 運算有問題。

由於這裡 value 型態為有號數，將使用算數位移。
算術位移與邏輯位移最大的差別在於，算術位移會考慮數值的正負號。
因此這裡 value 如果為負值，將會進入死迴圈。

[yanjiuntw]

剛剛查了一下C99規格書。
6.5.7 Bitwise shift operators 上寫道：
The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type
or if E1 has a signed type and a nonnegative value, the value of the result is the integral
part of the quotient of E1 divided by the quantity, 2 raised to the power E2. If E1 has a
signed type and a negative value, the resulting value is implementation-defined.

會不會是我的編譯器實作不一樣？

[LuisHsu]

的確這邊需要考慮有號數的問題，所以加個 unsigned 比較妥當