按位运算结果和布尔值

Question

让我感到沮丧的代码如下:

bool a = 0x00000FF0 & 0x00000FF0 == 0x00000FF0;
if (a) {
  Serial.println("True");
} else {
  Serial.println("False");
}

这打印"假".我真的不明白为什么.更多测试:

bool a = 0x00000FF0 & 0x00000FF0 == 0x00000FF0;
Serial.println(a);

打印0.

和:

unsigned long a = 0x00000FF0 & 0x00000FF0;
Serial.println(a, HEX);

打印FF0.

Answer 1

运算符优先级,编译时带有警告:

warning: suggest parentheses around comparison in operand of ‘&’ [-Wparentheses]

改成

bool a = (0x00000FF0 & 0x00000FF0) == 0x00000FF0;

Answer 2

查看运算符优先级,解析为

0x00000FF0 & (0x00000FF0 == 0x00000FF0)

当你想要的时候

(0x00000FF0 & 0x00000FF0) == 0x00000FF0

添加括号,您应该得到预期的结果.

Answer 3

这是一个简单的优先事项.

从事物的声音来看,你假设:0x00000FF0 & 0x00000FF0 == 0x00000FF0;将被视为:(0x00000FF0 & 0x00000FF0) == 0x00000FF0;,但实际上它与:相同:0x00000FF0 & (0x00000FF0 == 0x00000FF0);.

后者是情况下,结果是明显false-- ==或者产生0或1,和0xff0 & 1和0xff0 & 0或两者0.