由于环绕式工程有效，因此x - y + 1 <= 2怎么样.而且由于-n - 1是一个补码，即~，并且Solidity整数像C无符号整数一样具有环绕效果，因此您还可以使用-~x - y <= 2，它的常数要少一个.

Since wraparound works, how about just x - y + 1 <= 2. And since -n - 1 is one's complement, i.e ~, and Solidity integers have wraparound just like C unsigned integers, then you could also use -~x - y <= 2 which has one constant less.

#include <stdio.h>

void test(unsigned int a, unsigned int b) {
    printf("%d and %d: ", a, b);
    printf("method1 %d ", a - b + 1 <= 2);
    printf("method2 %d\n", -~a - b <= 2);
}

int main(void) {
    test(1, 1);
    test(1, 2);
    test(1, 3);
    test(1, 4);
    test(2, 1);
    test(3, 1);
    test(4, 1);
}

输出:

1 and 1: method1 1 method2 1
1 and 2: method1 1 method2 1
1 and 3: method1 0 method2 0
1 and 4: method1 0 method2 0
2 and 1: method1 1 method2 1
3 and 1: method1 0 method2 0
4 and 1: method1 0 method2 0

在Solidity中，旋转一下并不能真正帮到您-您需要查看汽油费用表用于不同的操作.第一个执行3个非常低的算术运算和两次下推，第二个执行4个非常低的算术运算和一次下推，因此它们应该或多或少具有可比性.

In Solidity, bit twiddling does not really help you that much - you need to look into the gas cost table for different operations. The first does 3 verylow arithmetic operations and two pushes, the second one does 4 verylow arithmetic operations and one push, so they should be more or less comparable.