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收到一只叮咚
您可以应用位掩码以仅保留您感兴趣的位。在这种情况下,如果您只想要最后一位,则应用位掩码0b0000000001https://go.dev/play/p/RNQEcON7sw1 // 'x' is your value x := 0b1000000001 // Make the bitmask mask := 0b0000000001 // Apply the bitmask with bitwise AND output := x&mask fmt.Println("This value == 1: ", output) 解释&是“AND”的位运算符。这意味着它逐位遍历这两个值,并将结果位设置为1当且仅当两个输入位都是1. 我在下面为 AND 运算符提供了一个真值表。+-----------+----------+--------------+| Input Bit | Mask Bit | Input & Mask |+-----------+----------+--------------+| 0 | 0 | 0 || 0 | 1 | 0 || 1 | 0 | 0 || 1 | 1 | 1 |+-----------+----------+--------------+因为我的掩码函数1在最后一个位置只有a,所以只保留原始输入的最后一个位置。所有前面的位将始终为0.
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动漫人物
构建一个面具,在您想要操作的每个地方都有一个面具使用按位或设置位。使用按位 AND 和反向掩码来清除位。使用 XOR 切换位package mainimport "fmt"func main() { k := 3 // manipulate the 3rd bit ... mask := uint8(1) << (k - 1) // ... using 0b00000100 as a mask var n uint8 = 0b10101010 fmt.Printf("0b%08b\n", n) // 0b10101010 // set kth bit n |= mask fmt.Printf("0b%08b\n", n) // 0b10101110 // clear kth bit n &^= mask // &^ is Go's AND NOT operator fmt.Printf("0b%08b\n", n) // 0b10101010 // toggle kth bit n ^= mask fmt.Printf("0b%08b\n", n) // 0b10101110}
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catspeake
func test() { i := 1 << 9 //1000000000 i = i | (1 << 8) //1000000000 | 0100000000 == 1100000000 i = i | (1 << 7) //1100000000 | 0010000000 == 1110000000 i = i | (1 << 0) //1110000000 | 0000000001 == 1110000001 fmt.Printf("BEFORE: %010b\n", i) // 1110000001 i = i & ((1 << 9) - 1) // 1110000001 & ((1000000000) - 1) == 1110000001 & (0111111111) == 0110000001 fmt.Printf("AFTER: %010b\n", i) // 0110000001}