检查 2 的 pow 有限制

Question

我需要编写一个条件，这样如果参数为 1 （即 2 ^ 0 = 1 ），它将跳过它。例如：

powsnsums 1 2 8 

2

powsnsums 1 16

1

#!/bin/bash

# loop over all numbers on the command line
# note: we don't verify that these are in fact numbers
for number do
    w=0         # Hamming weight (count of bits that are 1)
    n=$number   # work on $n to save $number for later

    # test the last bit of the number, and right-shift once
    # repeat until number is zero
    while (( n > 0 )); do
        if (( (n & 1) == 1 )); then
            # last bit was 1, count it
            w=$(( w + 1 ))
        fi

        if (( w > 1 )); then
            # early bail-out: not a power of 2
            break
        fi

        # right-shift number
        n=$(( n >> 1 ))
    done

    if (( w == 1 )); then
        # this was a power of 2
        printf '%d\n' "$number"
    fi
done

Answer 1

if [ "$number" -eq 1 ]; then
    continue
fi

或者

if (( number == 1 )); then
    continue
fi

$number如果数值等于 1，这将使循环跳到下一次迭代。测试将for number do在线下进行。