除法结果在bash中为'0‘，但分子和分母的值应导致不同的结果

Question

我正在尝试使用以下代码来查找我的android应用程序的平均CPU利用率

#!/bin/bash
counter=1
while [ $counter -le 10 ]
    do
        current_cpu=$(adb shell top -n 1 | grep org.carleton.iot.mobile_cep | awk '{print $5}' | sed 's|%||g')
        echo "current_cpu = "$current_cpu
        total_cpu=$((total_cpu + current_cpu))
        echo "total_cpu = "$total_cpu
        echo "counter = "$counter
        average_cpu=$(((totalMemory / counter)))
        echo "average_cpu = "$average_cpu
        echo "\n"
        ((counter++))

        sleep 1
    done
echo  done

它提供了以下结果

current_cpu = 7
total_cpu = 7
counter = 1
average_cpu = 0

current_cpu = 8
total_cpu = 15
counter = 2
average_cpu = 0


current_cpu = 6
total_cpu = 21
counter = 3
average_cpu = 0

current_cpu = 8
total_cpu = 29
counter = 4
average_cpu = 0

但是，average_cpu的值应该等于total_cpu/counter的值。

Answer 1

echo "scale=2 ; $totalmemory / $counter" | bc

在这里，小数位数的值是精度，如果它是2，它将返回小数点后2位的值，如.55