Question

我想知道是否可以为我想在代码中评估的字符串数组创建一个循环。我想一次做多个二进制数。到目前为止，我的工作正常，但我不知道如何让它一次评估多个二进制数。谢谢。

 package twoComplement;

public class bintodec {

public static void main (String[] args)throws java.io.IOException {

int number,
    digit1,
    digit2,
    digit3,
    digit4,
    digit5,
    digit6,
    digit7,
    digit8,
    result;

String num = "11100111";

number = Integer.parseInt(num);

digit1 = ((number % 100000000) - (number % 10000000 % 10000000)) / 10000000;
digit2 = ((number % 10000000) - (number % 10000000 % 1000000)) / 1000000;
digit3 = ((number % 1000000) - (number % 1000000 % 100000)) / 100000;
digit4 = ((number % 100000) - (number % 100000 % 10000)) / 10000;
digit5 = ((number % 10000) - (number % 10000 % 1000)) / 1000;
digit6 = ((number % 1000) - (number % 1000 % 100)) / 100;
digit7 = ((number % 100) - (number % 100 % 10)) / 10;
digit8 = (number % 10);

result = (digit1 * -128) + (digit2 * 64) + (digit3 * 32) + (digit4 * 16) + (digit5 * 8) + (digit6 * 4) + (digit7 * 2) + (digit8 * 1);

System.out.println ( "Binary number: " + num + "\nDecimal Number: " + result);
System.out.println();
System.exit( 0 );

}
}

Answer 1

是的，这是循环的目的 ;-) 高级循环是迭代数组的最佳方式。您还可以遍历集合（IE：ArrayList），这样可以更轻松地添加新项目。

String[] numbersToEvaluate = new String[]{"11100111", "100101", "10101101"}; for (String num: numbersToEvaluate) { number = Integer.parseInt(num); digit1 = ((number % 100000000) - (number % 10000000 % 10000000)) / 10000000; digit2 = ((number % 10000000) - (number % 10000000 % 1000000)) / 1000000; digit3 = ((number % 1000000) - (number % 1000000 % 100000)) / 100000; digit4 = ((number % 100000) - (number % 100000 % 10000)) / 10000; digit5 = ((number % 10000) - (number % 10000 % 1000)) / 1000; digit6 = ((number % 1000) - (number % 1000 % 100)) / 100; digit7 = ((number % 100) - (number % 100 % 10)) / 10; digit8 = (number % 10); result = (digit1 * -128) + (digit2 * 64) + (digit3 * 32) + (digit4 * 16) + (digit5 * 8) + (digit6 * 4) + (digit7 * 2) + (digit8 * 1); System.out.println ( "Binary number: " + num + "\nDecimal Number: " + result); }

Answer 2

这是定义字符串数组的方法：

int n = 10;
String[] arr = new String[n];

以下是如何迭代数组：

for (int i = 0; i < n; i++) {
    arr[i] = "element number " + i;
}

或者这样：

for (String element : arr) {
    System.out.println(element);
}

Answer 3

如果您至少运行Java 5，则可以使用for-each构造来遍历任意数组或集合。

String[] nums = //init
for (String num:nums) {
    // Do work.
}

但是，如果您使用的是Java 1.4或更早版本，或者您关心数组的索引，则需要使用传统的for循环

String[] nums = //init
for (int i = 0; i<nums.length; i++) {
    String num = nums[i];
    // Do work.
}

Answer 4

你的意思是

for (String num : "11100111,1010,11111,110101010101010101011101010101010101010101".split(",")) {
    long result = Long.parseLong(num, 2);
    System.out.println("Binary number: " + num + ", Decimal Number: " + result);
}

打印

Binary number: 11100111, Decimal Number: 231
Binary number: 1010, Decimal Number: 10
Binary number: 11111, Decimal Number: 31
Binary number: 110101010101010101011101010101010101010101, Decimal Number: 3665040856405

Answer 5

是的，这很容易。

String[] array = {"first", "second", "third"};
for (String element : array) {
   // do whatever
}

但是，除此之外，您可能希望考虑以下内容以获得更清晰的代码。

String num = "1000";
int digit1 = Character.digit(num.charAt(0), 2);
// if you are only expecting a binary number then the second argument (2)
// tells the digit method to throw an error if it gets a digit other than 0 or 1
...

如果这是针对学校练习或类似练习，您可能希望针对内置的Integer解析器测试您的代码。

int numFromBinaryString = Integer.parseInt(numberString, 2);
// again, here the second argument tells parseInt to interpret numberString as
// binary string

字符串数组以及如何循环它们

5 个答案: