問題描述

我也想將我的二進(jìn)制(在字符串中)轉(zhuǎn)換為十六進(jìn)制字符串，這只是一個程序片段，因為這個程序只是另一個更大程序的一部分:

I want to convert my binary(which is in string) to hexadecimal string also, this is just a program fragment since this program is just a part of another bigger program:

//the variable name of the binary string is: "binary"
int digitNumber = 1;
    int sum = 0;
    int test = binary.length()%4;
    if(test!=0) {
        binary = padLeft(binary, test);
    }
    for(int i = 0; i < binary.length(); i++){
        if(digitNumber == 1)
            sum+=Integer.parseInt(binary.charAt(i) + "")*8;
        else if(digitNumber == 2)
            sum+=Integer.parseInt(binary.charAt(i) + "")*4;
        else if(digitNumber == 3)
            sum+=Integer.parseInt(binary.charAt(i) + "")*2;
        else if(digitNumber == 4 || i < binary.length()+1){
            sum+=Integer.parseInt(binary.charAt(i) + "")*1;
            digitNumber = 0;
            if(sum < 10)
                System.out.print(sum);
            else if(sum == 10)
                System.out.print("A");
            else if(sum == 11)
                System.out.print("B");
            else if(sum == 12)
                System.out.print("C");
            else if(sum == 13)
                System.out.print("D");
            else if(sum == 14)
                System.out.print("E");
            else if(sum == 15)
                System.out.print("F");
            sum=0;
        }
        digitNumber++;  
    }
    public static String padLeft(String s, int n) {
        return String.format("%0$"+n+"s", s);
    }//i added this for padding

問題是我不知道填充是否有效，但我確信這個程序返回錯誤的二進(jìn)制字符串的十六進(jìn)制轉(zhuǎn)換我正在嘗試這樣做:

the problem is that i dont know if the padding works but i am sure that this program return a wrong hexadecimal conversion of the binary string I am trying to do this:

http://www.wikihow.com/Convert-Binary-to-Hexadecimal

PS:我需要實現(xiàn)它(不使用任何內(nèi)置函數(shù))

推薦答案

如果您不必自己實現(xiàn)該轉(zhuǎn)換，則可以使用現(xiàn)有代碼:

If you don't have to implement that conversion yourself, you can use existing code :

int decimal = Integer.parseInt(binaryStr,2);
String hexStr = Integer.toString(decimal,16);

如果你必須自己實現(xiàn)，你的代碼有幾個問題:

If you must implement it yourself, there are several problems in your code :

1. 循環(huán)應(yīng)該從 0 迭代到 binary.length()-1(假設(shè)字符串的第一個字符代表最高有效位).
2. 您隱含地假設(shè)您的二進(jìn)制字符串對于某個整數(shù) x 有 4*x 個字符.如果這不是真的，那么您的算法就會中斷.你應(yīng)該用零填充你的字符串以獲得這樣長度的字符串.
3. sum 必須在您輸出每個十六進(jìn)制數(shù)字后重置為 0.
4. System.out.print(digitNumber); - 這里應(yīng)該打印 sum，而不是 digitNumber.

1. The loop should iterate from 0 to binary.length()-1 (assuming the first character of the String represents the most significant bit).
2. You implicitly assume that your binary String has 4*x charcters for some integer x. If that's not true, your algorithm breaks. You should left pad your String with zeroes to get a String of such length.
3. sum must be reset to 0 after each hex digit you output.
4. System.out.print(digitNumber); - here you should print sum, not digitNumber.

這是大部分固定代碼的外觀:

Here's how the mostly fixed code looks :

    int digitNumber = 1;
    int sum = 0;
    String binary = "011110101010";
    for(int i = 0; i < binary.length(); i++){
        if(digitNumber == 1)
            sum+=Integer.parseInt(binary.charAt(i) + "")*8;
        else if(digitNumber == 2)
            sum+=Integer.parseInt(binary.charAt(i) + "")*4;
        else if(digitNumber == 3)
            sum+=Integer.parseInt(binary.charAt(i) + "")*2;
        else if(digitNumber == 4 || i < binary.length()+1){
            sum+=Integer.parseInt(binary.charAt(i) + "")*1;
            digitNumber = 0;
            if(sum < 10)
                System.out.print(sum);
            else if(sum == 10)
                System.out.print("A");
            else if(sum == 11)
                System.out.print("B");
            else if(sum == 12)
                System.out.print("C");
            else if(sum == 13)
                System.out.print("D");
            else if(sum == 14)
                System.out.print("E");
            else if(sum == 15)
                System.out.print("F");
            sum=0;
        }
        digitNumber++;  
    }

輸出:

7AA

這僅在二進(jìn)制位數(shù)可被 4 整除時才有效，因此您必須添加左側(cè) 0 填充作為 preliminray 步驟.

This will work only if the number of binary digits is divisable by 4, so you must add left 0 padding as a preliminray step.

這篇關(guān)于將二進(jìn)制字符串轉(zhuǎn)換為十六進(jìn)制字符串 JAVA的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！