Program penukaran heksadesimal kepada perlapanan dalam program C

我们得到一个十六进制数字作为字符串；任务是将其转换为八进制。要将十六进制数转换为八进制数，我们必须 -

• 找到与十六进制数等效的二进制数。
• 将二进制数转换为八进制数。

什么是十六进制数

十六进制数是以16为基数的数字，数字从0到9，从10开始数字表示为A其中代表 10，B 代表 11，C 代表 12，D 代表 13，E 代表 14，F 代表 15。

要将十六进制数转换为二进制数，每个数字都会转换为 4 位的二进制数

什么是八进制

计算机中的八进制以8为基数表示，即0-7的八进制数由三个二进制数或三个二进制数字组成。

我们必须做什么

就像我们有一个十六进制数 1A6，所以它现在对于十六进制表示 1、10 和 6首先，我们必须找到十六进制数的二进制等价物，即，

因此，1A6 的二进制 = 0001 1010 0110

现在找到十六进制数的二进制后，下一个任务是找到八进制

在此之前，我们将二进制数分为三组。分组为 3 后，我们将得到 000 110 100 110

其八进制表示形式为 -

因此十六进制数 1A6 的八进制表示为 − 646

示例

Input: 1A6
Output: Octal Value = 646
Explanation:

Input: 1AA
Output: 652

我们将用来解决给定问题的方法 -

• 获取输入并将其存储为字符串。
• 转换十六进制数或表达式转换为二进制，按照以下方法 -
  • 通过添加各自的二进制表示来检查所有 16 种十六进制情况。
  • 返回结果。
• 按照以下步骤将二进制数转换为八进制数 -
  • 通过比较二进制数与八进制数的所有可能情况，取 3 个位置.
  • 设置八进制的值=（val * place）+八进制；
  • 二进制数除以1000
  • place *= 10
  < /ul>
  • 返回结果。

  算法

  Start
  Step 1-> In function long long int hexa_binary(char hex[])
     Declare variables binary, place
     Declare and initialize i = 0, rem, val
     Initialize t n = strlen(hex)
     Initialize binary = 0ll and place = 0ll
     Loop For i = 0 and hex[i] != '\0' and i++ {
        binary = binary * place;
        switch (hex[i]) {
           case '0':
              binary += 0
           case '1':
              binary += 1
           case '2':
              binary += 10
           case '3':
              binary += 11
           case '4':
              binary += 100
           case '5':
              binary += 101
           case '6':
              binary += 110
           case '7':
              binary += 111
           case '8':
              binary += 1000
           case '9':
              binary += 1001
           case 'a':
           case 'A':
              binary += 1010
           case 'b':
           case 'B':
              binary += 1011
           case 'c':
           case 'C':
              binary += 1100
           case 'd':
           case 'D':
              binary += 1101;
              break;
           case 'e':
           case 'E':
              binary += 1110;
              break;
           case 'f':
           case 'F':
              binary += 1111;
              break;
           default:
              printf("Invalid hexadecimal input.");
        }
        place = 10000;
     }
     return binary;
  }
  long long int binary_oct(long long binary) {
     long long int octal, place;
     int i = 0, rem, val;
     octal = 0ll;
     place = 0ll;
     place = 1;
     while (binary > 0) {
        rem = binary % 1000;
        switch (rem) {
        case 0:
           val = 0;
           break;
        case 1:
           val = 1;
           break;
        case 10:
           val = 2;
           break;
        case 11:
           val = 3;
           break;
        case 100:
           val = 4;
           break;
        case 101:
           val = 5;
           break;
        case 110:
           val = 6;
           break;
        case 111:
           val = 7;
           break;
        }
        octal = (val * place) + octal;
        binary /= 1000;
        place *= 10;
     }
     return octal;
  }
  long long int hexa_oct(char hex[]) {
     long long int octal, binary;
     // convert HexaDecimal to Binary
     binary = hexa_binary(hex);
     // convert Binary to Octal
     octal = binary_oct(binary);
     return octal;
  }
  int main() {
     char hex[20] = "1a99";
     printf("Octal Value = %lld", hexa_oct(hex));
     return 0;
  }

  示例

  #include <stdio.h>
  #include <string.h>
  #include <math.h>
  //To convert hex to binary first
  long long int hexa_binary(char hex[]) {
     long long int binary, place;
     int i = 0, rem, val;
     int n = strlen(hex);
     binary = 0ll;
     place = 0ll;
     for (i = 0; hex[i] != &#39;\0&#39;; i++) {
        binary = binary * place;
        switch (hex[i]) {
        case &#39;0&#39;:
           binary += 0;
           break;
        case &#39;1&#39;:
           binary += 1;
           break;
        case &#39;2&#39;:
           binary += 10;
           break;
        case &#39;3&#39;:
           binary += 11;
           break;
        case &#39;4&#39;:
           binary += 100;
           break;
        case &#39;5&#39;:
           binary += 101;
           break;
        case &#39;6&#39;:
           binary += 110;
           break;
        case &#39;7&#39;:
           binary += 111;
           break;
        case &#39;8&#39;:
           binary += 1000;
           break;
        case &#39;9&#39;:
           binary += 1001;
           break;
        case &#39;a&#39;:
        case &#39;A&#39;:
           binary += 1010;
           break;
        case &#39;b&#39;:
        case &#39;B&#39;:
           binary += 1011;
           break;
        case &#39;c&#39;:
        case &#39;C&#39;:
           binary += 1100;
           break;
        case &#39;d&#39;:
        case &#39;D&#39;:
           binary += 1101;
           break;
        case &#39;e&#39;:
        case &#39;E&#39;:
           binary += 1110;
           break;
        case &#39;f&#39;:
        case &#39;F&#39;:
           binary += 1111;
           break;
        default:
           printf("Invalid hexadecimal input.");
        }
        place = 10000;
     }
     return binary;
  }
  //To convert binary to octal
  long long int binary_oct(long long binary) {
     long long int octal, place;
     int i = 0, rem, val;
     octal = 0ll;
     place = 0ll;
     place = 1;
     // giving all binary numbers for octal conversion
     while (binary > 0) {
        rem = binary % 1000;
        switch (rem) {
        case 0:
           val = 0;
           break;
        case 1:
           val = 1;
           break;
        case 10:
           val = 2;
           break;
        case 11:
           val = 3;
           break;
        case 100:
           val = 4;
           break;
        case 101:
           val = 5;
           break;
        case 110:
           val = 6;
           break;
        case 111:
           val = 7;
           break;
        }
        octal = (val * place) + octal;
        binary /= 1000;
        place *= 10;
     }
     return octal;
  }
  // to convert the hexadecimal number to octal
  long long int hexa_oct(char hex[]) {
     long long int octal, binary;
     // convert HexaDecimal to Binary
     binary = hexa_binary(hex);
     // convert Binary to Octal
     octal = binary_oct(binary);
     return octal;
  }
  //main function
  int main() {
     char hex[20] = "5CD";
     printf("Octal Value = %lld", hexa_oct(hex));
     return 0;
  }

  输出

  Octal Value = 2715

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