C 프로그램의 16진수를 8진수로 변환하는 프로그램

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
• 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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