二进制转换为八进制字符串使用 [英] Converting binary to octal using strings

问题描述

我必须通过转换十六进制数为二进制先十六进制数八进制数转换，并且从二元到八进制。

 的#include＆LT;＆stdio.h中GT;
＃包括LT＆;＆string.h中GT;
诠释主（）{
    炭binarni_brojevi [16] [5] = {0000，0001，0010，0011，0100，0101，0110，0111，1000，1001， 1010，1011，1100，1101，1110，1111};
    烧焦heksadekadni_broj [] =AF1;
    INT I，vrednost;
    焦炭binarni [50];
    binarni [0] ='\\ 0';
    对于（i = 0; heksadekadni_broj [I];我++）{
        如果（因而isalpha（heksadekadni_broj [I]））
            vrednost =（heksadekadni_broj [I]  - 'A'+ 10）;
        其他
            vrednost =（heksadekadni_broj [I]  - '0'）;
        strcat的（binarni，binarni_brojevi [vrednost]）;
    }
    //我该怎么在这里做什么？我应该怎么组由3
    返回0;
}
 

解决方案

要字符组3，先算有多少：

  INT num_of_binary_digits = strlen的（binarni）;
 

这可能不会被3整除例如：

 二进制字符串：00001111
分为3组：00 | 001 | 111
 

要与四舍五入算八进制数字3 数，除以：

  INT num_of_octal_digits =（num_of_binary_digits + 2）/ 3;
 

要确定有多少个二进制数字有在第一组中，使用一些基本的算术（我把它出简洁）。

然后做嵌套循环：

 的for（int OD = 0; OD＆LT; num_of_octal_digits ++ OD）
{
    INT bits_in_group =（OD == 0）？ digits_in_first_group：3;
    对于（INT BD = 0; BD＆LT; bits_in_group ++ BD）
    {
        ...
    }
}
 

内环内，您将不得不像11或110字符的字符串转换为数字像3或6。这应该很容易。

I have to convert hexadecimal numbers to octal numbers by converting hex numbers to binary first, and from binary to octal.

#include <stdio.h>
#include <string.h>
int main(){
    char binarni_brojevi[16][5] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    char heksadekadni_broj[] = "AF1";
    int i, vrednost;
    char binarni[50];
    binarni[0] = '\0';
    for(i = 0; heksadekadni_broj[i]; i++) {
        if(isalpha(heksadekadni_broj[i]))
            vrednost = (heksadekadni_broj[i] - 'A' + 10);
        else
            vrednost = (heksadekadni_broj[i] - '0');
        strcat(binarni, binarni_brojevi[vrednost]);
    }
    // what do I do from here? How should I group by 3
    return 0;
}

解决方案

To group characters by 3, first count how many there are:

int num_of_binary_digits = strlen(binarni);

This may not be divisible by 3. For example:

Binary string: 00001111
Subdivided into groups of 3: 00|001|111

To count the number of octal digits, divide by 3 with rounding up:

int num_of_octal_digits = (num_of_binary_digits + 2) / 3;

To determine how many binary digits there are in the first group, use some basic arithmetic (I left it out for brevity).

Then do nested loops:

for (int od = 0; od < num_of_octal_digits; ++od)
{
    int bits_in_group = (od == 0) ? digits_in_first_group : 3;
    for (int bd = 0; bd < bits_in_group; ++bd)
    {
        ...
    }
}

Inside the inner loop you will have to convert a string of characters like "11" or "110" into a number like 3 or 6. This should be easy.

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