C 语言实例 - 八进制与二进制相互转换
C 语言实例八进制与二进制相互转换。
实例 - 二进制转换为八进制
#include <stdio.h>
#include <math.h>
int convertBinarytoOctal(long long binaryNumber);
int main()
{
long long binaryNumber;
printf("输入一个二进制数: ");
scanf("%lld", &binaryNumber);
printf("二进制数 %lld 转换为八进制为 %d", binaryNumber, convertBinarytoOctal(binaryNumber));
return 0;
}
int convertBinarytoOctal(long long binaryNumber)
{
int octalNumber = 0, decimalNumber = 0, i = 0;
while(binaryNumber != 0)
{
decimalNumber += (binaryNumber%10) * pow(2,i);
++i;
binaryNumber/=10;
}
i = 1;
while (decimalNumber != 0)
{
octalNumber += (decimalNumber % 8) * i;
decimalNumber /= 8;
i *= 10;
}
return octalNumber;
}
输出结果为：
输入一个二进制数: 101001 二进制数 101001 转换为八进制为 51
实例 - 八进制转换为二进制
#include <stdio.h>
#include <math.h>
long long convertOctalToBinary(int octalNumber);
int main()
{
int octalNumber;
printf("输入一个八进制数: ");
scanf("%d", &octalNumber);
printf("八进制数 %d 转二进制为 %lld", octalNumber, convertOctalToBinary(octalNumber));
return 0;
}
long long convertOctalToBinary(int octalNumber)
{
int decimalNumber = 0, i = 0;
long long binaryNumber = 0;
while(octalNumber != 0)
{
decimalNumber += (octalNumber%10) * pow(8,i);
++i;
octalNumber/=10;
}
i = 1;
while (decimalNumber != 0)
{
binaryNumber += (decimalNumber % 2) * i;
decimalNumber /= 2;
i *= 10;
}
return binaryNumber;
}
输出结果为：
输入一个八进制数: 51 八进制数 51 转二进制为 10100
文人墨客
//分别定义函数实现八进制转二进制 #include <stdio.h> int octTodeci(int num);//声明八进制转十进制函数 long long deciTobina(int num);//声明十进制转二进制函数 int powNum(int base,int exp);//声明求base的exp次方函数 void main() { int octNum, deciNum; long long binaNum; printf("Please enter an octal number:"); scanf("%d", &octNum); deciNum = octTodeci(octNum); binaNum = deciTobina(deciNum); printf("The conversion from octal number %d to binary number is %lld.\n%", octNum, binaNum); } int octTodeci(int num)//定义八进制转十进制函数 { int n = 0,remainder,result = 0; while (num != 0) { remainder = num %10; result += remainder*powNum(8, n); num /= 10; ++n; } return result; } long long deciTobina(int num)//定义十进制转二进制函数 { int remainder,n=0; long long result = 0; while (num != 0) { remainder = num % 2; result += remainder*powNum(10, n); num /= 2; ++n; } return result; } int powNum(int base,int exp)//定义求base的exp次方函数 { int result = 1,k; for (k = exp; k >= 1;--k) result *= base; return result; }