C语言：八进制与二进制相互转换

C语言：八进制与二进制相互转换

问题：八进制与二进制相互转换。

代码1：二进制转换为八进制

#include 
#include 

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;
}

运行结果：

输入一个二进制数: 101001
二进制数 101001 转换为八进制为 51

代码2：八进制转换为二进制

#include 
#include 

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;
}

运行结果：

输入一个八进制数: 51
八进制数 51 转二进制为 101001

参考代码：

//分别定义函数实现八进制转二进制

#include 

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;
}