Binary to octal numbers in c plus plus codes?

Answer 1

The following code demonstrates one way to convert from binary to octal based on user input:

#include<iostream>

#include<string>

typedef unsigned int uint;

std::string input(std::string prompt)

{

std::cout<<prompt<<": ";

std::string s;

std::getline(std::cin,s);

return(s);

}

uint input_binary( uint min, uint max )

{

uint result=0;

bool repeat=true;

while(repeat)

{

result=0;

repeat=!repeat;

std::string s=input("Enter a binary value");

for(uint i=0; i<s.size() && !repeat; ++i)

{

const char c=s.at(i);

if(c<'0'c>'1')

{

std::cout<<"Binary values may only contain characters '0' or '1'\n"<<std::endl;

repeat=!repeat;

}

result<<=1;

result|=(c-'0');

}

if(!repeat)

{

if(result<min result>max)

{

std::cout<<"The binary value must be in the decimal range "<<min<<".."<<max<<"\n"<<std::endl;

repeat=!repeat;

}

}

}

return( result );

}

int main()

{

uint binary = input_binary(0x1,0xffffff);

uint temp=binary;

uint len=1;

while(temp>>=3)

++len;

std::string octal(len, '0');

for( std::string::reverse_iterator i=octal.rbegin(); i!=octal.rend(); ++i )

{

*i=(binary & 0x7)+'0';

binary>>=3;

}

std::cout<<"Octal: "<<octal.c_str()<<"\n"<<std::endl;

return(0);

}