SUM character to Binary

This is a discussion on SUM character to Binary within the C++ Programming forums, part of the General Programming Boards category; i am definitely a newbie here, i being thinking of this code for 2 hours and that kill my brain.. ...

  1. #1
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    SUM character to Binary

    i am definitely a newbie here, i being thinking of this code for 2 hours and that kill my brain.. so i reach here to ask for help.

    my problem is i couldnt sum up the binary number , for example

    character 'd' = 1100100 and 'e' = 1100101
    the answer is add up the one which is d = 3 + e = 4 sum = 7
    so the sum (cout<<sum will be 7

    and can some one explain this code wich i got it in this forum

    Code:
    #include <iostream>
    #include <cstdlib>
    #include <cstring>
    #include <cctype>
    #include <bitset>
    using namespace std;
    
    int main()
    {
        string name;
        string lowerCase;
        int sum = 0;
        int total = 0;
        //prompt user input
        
        cout<<"Please enter your name to generate password: ";
        cin>>name;
    
        for( string::iterator it = name.begin(); it != name.end(); ++it )
            cout << bitset<7>((unsigned long)*it) <<" ";
    
        system("pause");
        return 0;
    }
    anyone can give another example similar code that is easy to understand?

  2. #2
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    I can't make any sense of this:
    my problem is i couldnt sum up the binary number , for example

    character 'd' = 1100100 and 'e' = 1100101
    the answer is add up the one which is d = 3 + e = 4 sum = 7
    so the sum (cout<<sum will be 7
    This:
    Code:
    for( string::iterator it = name.begin(); it != name.end(); ++it )
            cout << bitset<7>((unsigned long)*it) <<" ";
    displays the binary representation of every character in a string variable called 'name'. An iterator is like a pointer, and name.begin() is a pointer to the first char in name, and name.end() is a pointer to one past the last char in name. *it is the character. This part:
    Code:
    (unsigned long)*it
    converts the char to it's ascii code. It would more properly be written as:
    Code:
    static_cast<unsigned long>(*it)
    Last edited by 7stud; 12-03-2005 at 08:22 PM.

  3. #3
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    ok for example
    character 'd' = 1100100 AND // 3's 1
    character 'e' = 1100101 ADD TOGETHER // 4's 1
    TOTAL = 7 // 3 + 4 = 7

    not adding by using binary method and just sum all one together..

  4. #4
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    1) You have to convert the char to its ascii code
    2) Then you need to obtain a binary representation of the ascii code
    3) Then you can check each char of the binary representation and see if it is a '1'. If it is, then add 1 to a counter variable.

  5. #5
    ZuK
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    Your question is a little bit hard to understand and the code you are showing does not have much to do with what I understand.
    Could this be what you are looking for ?
    Code:
    #include <iostream>
    
    int num_set_bits( unsigned char c ) {
       int ret = 0;
       unsigned char mask = 0x01;
       int i;
       for ( i = 0; i < 8; ++i ) {
          if ( c & mask ) ret++;
          mask <<= 1;      
       }
       return ret;
    }
    
    int main() {
       unsigned char e = 'e';
       unsigned char d = 'd';
       std::cout << num_set_bits(e) + num_set_bits(d) << std::endl;
    }
    Kurt

  6. #6
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    Nice.
    Last edited by 7stud; 12-04-2005 at 05:20 AM.

  7. #7
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    Question

    yeah tat is the solution that i want but what if the input is a string of name = John

    the binary for John is 1001010 1101111 1101000 1101110
    TO ADD THOSE 1s = 3 + 6 + 3 + 5
    EQUAL = 17




  8. #8
    ZuK
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    Code:
       string name = "John";
       int sum = 0;
       for ( int i = 0; i < name.length(); i++ )
          sum += num_set_bits(name[i]);
       std::cout << sum << std::endl;
    Kurt

  9. #9
    Registered User hk_mp5kpdw's Avatar
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    bitsets also offer a count member function that returns the number of bits set to 1 which could be used to accomplish this:

    Code:
    string str = "John";
    int sum = 0;
    
    for( string::const_iterator it = str.begin(); it != str.end(); ++it )
    {
        bitset<7> bits((unsigned long)*it);
        cout << *it << " = " << bits << endl;
        sum += bits.count();
    }
    
    cout << "Sum is: " << sum << endl;
    Output:
    Code:
    J = 1001010
    o = 1101111
    h = 1101000
    n = 1101110
    Sum is: 17
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
    -Christopher Hitchens

  10. #10
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    The STL master--impressive.

    Unfortunately, it looks like gtr_s15 hasn't written a single line of code in his last 4 or 5 posts--all related to this program.
    Last edited by 7stud; 12-04-2005 at 10:13 AM.

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