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:

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#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;
}

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anyone can give another example similar code that is easy to understand?

I can't make any sense of this:

Quote:

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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:

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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:

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(unsigned long)*it

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converts the char to it's ascii code. It would more properly be written as:

Code:

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static_cast<unsigned long>(*it)

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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..

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.

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:

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#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;
}

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Kurt

Nice. :)

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



:eek: :eek:

Code:

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  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

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:

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J = 1001010
o = 1101111
h = 1101000
n = 1101110
Sum is: 17

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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. :eek: