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Bitwise more than 8? or less than 8?

This is a discussion on Bitwise more than 8? or less than 8? within the C++ Programming forums, part of the General Programming Boards category; hey all... i successfully learnt everything in the bitwise lesson and made a program: Code: #include <iostream> #include <cstdlib> using ...

  1. #1
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    Bitwise more than 8? or less than 8?

    hey all...
    i successfully learnt everything in the bitwise lesson and made a program:
    Code:
    #include <iostream>
    #include <cstdlib>
    using namespace std;
    
    
    char in_use = 0;
    int *b = new int;
    int *c = new int;
    
    
    void check(){
    
    
        for(int a = 0; a <= 8; a++){
    
    
            if((in_use & (1 << a)) != 0){
    
    
                cout << "The car " << a + 1 << " is in use...\n";
    
    
            }else{
    
    
                cout << "The car " << a + 1 << " is not in use...\n";
    
    
            }
    
    
        }
    
    
    }
    
    
    void set(int pos){
    
    
        in_use = (in_use | (1 << pos));
        cout << "The car is now in use...";
    
    
    }
    
    
    void unset(int pos){
    
    
        in_use = (in_use & ~(1 << pos));
        cout << "The car is now not in use...";
    
    
    }
    
    
    int main(){
    
    
        char x = 'y';
        while(x == 'Y' || x == 'y'){
    
    
            cout << "Welcome to tennisstar's Car Company...";
            cout << "\n\n1)Take a Car for rent...";
            cout << "\n2)Give a rented car back...";
            cout << "\n3)See availability...";
            cout << "\n4)Exit...";
            cout << "\n\nEnter choice: ";
            int *a = new int;
            cin >> *a;
            switch(*a){
    
    
                case 1:
                    cout << "\n===\n\nChoose a car from 1 - 7 which you want to take: ";
                    cin >> *b;
                    *b -= 1;
                    set(*b);
                    break;
                case 2:
                    cout << "\n===\n\nChoose a car from 1 - 7 which you want to give: ";
                    cin >> *c;
                    *c -= 1;
                    unset(*c);
                    break;
                case 3:
                    cout << "\n===\n\n";
                    check();
                    break;
                case 4:
                    cout << "\n";
                    system("pause");
                    return 0;
                default:
                    cout << "\n===\n\nUNKNOWN OPERATION!!!";
                    
    
    
            }
            cout << "\n\n";
            system("pause");
            system("cls");
    
    
        }
    
    
    
    
    }
    the program runs successfully..
    but my question is: can we use a byte (i.e a char) to store something more than 8 values?
    or maybe less?
    can we use an encoding in which its 5 bits = 1 byte
    or 9 bits = 1 byte

    if yes...
    how can we do so?

    kindly help...
    -tennisstar

  2. #2
    The superhaterodyne twomers's Avatar
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    You can use any integral type for bitwise operations. You can use unsigned ints, for example, and on a 64-bit machine you have 64 bits that you can play around with.

    To use fewer than 8, or 32, or 64 bits, you can use a bit mask. This acts as a... stencil which doesn't allow you to overstep the boundaries of a pre-designated range. If, as you said, you want to only use 5 bits, you can do the following:
    Code:
    const int bitMask = (1 << 5 ) - 1; // 1<<5=100000, (1<<5)-1: 11111
    in_use = (in_use | (1<<bit)) & bitMask;
    and if your bit variable was, say, 9, the corresponding bit of the in_use variable doesn't get touched. You can also use some other operators for more convenient bitwise manipulation.
    Code:
    in_use |= (1<<bit) & bitMask;
    Lesshardtofind likes this.

  3. #3
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    Given that any char type can represent 256 distinct values, it is a fair bit it can represent 8.

    Consider how many values can be represented using two bits. The distinct combinations are 00, 01, 10, 11 - i.e. 4 distinct values. With 3 bits, the distinct combinations are 000, 001, 010, 011, 100, 101, 110, 111 - i.e. 8 distinct values. You can keep doig the math, and work out that - with 5 bits - it is possible to represent 32 distinct values.
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy or unhelpful in reply to you, or tell you you need to demonstrate more effort before you can expect help, it is likely you deserve it. Suck it up, Sunshine, and read this, this, and this before posting again.

  4. #4
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    Quote Originally Posted by grumpy View Post
    Given that any char type can represent 256 distinct values, it is a fair bit it can represent 8.

    Consider how many values can be represented using two bits. The distinct combinations are 00, 01, 10, 11 - i.e. 4 distinct values. With 3 bits, the distinct combinations are 000, 001, 010, 011, 100, 101, 110, 111 - i.e. 8 distinct values. You can keep doig the math, and work out that - with 5 bits - it is possible to represent 32 distinct values.
    not that...
    my question is that:
    in my example of the car company which can rent 8 cars
    (since 1 char = 1 byte = 8 bits)
    how do i make it rent 9 or 10 cars
    for that i have to use an encoding where 1 byte = 9 bits or 10 bits respectively...

    how to do that?

  5. #5
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    If one char can be used to keep track of whether 8 cars are individually rented out, then 2 chars can be used to keep track of 16 .....

    Alternatively, an integral type for which sizeof(type) is 2 or more, can be used .....
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy or unhelpful in reply to you, or tell you you need to demonstrate more effort before you can expect help, it is likely you deserve it. Suck it up, Sunshine, and read this, this, and this before posting again.

  6. #6
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    thanx a lot grumpy..
    that was basic thinkin' which didn;t come to my head while thinking for all complicated stuff

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