Program for conversion of a number from one form to other using C++

void decToBinary(long long int n) 
{  
    int binaryNum[32]; 
    int i = 0; 
    while (n > 0) { 
        binaryNum[i] = n % 2; 
        n = n / 2; 
        i++; 
    } 
    for (int j = i - 1; j >= 0; j--) 
        cout << binaryNum[j]; 
} 

Steps for the conversion of decimal number to their Octal form:

First, divide the decimal number by 8 till the quotient is zero and get the remainder for each division. Write the sequence of reminder from bottom to top fashion.

Example :

Convert (57)₁₀ to octal number.

Division by 8	Quotient	Remainder (Digit)	Digit #
(57)/8	7	1	0
(7)/8	0	7	1

So (57)₁₀= (71)₈

Code :

void decToOctal(long long int n)
{
    int octalNum[100];
    int i = 0;
    while (n != 0) {
        octalNum[i] = n % 8;
        n = n / 8;
        i++;
    }
    for (int j = i - 1; j >= 0; j--)
        cout << octalNum[j];
}

 

Steps for the conversion of decimal number to their Hexadecimal form:

First, divide the decimal number by 16 till the quotient is zero and get the remainder for each division to replace the decimals in hexadecimal form from the table. Write the sequence of reminder from bottom to top fashion.

Example:

Convert 7562₁₀ to hex:

Division by 16	Quotient (integer)	Remainder (decimal)	Remainder (hex)	Digit #
7562/16	472	10	A	0
472/16	29	8	8	1
29/16	1	13	D	2
1/16	0	1	1	3

So 7562₁₀ = 1D8A₁₆

Code:

void decToHexa(long long int n) 
{
    char hexaDeciNum[100];
    int i = 0; 
    while(n != 0) 
    {    
        int temp  = 0; 
        temp = n % 16; 
        if (temp < 10) 
        { 
            hexaDeciNum[i] = temp + 48-32; 
            i++; 
        } 
        else
        { 
            hexaDeciNum[i] = temp + 87-32; 
            i++; 
        } 
        n = n / 16; 
    } 
    for(int j = i - 1; j >= 0; j--) 
    {
        cout<<hexaDeciNum[j]; 
    }
} 

Steps for the conversion of Binary number to decimal form:

For binary number with n digits:

d_n-1 ... d₃ d₂ d₁ d₀

The decimal number is equal to the sum of binary digits (d_n) times their power of 2 (2ⁿ):

decimal = d₀×2⁰ + d₁×2¹ + d₂×2² + ... so on

Example

Find the decimal value of 111001₂:

binary number:	1	1	1	0	0	1
power of 2:	25	24	23	22	21	20

So,111001₂ = 1⋅2⁵+1⋅2⁴+1⋅2³+0⋅2²+0⋅2¹+1⋅2⁰ = 57₁₀

Code :

long long  binaryToDecimal(long long int n)
{
    long long int num = n;
    long long int dec_value = 0;
    long long int base = 1;
    long long int temp = num;
    while (temp) {
        int last_digit = temp % 10;
        temp = temp / 10;
 
        dec_value += last_digit * base;
 
        base = base * 2;
    }
    return dec_value;
}

Steps for the conversion of Binary number to their hexadecimal form:

Consider a binary number, group them in a group of 4 digits from the end and write the hexadecimal equivalent of that number from the table given above.

Example 

Convert (01001110)₂ to hex:

(0100)₂ = (4)₁₆

(1110)₂ = (E)₁₆

_So

(01001110)₂ = (4E)₁₆

Code:

In this code, I have just converted binary to decimal form and then decimal to hexadecimal but one can code using the above logic.

 

Steps for the conversion of Binary number to their Octal form:

Consider a binary number, group them in a group of 3 digits from the end and write the hexadecimal equivalent of that number from the table given above.

Binary Number	Octal Number
0	0
1	1
10	2
11	3
100	4
101	5
110	6
111	7
1000	10
1001	11
1010	12
1011	13
1100	14
1101	15
1110	16
1111	17
10000	20
10001	21
10010	22
10011	23
10100	24
10101	25
10110	26
10111	27
11000	30
11001	31
11010	32
11011	33
11100	34
11101	35
11110	36
11111	37
100000	40
1000000	100
10000000	200
100000000	400

Example :

Convert binary number 110110001010₂ into an octal number.

110110001010₂ = 110 110 001 010 = 110₍₌₆₎ 110₍₌₆₎ 001₍₌₁₎ 010₍₌₂₎ = 6612₈

Code :

long long int bin_to_oct(long long bin)
{
    long long int octal, place;
    int i = 0, rem, val;
    octal = 0ll;
    place = 0ll;
    place = 1;
    while (bin > 0) {
        rem = bin % 1000;
 
        switch (rem) {
        case 0:
            val = 0;
            break;
        case 1:
            val = 1;
            break;
        case 10:
            val = 2;
            break;
        case 11:
            val = 3;
            break;
        case 100:
            val = 4;
            break;
        case 101:
            val = 5;
            break;
        case 110:
            val = 6;
            break;
        case 111:
            val = 7;
            break;
        }
 
        octal = (val * place) + octal;
        bin /= 1000;
 
        place *= 10;
    }
 
    return octal;
}

 

Steps for the conversion of Hexadecimal number to their Decimal form:

Convert a Hexadecimal number to their decimal form:

A regular decimal number is the sum of the digits multiplied with a power of 10.

137 in base 10 is equal to each digit multiplied with its corresponding power of 10:

137₁₀ = 1×10²+3×10¹+7×10⁰ = 100+30+7

Hex numbers are read the same way, but each digit counts the power of 16 instead of the power of 10.

For hex number with n digits:

d_n-1 ... d₃ d₂ d₁ d₀

Multiply each digit of the hex number with its corresponding power of 16 and sum:

decimal = d_n-1×16^n-1 + ... + d₃×16³ + d₂×16² + d₁×16¹+d₀×16⁰

Example:

3B in base 16 is equal to each digit multiplied with its corresponding 16ⁿ:

3B₁₆ = 3×16¹+11×16⁰ = 48+11 = 59₁₀

Code:

In this code, we have used the shortcut technique firstly we have converted hexadecimal numbers to binary numbers using the reverse of grouping technique and then binary to decimal number.

Steps for the conversion of Hexadecimal number to their binary form:

Convert every hex digit (start the lowest digit) to 4 binary digits, with the above table.

Example:

So 4CD₁₆ = (0100)(1100)(1101)₂

Code:

long long int hex_to_bin(string hex)
{
    long long int bin, place;
    int i = 0, rem, val;
    bin = 0ll;
    place = 0ll;
    for (i = 0; i<hex.size(); i++) {
        bin = bin * place;
        switch (hex[i]) {
        case '0':
            bin += 0;
            break;
        case '1':
            bin += 1;
            break;
        case '2':
            bin += 10;
            break;
        case '3':
            bin += 11;
            break;
        case '4':
            bin += 100;
            break;
        case '5':
            bin += 101;
            break;
        case '6':
            bin += 110;
            break;
        case '7':
            bin += 111;
            break;
        case '8':
            bin += 1000;
            break;
        case '9':
            bin += 1001;
            break;
        case 'a':
        case 'A':
            bin += 1010;
            break;
        case 'b':
        case 'B':
            bin += 1011;
            break;
        case 'c':
        case 'C':
            bin += 1100;
            break;
        case 'd':
        case 'D':
            bin += 1101;
            break;
        case 'e':
        case 'E':
            bin += 1110;
            break;
        case 'f':
        case 'F':
            bin += 1111;
            break;
        default:
            cout << "Invalid hexadecimal input.";
        }
 
        place = 10000;
    }
 
    return bin;
}

Steps for the conversion of Hexadecimal number to their Octal form:

First, convert Hexadecimal number to binary number using the reverse of grouping technique and then convert binary number to octal number by grouping in a group of 3.

Example

So, the binary of (1A6)₁₆ = 0001 1010 0110

Now after finding the binary of the hexadecimal number now the next task is to find the octal of the binary number.

Before that, we will group the binary number into three. After grouping in 3, we will get 000 110 100 110

So the octal representation of a hexadecimal number (1A6)₁₆ is (646)_8.

Code:

long long int hex_to_oct(string hex)
{
    long long int octal, bin;
    bin = hex_to_bin(hex);
    octal = bin_to_oct(bin);
 
    return octal;
}

Steps for the conversion of Octal number to their Decimal form:

Let see how we can convert an octal number to a decimal number. For that, we just start with the digit at the unit's place and keep going from right to left. Multiply every digit with raised power of 8 to the position of the digit (starting from zero) from right to left and sum up. The resulting number so formed is in Decimal representation.

Example:

So (345)₈= (3 * 8²) + (4 * 8¹) + (5 * 8⁰) = (3 * 64) + (4 * 8) + (5 * 1) = (229)₁₀

Code:

long long int octalToDecimal(long long int n) 
{ 
    int num = n; 
    int dec_value = 0; 
    int base = 1; 
    int temp = num; 
    while (temp) { 
        int last_digit = temp % 10; 
        temp = temp / 10;
        dec_value += last_digit * base; 
  
        base = base * 8; 
    } 
    return dec_value; 
} 

 

Steps for the conversion of Octal number to their Binary form:

We will see the steps for the conversion of Octal number to binary number. We first write each octal digit to its 3-digit binary representation. Each of the digits must be treated as a decimal value. Then combine these binary representations to form a single binary number.

In the code, first, the number is converted to a decimal number and then the decimal number is converted to a binary number.

Example: