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+ 2

how to convert binary number to decimal number?

i face problem in dealing with point binary number to decimal number

26th Oct 2017, 1:44 PM
Rao Kamran
Rao Kamran - avatar
5 Answers
+ 9
int n; string s = ""; cin >> n; while(n != 0) { s = n % 2 + s; n /= 2; } cout << s
26th Oct 2017, 1:59 PM
Vukan
Vukan - avatar
+ 3
@Martin Taylor, yeah I wondered if it would be best. Fixed.
26th Oct 2017, 4:57 PM
Kirk Schafer
Kirk Schafer - avatar
+ 1
In decimal (base 10), 1234 stands for (1000 * 1) + (100 * 2) + (10 * 3) + (1 * 4), or (10^3 * 1) + (10^2 * 2) + (10^1 * 3) + (10^0 * 4). In binary (base 2), 1101 stands for (8 * 1) + (4 * 1) + (2 * 0) + (1 * 1), or (2^3 * 1) + (2^2 * 1) + (2^1 * 0) + (2^0 * 1). which is 13. And that's how you do the conversion too. For each binary digit in your number starting from the right, you add 2^i to a total if it's 1 and 0 otherwise.
26th Oct 2017, 1:54 PM
Schindlabua
Schindlabua - avatar
+ 1
3210 - bit order (index) 8420 - multiples of 2 1010 - binary number (10 in decimal) Those are the steps (if you want to implement it by yourself): 1. You should store the binary number as String. 2. Declare an integer variable 3. Loop through all bits On every iteration: - get current bit and convert it to integer - multiply the bit * 2^(bit index times) - add the value to the integer variable 4. Print the integer variable. Here is my Java implementation of above: https://code.sololearn.com/ckuUA3RJhAqn/?ref=app
26th Oct 2017, 2:36 PM
Boris Batinkov
Boris Batinkov - avatar
+ 1
You can test your implementation is working by comparing to... #include <bitset> cout << dec << 0b00101010 << endl; // binary to decimal cout << bitset<8>(42) << endl; // 8 bits, 16 etc... cout << hex << 42 << endl; // fyi // 2a == 2*16¹ + 10*16° == 32 + 10 == 42
26th Oct 2017, 4:22 PM
Kirk Schafer
Kirk Schafer - avatar