Binary number function

Question

#include <iostream>
using namespace std;
void binary(unsigned  a) {
    int i;
    cout << "0" << endl;
    do {
        for (i = 1; i < a; i=i/2) {
            if ((a & i) != 0) {
                cout << "1" << endl;
            }
            else {
                cout << "0" << endl;
            
            }
    
        }
    }
    while (1 <= a && a <= 10);
}

int main(void) {
    binary(4);
    cout << endl;
}

I wrote a code about binary numbers. İt should give bits respect to entering number like for 4 (0100) for 2 (10). However my code goes infinity could you explain. I wrote in visual studio and I cannot use <bits/stdc++.h> because there is no such a library in visual studio

***I wrote in visual studio and I cannot use because there is no such a library in visual studio*** That is a very good thing. Never use this header ever. Related: [https://stackoverflow.com/Questions/31816095/Why-Should-I-Not-Include-Bits-Stdc-H.](https://stackoverflow.com/Questions/31816095/Why-Should-I-Not-Include-Bits-Stdc-H.) — drescherjm, Oct 11 '21 at 12:46
Ask yourself: If `a` is `4`, when does `while (1 <= a && a <= 10)` end? — NathanOliver, Oct 11 '21 at 12:47
@NathanOliver it can change I just wrote . İt should be in function and our teacher can give another number too but according to constraints it should be between 1 and 10 — Sude Şahin, Oct 11 '21 at 12:51
*because there is no such a library in visual studio* -- Hurray for Visual Studio. -- *and our teacher* -- Your teacher is giving you `#include `? I hope not. — PaulMcKenzie, Oct 11 '21 at 12:52
You should search the internet for "c++ binary number cout" or something similar. There are a lot of similar questions on both StackOverflow and the internet. — Thomas Matthews, Oct 11 '21 at 14:48
In your loop, `i` has an initial value of `1` (actually `1u`) and the loop keeps dividing `i` by `2` (via `i = i/2`). Dividing `1` by `2` produces `0` (rounding toward zero). Dividing `0` by `2` also produces `0`. So the only possible values of `i` in that loop are `1` and `0`. If `a` has a value of `4`, the test `i < a` will always be true. Hence an infinite loop. — Peter, Oct 11 '21 at 16:12

Bathsheba · Answer 1 · 2021-10-11T13:00:43.447

2

Initially i is 1 but i = i / 2 sets i to 0, where it remains. The inner loop, therfore, loops for ever.

To output an unsigned number a in binary, use

#include <bitset>
#include <climits>
std::cout << std::bitset<sizeof(a) * CHAR_BIT>(a) << '\n';

(There is, at the time of writing no std::bin i/o manipulator cf. std::hex.)

edited Oct 11 '21 at 13:00

answered Oct 11 '21 at 12:46

Bathsheba

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1

`sizeof(a)` -- Wouldn't you need to multiply by the number of bits in a byte? – PaulMcKenzie Oct 11 '21 at 12:55
@PaulMcKenzie: Oops. – Bathsheba Oct 11 '21 at 12:55
thank you very much but our teacher said you can not use built in function. – Sude Şahin Oct 11 '21 at 12:56
@SudeŞahin *but our teacher said you can not use built in function* -- Please post up-front what you can or cannot use. Otherwise you have persons wasting time putting up legitimate answers that will not be accepted (at least won't get the green tick mark). – PaulMcKenzie Oct 11 '21 at 12:57
We'll you're using `std::cout`? – Bathsheba Oct 11 '21 at 12:57
@Bathsheba thank you but I have already wrote using namespace std . So, I do not need std::cout I can write just cout – Sude Şahin Oct 11 '21 at 13:00
1

Indeed, and no professional uses `using namespace std;` as it causes namespace pollution for one thing. You're still obviously using `std::cout`. – Bathsheba Oct 11 '21 at 13:01

east1000 · Answer 2 · 2021-10-11T13:44:42.950

Without using a built-in function, you can write your own function and perform your operation as follows.

Solution-1

#include <iostream>
void binary(unsigned int number)
{
    if (number / 2 != 0) {
        binary(number / 2);
    }
    std::cout << number % 2;
}

int main() {
    binary(10);
}

Solution-2

#include <iostream>
#include<string>

void binary(unsigned int number)
{
    std::string str = "";
    while (number != 0) { 
        str = (number % 2 == 0 ? "0" : "1") + str;
        number /= 2; 
    }
    std::cout << str;
}
int main()
{
   binary(4);
}

Note : Don't use using namespace std; . Why is "using namespace std;" considered bad practice?

Binary number function

2 Answers2

Solution-1

Solution-2