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how can I convert the x position digit of binary number to for example 1 or 0 without using loop... x is an integer number and position is where you change the binary number

royan maeri
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  • You could use bit-shift; `unsigned value = (x >> position) & 1;`. – trojanfoe Nov 21 '21 at 16:29
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    An efficient way is [bit masking](https://stackoverflow.com/q/10493411/2472827). – Neil Nov 21 '21 at 16:31
  • Or if you don't want to write the bitmasking code yourself : https://en.cppreference.com/w/cpp/utility/bitset – Pepijn Kramer Nov 21 '21 at 16:36
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    It's important to understand that there isn't such a thing as a "binary number". Binary is a *way to represent* the number. If I write `23` or `0x17` or `0x10111` or `twenty-three` I am talking about the same thing; and by writing it in different ways, I *didn't change* the thing. – Karl Knechtel Nov 21 '21 at 16:46
  • @royanmaeri Did the answer help or do you want me to explain anything in more detail? – Ted Lyngmo Nov 21 '21 at 18:35

2 Answers2

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  • Toggle the bit pos in num: 
    num ^= 1ULL << pos;       // ^ is xor
  • Set the bit pos in num: 
    num |= 1ULL << pos;       // | is bitwise or
  • Clear the bit pos in num: 
    num &= ~(1ULL << pos);    // & is bitwise and, ~ is bitwise not
Ted Lyngmo
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You can use this code to change the x-th digit.

#include <stdio.h>

int main()
{
    int input; 
    int x=2;//number of the digit that you want to change
    int y=0;//x-th digit will be changed to this 
    scanf("%d",&input);
    int left_digits; //the digits left to the (x-1)-th digit 
    int right_digits;//the x-1 first digits 
    left_digits=input>>x-1;
    right_digits=input-(left_digits<<x-1);
    left_digits-=left_digits%2;
    left_digits+=y;
    int ans=(left_digits<<x-1)+right_digits;
    printf("%d",ans);
}
Ashkan Majidi
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  • This is unnecessarily complicated. `if (y) input |= 1 << x; else input &= ~1 << x;` (suggested in the other answer) should do the same thing. – HolyBlackCat Nov 29 '21 at 10:42