Operadores bit a bit en C/C++

En C, los siguientes 6 operadores son operadores bit a bit (funcionan a nivel de bit) 
 

Bitwise Operators in C/C++

C++

#include <iostream>
using namespace std;
  
int main() {
      // a = 5(00000101), b = 9(00001001)
    int a = 5, b = 9;
  
    // The result is 00000001
    cout<<"a = " << a <<","<< " b = " << b <<endl;
    cout << "a & b = " << (a & b) << endl;
  
    // The result is 00001101
    cout << "a | b = " << (a | b) << endl;
  
    // The result is 00001100
    cout << "a ^ b = " << (a ^ b) << endl;
  
    // The result is 11111010
    cout << "~a = " << (~a) << endl;
  
    // The result is 00010010
    cout<<"b << 1" <<" = "<< (b << 1) <<endl;
  
    // The result is 00000100
    cout<<"b >> 1 "<<"= " << (b >> 1 )<<endl;
  
    return 0;
}
  
// This code is contributed by sathiyamoorthics19

C

// C Program to demonstrate use of bitwise operators
#include <stdio.h>
int main()
{
    // a = 5(00000101), b = 9(00001001)
    unsigned char a = 5, b = 9;
  
    // The result is 00000001
    printf("a = %d, b = %d\n", a, b);
    printf("a&b = %d\n", a & b);
  
    // The result is 00001101
    printf("a|b = %d\n", a | b);
  
    // The result is 00001100
    printf("a^b = %d\n", a ^ b);
  
    // The result is 11111010
    printf("~a = %d\n", a = ~a);
  
    // The result is 00010010
    printf("b<<1 = %d\n", b << 1);
  
    // The result is 00000100
    printf("b>>1 = %d\n", b >> 1);
  
    return 0;
    
}

C++

#include <iostream>
using namespace std;
  
// Function to return the only odd
// occurring element
int findOdd(int arr[], int n)
{
    int res = 0, i;
    for (i = 0; i < n; i++)
        res ^= arr[i];
    return res;
}
  
// Driver Method
int main(void)
{
    int arr[] = { 12, 12, 14, 90, 14, 14, 14 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "The odd occurring element is  "<< findOdd(arr, n);
    return 0;
}
  
// This code is contributed by shivanisinghss2110

C

#include <stdio.h>
  
// Function to return the only odd
// occurring element
int findOdd(int arr[], int n)
{
    int res = 0, i;
    for (i = 0; i < n; i++)
        res ^= arr[i];
    return res;
}
  
// Driver Method
int main(void)
{
    int arr[] = { 12, 12, 14, 90, 14, 14, 14 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printf("The odd occurring element is %d ",
           findOdd(arr, n));
    return 0;
}

C++

#include <iostream>
using namespace std;
  
int main()
{
    int x = 2, y = 5;
    (x & y) ? cout <<"True " : cout <<"False ";
    (x && y) ? cout <<"True " : cout <<"False ";
    return 0;
}
  
  
  
// This code is contributed by shivanisinghss2110

C

#include <stdio.h>
  
int main()
{
    int x = 2, y = 5;
    (x & y) ? printf("True ") : printf("False ");
    (x && y) ? printf("True ") : printf("False ");
    return 0;
}

C++

#include <iostream>
using namespace std;
  
int main() {
  
    int x = 19;
    cout<<"x << 1 = "<< (x << 1) <<endl;
    cout<<"x >> 1 = "<< (x >> 1) <<endl;
    return 0;
}
  
// This code is contributed by sathiyamoorthics19

C

#include <stdio.h>
  
int main()
{
    int x = 19;
    printf("x << 1 = %d\n", x << 1);
    printf("x >> 1 = %d\n", x >> 1);
    return 0;
}

C++

#include <iostream>
using namespace std;
  
int main() {
  
    int x = 19 ;
    (x & 1) ? cout<<"Odd" : cout<< "Even" ;
        
    return 0;
}
  
// This code is contributed by sathiyamoorthics19

C

#include <stdio.h>
  
int main()
{
    int x = 19;
    (x & 1) ? printf("Odd") : printf("Even");
    return 0;
}

C++

#include <iostream>
using namespace std;
  
int main() {
  
    unsigned int x = 1;
    signed int a = 1;
    cout<<"Signed Result "<< ~a <<endl ;
    cout<<"Unsigned Result "<< ~x ;
    return 0;
}
// This code is contributed by sathiyamoorthics19

C

// Note that the output of the following
// program is compiler dependent
#include <stdio.h>
  
int main()
{
    unsigned int x = 1;
    printf("Signed Result %d \n", ~x);
    printf("Unsigned Result %ud \n", ~x);
    return 0;
}

Publicación traducida automáticamente

Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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