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C/C++ Tricky Programs

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  • Difficulty Level : Medium
  • Last Updated : 27 Jun, 2022
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We may come across various tricky programs in our day-to-day life. Maybe in technical interviews, coding tests, or C/C++ classrooms. 

Here is a list of such programs:- 

  • Print text within double quotes (” “). 
    This may seem easy, but beginners may get puzzled while printing text within double quotes.

C++




// CPP program to print double quotes
#include<iostream>
 
int main()
{
   std::cout << "\"geeksforgeeks\"";
   return 0;
}

Output

"geeksforgeeks"

Time Complexity: O(1)
Auxiliary Space: O(1)

  • To check if two numbers are equal without using arithmetic operators or comparison operators. 
    The simplest solution for this is using the Bitwise XOR operator (^). We know that for two equal numbers XOR operator returns 0. We will use the XOR operator to solve this problem.

C++




// C++ program to check if two numbers are equal
// without using arithmetic operators or
// comparison operators
#include <iostream>
using namespace std;
 
int main()
{
   int x = 10;
   int y = 10;
    
   if (!(x ^ y))
      cout << " x is equal to y ";
   else
      cout << " x is not equal to y ";
       
   return 0;
}
 
// This code is contributed by shivani

C




// C program to check if two numbers are equal
// without using arithmetic operators or
// comparison operators
#include<stdio.h>
 
int main()
{
   int x = 10;
   int y = 10;
   if ( !(x ^ y) )
      printf(" x is equal to y ");
   else
      printf(" x is not equal to y ");
   return 0;
}

Output

 x is equal to y 

Time Complexity: O(1)
Auxiliary Space: O(1)

  • Print all natural numbers up to N without using a semi-colon. 
    We use the idea of recursively calling the main function.

C++




// C++ program to print all natural numbers upto
// N without using semi-colon
#include<iostream>
 
using namespace std;
int N = 10;
 
int main()
{
  static int x = 1;
  if (cout << x << " " && x++ < N && main())
  { }
  return 0;
}

Output

1 2 3 4 5 6 7 8 9 10 

Time Complexity: O(1)
Auxiliary Space: O(1)

  • To Swap the values of two variables without using any extra variable.

C++




// C++ program to check if two numbers are equal
#include<bits/stdc++.h>
using namespace std;
 
int main()
{
   int x = 10;
   int y = 70;
 
   x = x + y;
   y = x - y;
   x = x - y;
 
   cout << "X : " << x << "\n";
   cout << "Y : " << y << "\n";
    
   return 0;
}

Output

X : 70
Y : 10

Time Complexity: O(1)
Auxiliary Space: O(1)

  • Program to find the Maximum and minimum of two numbers without using any loop or condition. 
    The simplest trick is-

C++




// C++ program to find maximum and minimum of
// two numbers without using loop and any
// condition.
#include<bits/stdc++.h>
 
int main ()
{
   int a = 15, b = 20;
   printf("max = %d\n", ((a + b) + abs(a - b)) / 2);
   printf("min = %d", ((a + b) - abs(a - b)) / 2);
   return 0;
}

Output

max = 20
min = 15

Time Complexity: O(1)
Auxiliary Space: O(1)

  • Print the maximum value of an unsigned int using One’s Complement (~) Operator in C. 
    Here is a trick to find the maximum value of an unsigned int using one’s complement operator:

C++




// C++ program to print maximum value of
// unsigned int.
 
#include <iostream>
 
int main()
{
   unsigned int max;
   max = 0;
   max = ~max;
   
   std::cout << "Max value : " << max;
   
   return 0;
}   
 
// This code is contributed by sarajadhav12052009

C




// C program to print maximum value of
// unsigned int.
#include<stdio.h>
 
int main()
{
   unsigned int max;
   max = 0;
   max = ~max;
   printf("Max value : %u ",  max);
  return 0;
}   

Output

Max value : 4294967295

Time Complexity: O(1)
Auxiliary Space: O(1)

  • To find the sum of two integers without using ‘+’ operator. 
    This is a very easy mathematics trick. 
    We know that a + b = – (-a-b). So this will work as a trick for us.

C++




// CPP program to print sum of two integers
// without +
#include<iostream>
 
using namespace std;
int main()
{
  int a = 5;
  int b = 5;
  int sum = -( -a-b );
  cout << sum;
  return 0;
}

Output

10

Time Complexity: O(1)
Auxiliary Space: O(1)

  • Program to verify the condition inside if block.

C++




// CPP program to verifies the condition inside if block
// It just verifies the condition inside if block,
// i.e., cout << "geeks" which returns a non-zero value,
// !(non-zero value) is false, hence it executes else
// Hence technically it only executes else block
#include<iostream>
 
using namespace std;
int main()
{
    if (!(cout << "geeks"))
    cout <<" geeks ";
    else
    cout << "forgeeks ";
     
    return 0;
}

Output

geeksforgeeks 

Time Complexity: O(1)
Auxiliary Space: O(1)

  • Program to divide an integer by 4 without using ‘/’ operator. 
    One of the most efficient ways to divide an integer by 4 is to use right shift operator (“>>”).

C++




// CPP program to divide a number by 4
// without using '/'
#include<iostream>
 
using namespace std;
int main()
{
   int n = 4;
   n = n >> 2;
   cout << n;
   return 0;

Output

1

Time Complexity: O(1)
Auxiliary Space: O(1)

  • Program to check endianness of the computer.

C++




// C++ program to find if machine is little
// endian or big endian.
 
#include <iostream>
 
int main()
{
   unsigned int n = 1;
   char *c = (char*)&n;
   
   if (*c)   
       std::cout << "LITTLE ENDIAN";
   else
       std::cout << "BIG ENDIAN";
   
   return 0;
}
 
// This code is contributed by sarajadhav12052009

C




// C program to find if machine is little
// endian or big endian.
#include <stdio.h>
 
int main()
{
   unsigned int n = 1;
   char *c = (char*)&n;
   if (*c)   
       printf("LITTLE ENDIAN");
   else
       printf("BIG ENDIAN");
   return 0;
}

Output

LITTLE ENDIAN

Time Complexity: O(1)
Auxiliary Space: O(1)

This article is contributed by Smitha Dinesh Semwal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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