Question

Write a C++ main program that has the following 5 short independent segments.

1. Show that for unsigned int a,b and a>0, b>0, we can get a+b < a

2. Show that for int a,b and a>0, b>0, we can get a+b < 0

3. Show that for int a,b and a 0

4. Show that for double x and x>0 we can get 1. + x = = 1.

5. Show that for double a,b,c in some cases (a+b)+c != (c+b)+a

Answer 1

SOURCE CODE:

*Please follow the comments to better understand the code.

**Please look at the Screenshot below and use this code to copy-paste.

***The code in the below screenshot is neatly indented for better understanding.

#include <iostream>
using namespace std;

int main() {
  
{// SEGMENT 1
// 1. example
   unsigned int a=4294967295;
   unsigned int b=100;
   cout<<"\na="<<a<<" ; b="<<b<<" ; a+b="<<a+b<<endl;
   cout<<"Hence, a+b can be less than a for some values of unsigned int."<<endl;
}
  
{// SEGMENT 2
  
   // 2. example
   int a=2147483647;
   int b=100;
   cout<<"\na="<<a<<" ; b="<<b<<" ; a+b="<<a+b<<endl;
   cout<<"Hence, a+b can be negative for some positive values of int."<<endl;
}
  
{// SEGMENT 3
  
   // 3. example
   int a=-2147483647;
   int b=-100;
   cout<<"\na="<<a<<" ; b="<<b<<" ; a+b="<<a+b<<endl;
   cout<<"Hence, a+b can be positive for some negative values of int."<<endl;
}

{// SEGMENT 4
  
   // 4. example
   double x=0.0000000007;
   cout<<"\nx="<<x<<" ; 1. + x ="<<1.+x<<endl;
   cout<<"Hence, we can get (1.+x==1.) for insignificant value of double x."<<endl;
}

{// SEGMENT 5
  
   // 5. example
   double a=1;
   double b=1e20;
   double c=-1e20;
  
   double x= (a+b)+c;
   double y= (c+b)+a;
  
   cout<<"\n(a+b)+c = "<<x<<" ; (c+b)+a = "<<y<<endl;
   cout<<"Hence, (a+b)+c != (c+b)+a"<<endl;
}
  
  
   return 0;
}

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OUTPUT: