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What will be printed for the following code : if(x=0) cout<<”zero”; if(x>=0) cout<<”pos”; else cout<<”neg”;

Question

What will be printed for the following code : if(x=0) cout<<”zero”; if(x>=0) cout<<”pos”; else cout<<”neg”;

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Solution

The code you've written will actually result in a compilation error. In most programming languages, the "=" operator is an assignment operator, not a comparison operator.

If you want to compare x to 0, you should use the "==" operator. So, the correct code should be:

if(x == 0) 
    cout<<"zero";
if(x >= 0) 
    cout<<"pos";
else  	
    cout<<"neg";

Now, let's analyze this corrected code:

  1. If x is equal to 0, it will print "zero" and then "pos" because 0 is also greater than or equal to 0.
  2. If x is greater than 0, it will only print "pos".
  3. If x is less than 0, it will print "neg".

This problem has been solved

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