Key for Homework 4B on I/O and Selection in C++

 Q  A   Reason
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For questions 1 through 5, we start with the input stream (separated by tabs):

       1.2   49A    -46.32
       
 1  1   cin >> anInt    // reads the value 1 into anInt
            >> aChar;   // reads the char '.' into aChar

 2  4   See analysis for question 1.

 3  9   cin >> anInt    // reads the value 1 into anInt
            >> aDble;   // reads the value .2 into aDble

 4  2   cin >> aDble    // reads the value 1.2 into aDble
            >> aChar;   // discards the tab, reads the value '4' into aChar
 
 5  2   cin >> anInt;   // reads the value 1 into anInt
        cin.get(aChar); // reads the value '.' into aChar
        cin >> anInt;   // reads the value 2 into anInt
 
 6  3   1 would print 78 two columns too far to the right
        2 would print 78 two columns too far to the left
        3 is just right
 
 7  3   If Enter == 1, the switch would execute case 1, setting Enter to -4,
        then fall through into case 2 since there's no break in case 1.
        That would set Enter to -6, but then fall through into case 4 since
        there's no break in case 2 either.
        That would set Enter to -8 and the break would cause us to exit the
        switch.
 
 8  4   If Enter == 10, there's no case for that specific value, so the
        default case would be executed, setting Enter to -1.
 
 9  4   Apply DeMorgan's Law:  !(A || !B && C) == !( A || ( !B && C )
                                               == !A && !( !B && C )
                                               == !A && (!!B || !C)
                                               == !A && (B || !C)
                                               
        Note:  the parentheses DO matter because && has higher precedence than ||.