Key for Homework 3 on I/O Basics in C++

 Q  A   Reason
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For questions 1 through 6, we start with the input stream (separated by tabs):

       12   4.9    A    -46.32
       
 1  3   cin >> anInt    // reads the integer 12 into anInt
            >> aChar;   // discards the tab, reads the char '4' into aChar

 2  2   See analysis for question 1.

 3  3   cin >> aDble    // reads the integer 12 and stores the decimal value 12.0 into aDble
            >> aChar;   // discards the tab, reads the char '4' into aChar

 4  2   See analysis for question 3.
 
 5  4   cin >> aDble    // reads the integer 12 and stores the decimal value 12.0 into aDble
            >> anInt;   // discards the tab, reads the integer 4 into anInt
 
 6  4   See analysis for question 5.
 
 7  3   30 + 12 is evaluated (to 42) before being printed.
        The setw(3) takes care of the spacing after the word "is".
 
 8  4   The key here is that TestScore holds a 2-digit number and it must be printed
        right-justified in 4 columns in order to line things up correctly.
        The first code fragment does that by including 2 spaces in the label string
        "SCore:  " and letting the value of TestScore print in the next 2 spaces.
        The second code fragment does that by printing 2 spaces as a separate string,
        after the label, and then printing TestScore.
        The third string just uses setw(4) to print TestScore.

 9  6   We have an input stream containing:  298.173.41.142
        For 1:  cin >> A;              // reads the integer 298 into A
                cin.ignore(100, '.');  // gets rid of the first period
                cin >> B;              // reads the integer 173 into B
        For 2:  cin >> A;              // reads the integer 298 into A
                char ch;
                cin.get(ch);           // reads the period into ch
                cin >> B;              // reads the integer 173 into B
        For 3:  cin.ignore(100, '.');  // gets rid of everything through the first period
                cin >> B;              // reads the integer 173 into B
        For 4:  cin.get(A);            // reads the character '2' into A
                cin.get(B);            // reads the character '9' into B
          
10  9  According to the specification, we have an input stream of the form:

               <label of 25 or fewer chars><tab><integer><newline>

       For 1:  In.ignore(25, '\t');    // gets rid of everything up to the tab;
                                       // if label is less than 25 chars long,
                                       // gets rid of tab also
               In >> NetSize;          ?? if tab is preset, discards it;
                                       // reads the integer 293 into NetSize
       For 2:  In.ignore(30, '\t');    // gets rid of everything through the tab
               In >> NetSize;          // reads the integer 293 into NetSize
       For 3:  In.ignore(25, ':');     // gets rid of everything through the colon,
                                       //   BUT the specification doesn't guarantee
                                       //   that the label ends with a tab!
               In >> NetSize;          // reads the integer 293 into NetSize
       For 4:  In.ignore(50, '\t');    // gets rid of everything through the tab
               In >> NetSize;          // reads the integer 293 into NetSize

For questions 11 and 12, we have the input file:

1234567890
1234567890
1234567890
1234567890

11  6   In.ignore(5, '\n');  // gets rid of everything through the first '5',
                             //   leaving us right before the '6';
        In >> Value;         // reads the character '6' into Value
                             
20  1   In.ignore(5, '\n');  // gets rid of everything through the first '0'
                             //   leaving us at the newline at the end of
                             //   the first line;
        In.get(Value);       // discards the newline; 
                             // reads the character '1' into Value