Chapter 3.3, Problem 55E

Explanation of Solution

Given: The following lines of code that contains a list box in which the font of the output is set to ‘Courier New’:

Dim fmtStr As String = "{0,-7} {1,-7:N1} {2,-7:P1}" 'Two spaces IstOutput.Items.Clear()

IstOutput.Items.Add("12345678901234567890") IstOutput.Items.Add(String.Format(fmtStr, "Element", "Weight", "Percent")) IstOutput.Items.Add(String.Format(fmtStr, "Oxygen", 97.5, 97.5 / 150)) IstOutput.Items.Add(String.Format(fmtStr, "Carbon", 27, 27 / 150))

To find: The output produced by the given lines of code.

Dim fmtStr As String = "{0,-7} {1,-7:N1} {2,-7:P1}" 'Two spaces IstOutput.Items.Clear()

IstOutput.Items.Add("12345678901234567890") IstOutput.Items.Add(String.Format(fmtStr, "Element", "Weight", "Percent")) IstOutput.Items.Add(String.Format(fmtStr, "Oxygen", 97.5, 97.5 / 150)) IstOutput.Items.Add(String.Format(fmtStr, "Carbon", 27, 27 / 150))

Solution:

The given lines of code will show “123456789012345678901234567890 Element Weight Percent Oxygen 97.5 65.0 % Carbon 27.0 18.0 %� as output in the list box named “IstOutput� because the method named “Add� will add the items to the list. The value “123456789012345678901234567890� will be concatenated with “Element�, “Weight�, “Percent�, "Oxygen", "97...