Given: The following statement:

Dim dtl As Date = Today

Dim dt2 As Date = dtl.AddDays(5)

txtOutput.Text = CStr(DateDiff(Datelnterval.Day, dtl, dt2))

To find: The output of the following codes:

Dim dtl As Date = Today

Dim dt2 As Date = dtl.AddDays(5)

txtOutput.Text = CStr(DateDiff(Datelnterval.Day, dtl, dt2))

Solution:

The following lines of code will display the output: "5". The two variables termed "dt1" and "dt2" are declared of type â€œDateâ€�. The variable dt1 is initialized to "Today". The variable dt2 is initialized to "d1.AddDays(5)". The function: â€œAddDays()â€� adds the value of the parameter (which is passed) to the present value of day...

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Chapter 3.3, Problem 40E

Chapter 3.3, Problem 42E

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Chapter 3 Solutions

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Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E Ch. 3.1 - Prob. 7E Ch. 3.1 - Prob. 8E Ch. 3.1 - Prob. 9E Ch. 3.1 - Prob. 10E

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Solution Summary: The author explains how the following lines of code will display the output: "5". The two variables are declared of type "Date".

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Chapter 3 Solutions