The multiplication of loops is the trap of programming
In the previous article, we have made ??a password generator essentially consists of numbers based on two different approaches: hard-coding consists of many lines of code proved to be the fastest way. Code with recursion is made with fewer lines of code, but has been slower in the final. Although one may think that the hard coding is faster in terms of performance, it is actually the least flexible. Look more closely at the following piece of code. Go directly to the end of the computer code for an explanation.
Private Function CheckLetter1() Dim index1 As Integer Dim index2 As Integer Dim index3 As Integer Dim index4 As Integer Dim index5 As Integer Dim index6 As Integer Dim found As Boolean Dim date1 As Date Dim date2 As Date Dim checker As String date1 = DateTime.Now found = False If NumericUpDown1.Value >= 1 Then 'look for first draw For index1 = 0 To 25 Step 1 'a - z checker = Microsoft.VisualBasic.ChrW(97 + index1) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next For index1 = 0 To 9 Step 1 '0 - 9 checker = Microsoft.VisualBasic.ChrW(48 + index1) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next For index1 = 0 To 25 Step 1 'A - Z checker = Microsoft.VisualBasic.ChrW(65 + index1) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next End If If NumericUpDown1.Value >= 2 Then 'look for numbers 2nd level For index1 = 0 To 9 Step 1 ' 0 - 9 ''' For index2 = 0 To 25 Step 1 'a - z checker = Microsoft.VisualBasic.ChrW(48 + index1) & Microsoft.VisualBasic.ChrW(97 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next For index2 = 0 To 9 Step 1 '0 - 9 checker = Microsoft.VisualBasic.ChrW(48 + index1) & Microsoft.VisualBasic.ChrW(48 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next For index2 = 0 To 25 Step 1 'A - Z checker = Microsoft.VisualBasic.ChrW(48 + index1) & Microsoft.VisualBasic.ChrW(65 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next '''' Next For index1 = 0 To 25 Step 1 ' 0 - 9 ''' For index2 = 0 To 25 Step 1 'a - z checker = Microsoft.VisualBasic.ChrW(97 + index1) & Microsoft.VisualBasic.ChrW(97 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next For index2 = 0 To 9 Step 1 '0 - 9 checker = Microsoft.VisualBasic.ChrW(97 + index1) & Microsoft.VisualBasic.ChrW(48 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next For index2 = 0 To 25 Step 1 'A - Z checker = Microsoft.VisualBasic.ChrW(97 + index1) & Microsoft.VisualBasic.ChrW(65 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next '''' Next For index1 = 0 To 25 Step 1 ' 0 - 9 ''' For index2 = 0 To 25 Step 1 'a - z checker = Microsoft.VisualBasic.ChrW(65 + index1) & Microsoft.VisualBasic.ChrW(97 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next For index2 = 0 To 9 Step 1 '0 - 9 checker = Microsoft.VisualBasic.ChrW(65 + index1) & Microsoft.VisualBasic.ChrW(48 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next For index2 = 0 To 25 Step 1 'A - Z checker = Microsoft.VisualBasic.ChrW(65 + index1) & Microsoft.VisualBasic.ChrW(65 + index2) ListBox1.Items.Add(checker) If TextBox1.Text = checker Then TextBox2.Text = checker found = True End If Next '''' Next End If date2 = DateTime.Now ' date2.Subtract(date1).ToString ToolStripStatusLabel1.Text = "Finished. " & date2.Subtract(date1).ToString & " and ListBox1 Count: " & ListBox1.Items.Count Return 0 End Function |
Function to generate the possible combinations of 0-9 and the letters of the alphabet in upper and lower case. After the function product with the same principle combinations along a length of password of two characters. We note that the first part of the code correctly produce the 62 possible combinations. We also note that the second part of the code also produces all possible combinations of passwords to two characters, but with an impressive amount of computer code. We guess pretty quickly that the complexity becomes ultra impressive when it comes time to encode a password to 3 characters long.
So what should we learn from this approach? Hard-coding the code for calculations of this type is very heavy and the final few ingenious. It is unthinkable not to use the power of the computer to get the desired result.
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