vb.net read usb barcode scanner FIGURE 10.8 in Software

Encode QR Code 2d barcode in Software FIGURE 10.8

FIGURE 10.8
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The robot can write (well, only three letters so far).
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VECTOR GRAPHICS ROBOT
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FIGURE 10.9
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Writing robot.
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Ascii(). The ASCII code for A is 65, for B is 66, and so on, but we need to index into the array of letters at 0 for A , 1 for B , and so on. Can you see a pattern
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Letter_Number of any letter = Ascii(of the letter) -Ascii( A )
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The index position of any letter into the array of letter fonts can now be calculated. This array will have only capital letters. What if the message to be printed has lower case letters We will use the function Upper() to convert all the letters in a string to upper case letters. What if the message contains unsupported characters We will test to see if the character about to be printed is within the limits Ascii( A ) to Ascii( Z ) and if it is not, a space will be printed in its place. The number of instructions to draw each letter will vary from one letter to another. This means that we will need a way of knowing how many instructions there are in each row in the letters array. We could do this by having the number inside the array, or create another array that has this number for each letter. Here is a better method. The data pairs will have a number telling us the instruction whether to move, turn, or raise/lower the pen. We can add one more instruction to indicate the end of instructions. So, for example, we can have Forward 1, Turn 2, Pen 3, and End 1000. Now a set of instructions can look like this:
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Data Temp_Letter; A ,3,up, 1,20, 2,45,..........,-1000
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Notice the 1000 at the end of the data. It is to indicate the end of instructions for this letter. We chose 1000 because we want a number that cannot possibly be used in the normal instructions. Unfortunately the above is not very readable, you might confuse the instruction with its quantity and will have to constantly keep referring to the codes table to know what instruction is what. A better solution would be to do this:
Data Temp_Letter; A ,p,up, f,20, t,45,..........,e
Notice the use of variables p, f, t, and e. This is a lot easier to understand. All we need is to add a few statements that set a variable f 1, t 2, p 3, and e 1000. This is the same as using the constant red to stand for the number 4 (see Sec. B.7.6). We need to establish a convention for the fonts. We will assume that the letter prints starting at where the robot happens to be and at 90 orientation to the current robot heading, also that the pen is up. The drawing instructions will create the font for the letter and position the robot at the end of the letter facing the same direction it was before, and the pen in the up position. The font will occupy a 6 6 pixels area in its base scale, so a scale of ve will make it 30 30 and so on. To summarize; if the robot is at position 100, 100 facing east (90 ) and a scale of 10:
The letter printed will occupy the area bounded by the square 40, 40, 160, 100 and be oriented vertically (90 90 0). The robot will end up at position 160, 100 facing east at the end of the printing of the letter and the pen will be up.
Figure 10.10 shows a program that makes use of all the principles discussed above. The subroutine Create_Font is not shown because it will be discussed later. The main program sets up the robot and sets the colors, scale, and message to be printed (the orientation is de ned by the robot s initial heading). Then it calls the subroutine Print_Message. This subroutine makes use of all the conventions discussed so far. 10.2.1.1 Print_Message Subroutine Look at Fig. 10.10 carefully. You should be able to recognize most of what we have discussed in this routine. You cannot run this program without combining it with the code given in Fig. 10.11. First the message is converted to upper case. Then we iterate for each letter in the message with the for-loop. Notice how we use the function Substring() to obtain each letter from the message string. We nd the index into the array of letters by the same formula we discussed above. Any character that is not from A to Z will be made into the last letter in the array, which is the space character. This means that the routine will write a space for any character in the message that is not a valid letter. The while-loop will execute each instruction for printing the letter until it nds the instruction code of 1000, which is, by the above convention, the indicator of the end of the drawing instructions. Notice that no for-next loop is used to obtain the instructions since it is not known ahead of time how many there are. Also, notice the formula for indexing to obtain the instruction:
Letters[L,Inst_No*2 1]
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