C++ Programming Applied to Robotics, Mark Aull Lesson 2: Intro to Visual Studio, debugger Intro to C++, variables, conditionals, loops, strings

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C++ Programming Applied to Robotics, Mark Aull Lesson 2: Intro to Visual Studio, debugger Intro to C++, variables, conditionals, loops, strings As in the first lesson, open visual studio, start a new

1 C++ Programming Applied to Robotics, Mark Aull Lesson 2: Intro to Visual Studio, debugger Intro to C++, variables, conditionals, loops, strings As in the first lesson, open visual studio, start a new empty project named lesson2, add a new.cpp file to that project, and paste the code in the Appendix into the.cpp file. The header files indicate that functions involving standard input/output and string manipulation can be called from this.cpp file. Main is called with no parameters as before. Attempt to build the code. On most modern compilers, this code will produce an error on the line for (ii=4,jj=1;ii>1;jj*=ii,ii--);//null for loop complaining about jj not being defined. Notice that jj is defined in the previous loop: for (int jj=0;jj<2*ii;jj++)//scope of jj varies by compiler this shows the scope of variables: a variable declared inside of a block (a loop, if statement, etc) doesn't exist outside of it. Some older versions of visual studio did not enforce this rule correctly, so you will occasionally come across code where variables are declared in loops, and used outside of them. The fix is to move the declaration for jj to the top of the file: int ii=23; changes to int ii=23,jj; and for (int jj=0;jj<2*ii;jj++)//scope of jj varies by compiler changes to for (jj=0;jj<2*ii;jj++)//scope of jj varies by compiler Now the code should build correctly. Start the debugger by selecting step over or pressing F10.

Note the data types, integer, double precision floating point, and character array (string). Also note the call stack tab on the bottom right. It currently indicates the function main().

2 Pressing F10 one more time will move to the first line of main. The variables declared at the beginning of the function are created as seen in the locals tab in the variable window, which is hopefully on the bottom left. Note the data types, integer, double precision floating point, and character array (string). Also note the call stack tab on the bottom right. It currently indicates the function main(). This will be much more useful when multiple functions are in use; you can move to whichever level you wish by double-clicking on the desired function in that tab. Stepping few a few more lines (with F10), the variables are initialized to the values indicated. Then the values are displayed by the formatted print (printf()) statement. %s indicates a string value, %c a character, %d an integer (in decimal) and %f a floating point number. The numbers indicate the width allowed, and the number after the dot indicates the precision with which to display a floating point number.

3 The next few lines demonstrate some mathematical operations. Addition, subtraction, multiplication and division occur as you would expect, but additionally, the ++ operator adds 1 to any number, -- subtracts 1, x+=y is equivalent to x=x+y, and the same works with -=x, *=x, and /=. strcat() is a string concatenation function. It takes the second string and appends it to the first string.

$Also note that even though ii==24 is true, the second branch is not evaluated because the first already was. The printf() statement includes a few escape characters: \n is a newline, and \t is a tab.$

4 Note that the value if ii is currently 24, which is greater than 23, so the first branch of this if statement is evaluated. Also note that even though ii==24 is true, the second branch is not evaluated because the first already was. The printf() statement includes a few escape characters: \n is a newline, and \t is a tab. Similarly \b is a backspace, \r is a carriage return, \f is a form feed, \a is a beep. \\ generates a \ character. \' and \ are needed to produce those characters. Note: when using '\b', the character is not actually deleted, but the position where characters will be written is moved back, so that the next character will overwrite the current one.

5 Next a /= and a -= command are demonstrated. Note the new values of ii and p. Also, the string formatted print (sprintf()) command is introduced. It works like printf(), but its first parameter is a string, into which the output is stored, rather than being printed on the screen. The printf() command on the next line displays the output of the sprintf() command.

6 Next, predict which branch of the if statement will be executed given the current values of the variables, and what will be printed. A for loop in C++ contains three parameters: initialization, conditional and an increment. The initialization runs before the beginning of the loop, after a ; is the conditional which is tested before the code in the loop is run, and the increment step after the next ; is run after the code in the loop. Attempt to predict what the loop will display before running it. You should then use the debugger to slowly step through the code, verifying that the description below is correct. for (int jj=0;jj<2*ii;jj++)//scope of jj varies by compiler {//1 printf("jj=%d\n",jj);//2 if (jj<6)//3 jj++; }//4 jj at 1 jj displayed (value at 2) (jj at 3)<6 (true or false) jj at 4 0 First, ii is 6 from before, and jj is set to 0. Since 0 is less that 2*6=12, the code between the { } is executed, displaying jj=0, then, since 0 is less than 6, incrementing jj to 1. Next the increment section of the for loop (containing a jj++) is run, increasing jj to 2. The conditional is then evaluated: 2 is less than 12, so the loop is run again, displaying jj==2, and since 2 is less than 6, increasing jj to 3. The increment section increases jj to 4. Once again, jj=4 is displayed, and by the next time the loop is evaluated, jj is increased to 6. After jj=6 is displayed, the if statement does not run because 6 is not less than 6. After this, the if statement is never true again, and the loop displays jj= for each value between 7 and 11. After this, jj++ is run, increasing jj to 12. Since now jj<2*ii is no longer true, the

7 code stops running the loop. The next line demonstrates an important fact to remember about C and C++: an operation involving a data type yields that data type as a result. A character plus a character is a character, a double times a double is a double, etc. When multiple data types are combined, the larger or more inclusive is used: an integer plus a character is an integer, an integer time a float is a float, etc. The only time this behavior is counter-intuitive is for integer division: an integer divided by an integer is the mathematical definition of a rational number, however according to the rules above, the result is an integer. Here, 5 and 2 are both read as integers, and the quotient is processed as 2, not 2.5, even though it is being stored into a floating point variable. Replacing 5/2 with double(5)/2 or even 5.0/2 or 5./2 would correct this behavior. The next line introduces the concept of a null for loop, i.e. a for loop terminated by a ; instead of being followed by a loop in { }. Null for loops therefore run in a single line; the line after is outside the loop. This line also shows that multiple initialization and increment steps are allowed, separated by commas. The conditional can be as complex as desired, incorporating multiple statements connected by and (&&) and or ( ), but only one decision value must be computed (the loop either continues to run or it does not). This null for loop runs like a function where ii is an input and jj is an output (f(ii)=jj). The reader's task is to fill in the table below without running the code, in order to predict the final value of jj. The initial values have been filled in to help you. for (ii=4,jj=1;ii>1;jj*=ii,ii--);//null for loop printf("f(4)=%d\n",jj);//displays ii jj ii>1 4 1 true Next, verify that your answer is correct. Looking back at the numbers above, inductively obtain the function (f) implemented by this loop and use that knowledge to predict the value of jj calculated by the next loop, which is identical except for the initial value of ii. The following is a short code demonstrating the getch() function (get character) in the conio (console input\output) library. This function grabs any key pressed on the keyboard, and stores it to a. It is then displayed on the screen as an integer and as a character. Note that a while loop is used, which has the conditional parameter like a for loop, but does not use the other sections as a for loop does. A while (1) loop runs forever, until the program is stopped externally, e.g. by clicking the x in the upper right of the window. Using this program, you can see what each keystroke outputs. The shift, ctrl, and alt keys modify these outputs. For example, note that ctrl-g produces a 7 and a beep. This is not so much useful as interesting.

8 #include <stdio.h> #include <conio.h> void main(void) {//main function int a=0; while(1) {//while loop (until ctrl-c or X) a=getch(); //get character from keyboard printf("%10d %5c\n",a,a);//display character as integer and character }//end while loop }//end main Escape returns 27, and special keys, such as function keys, arrow keys, etc are noted to produce two outputs: a zero or 224, followed by another character, i.e. the up arrow produces a 224 followed by an H. Compare this with shift-h. Therefore, the program can be modified to handle special characters differently, and exit on the escape key: #include <stdio.h> #include <conio.h> void main(void) {//main function int a=0; while(a!=27) {//while loop (until esc) a=getch(); printf("%10d %5c\n",a,a); if (a==224 a==0) {//control key a=getch(); printf("command: %10d %5c\n",a,a); }//end control key }//end while }//end main function The code available on the website stretches this a little further by including a large if-else statement displaying what keys are hit for selected special characters.

9 Assignment: Produce a program that drives the Galil motion controller using the libraries as demonstrated in lesson 1 and the keyboard, using the getch function. The program must: not move the robot when started until the user commands a motion, stop the robot when the program stops, display instructions describing all behavior on startup, send commands AC 80000,80000 and DC 80000,80000 to the Galil to limit acceleration, have a designated key to stop the robot (by sending ST to the Galil), have a designated key to exit the program, control both the speed and turning rate of the robot (using the JG command, as shown in the previous lesson), send a to each controller after one keystroke (This is a good choice for a lowest speed), and have comments describing its operation at each important step (using // or /* */ as seen above), i.e. I should be able to put the robot wherever I want, pointing in whatever direction I want fairly safely with your program. Reminders: after the JG command, the first number is the right motor, in reverse, and the second number is the left motor, forward. Also, you must send a BG command after the first JG command sent after startup or after the first JG command sent after an ST command. It is more reliable to send s BG command after each JG command.

10 Appendix: source code for part 1 /*comments, whitespace doesn't do anything, for readability only start lecture math ops, ++, --, +-* /= printf, sprintf, strcat conditional, loop (different from MATLAB)*/ #include <stdio.h> #include <string.h> void main(void) {//main function int ii=23; double p=3.1415; char string[80]="hello"; printf("%s,%10d,%11.7f\n",string,ii,p);//displays ii++;//increment to p*=2;//p= strcat(string," world");//concatenate printf("%s,%5d,%10.7f\n",string,ii,p);// displays if (ii>23)//displays { printf("ii >23\n\ttestf\b\n"); } else if(ii==24) printf("ii=24\n"); else printf("ii<=23\n"); ii/=4;//decrement to p-=ii;//p= sprintf(string,"%s,%10d,%11.7f\n",string); printf("%s",string,ii,p);//displays if (ii==6&&p==0)//displays { printf("ii is 6 and p is 0\n"); } else if(ii!=6) printf("ii is not 6\n"); else printf("p is not 0\n"); for (int jj=0;jj<2*ii;jj++)//scope of jj varies by compiler { printf("jj=%d\n",jj); if (jj<6) jj++; } p=5/2;// data type rules printf("p = %f\n",p); for (ii=4,jj=1;ii>1;jj*=ii,ii--);//null for loop printf("f(4)=%d\n",jj);//displays for (ii=6,jj=1;ii>1;jj*=ii,ii--); printf("f(6)=%d\n",jj);//f(x)= }// end main

UNIT - I. Introduction to C Programming. BY A. Vijay Bharath

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