Laboratory Exercise: Types and Variables

Incrementing

There are two ways to increment a variable: x++ and ++x. These should not be confused with each other.

10. Here is an example program. Examine the output to differentiate between these two incrementers.

      int a,b,c;
      a = 0;
      b = x++;
      c = ++x;
      printf(" a = %d b = %d c = %d\n", a,b,c);
      a = 0;
      b = ++x;
      c = x++;
      printf(" a = %d b = %d c = %d\n", a,b,c);	
    

Unexpected Results

Adding ints to chars

In C, it is possible to add an integer value to a char value.

11. Add the following lines of code after the /* Add code here */ comment:

                c = c + 7;
                printf("c + 7  = %c\n", c);
            

    1. Save, compile, and run the program again.
    2. What is the value of c?

12. Add 7 to c again (you can just copy the two lines again) and a print out the value.

    1. What is printed?
    2. Is that what you expected?

13. Set c back equal to zero and print out the result:

                c = 0;
                printf("char c = %c\n", c);
              

    You may notice the print statement doesn't print out c as being zero.

    Why is it blank to the right of the equals sign? Behind the scenes, a char is really just a small integer. This is why an expression such as: c + 7; is legal. The only difference is how printf interprets the value stored in the char variable c.

14. Set c equal to 48 and print it again.

    You will notice it is now a zero again.

Casting

As you may have encountered in the previous exercise, c is pretty forgiving as to what types can be assigned to each other. Whether you realized it or not, the assignment operator, =, converts between certain types silently because they are easily compatible -- such as chars and ints. Even in the case of types that are less compatible such as ints and doubles, if you try to assign a double value to an int variable, the compiler does not complain. It simply drops the floating point (the decimal).

15. If you didn't try that above, try it now.

                      i = 2.7;
                      printf("int i = %lf", i);
                    

    The process of converting between variable types is called casting. If you don't have to say anything for the process to take place, it is called implicit casting. This is the case with all the math operations in c. The +, -, *, /, and = operators all implicitly cast across primitive types. The rules for implicit casting can be found in K&R appendix A6 (specifically section 5).

16. Now try explicit casting. This is where you tell c exactly what you want to happen. The line:

                      int x = (int) 3.14;
                  

    tells the compiler (but more importantly anyone reading the code) that it is explicitly intended to convert the double (3.14) to an integer and store it in the space x which is available for integer data.
    

17. Why do you think explicit casting is necessary if the compiler already does it implicitly?

    As hinted at, it is usually an effective way of informing anyone reading your code that you mean to do something that may be questionable such as dropping the decimal from a number. The second reason is that casting is necessary when dealing with more complicated data such as pointers and structs. The compiler will not let you convert between certain types unless you explicitly tell it that is what you want to do.

    Finally, there is a danger in casting. As you experienced when playing with signed and unsigned characters, types have sizes. If you try to put a number that is bigger than the size of the type you are casting it to, you will get undesired results. Also, if you cast signed numbers to unsigned types, you can encounter odd results. Remember, a signed type has one less bit (to make room for the sign bit) to represent data with.

Experimenting with Casting

18. Declare a signed int and initialize it to a negative number greater than INT_MIN from the limits header (so between INT_MIN and 0).
19. Then, declare an unsigned int.
20. Cast your signed integer to an unsigned integer and assign its value to your unsigned int variable.
21. Finally, print the two values and examine the results.

    If you called your signed integer s, your explicit cast and print statements would look something like this:

                      unsigned int u = (unsigned int) s;
                      printf("signed int s   = %d\n", s);
                      printf("unsigned int s = %u\n", u);
                  

    * Note the use of %u to specify an unsigned int in the last printf statement.

22. Explain why the values print out as they do. Hint: Refer to the limits.h header file.

ASCII Table -- If time permits..

Recall earlier that setting a char equal to 0 is not the same as setting it equal to '0'. As you may know, this is because the zero in single quotes is actually interpreted as the character zero. The character zero has the integral value 48. That means setting a char equal to 48 is the same as setting it equal to '0'.

Here is a simple program, ascii-print.c , that prints out all the ASCII values along with their corresponding characters. It uses a for loop, which we will be considering later this module.* Note that while chars can range from 0 to 127, the printable characters start at 32.

23. What is the corresponding character to the ASCII value 85?
24. What about 32?
25. What is the ASCII value for the character 'A'?
26. What is the ASCII value for the character 'a'?
27. What is the integer difference between the values of 'A' and 'a' ?

Feedback Welcome

Development of laboratory exercises is an interative process. Prof. Walker welcomes your feedback! Feel free to talk to him during class or stop by his office.