Answers/Solutions to Exercises in Chapter 3, Exercise 6

E6: In a C program, we store two short integers in an int variable and retrieve them and display them later in the program:
    short* p;
    int x;
    ...
    p = (short*) &x;
    *p++ = 1;
    *p = 2;
    ...
    p = (short*) &x;      // this statement was missing in the book
    printf("first short=%d,second short=%d\n",*p, *(p+1));
During the execution we will see on the screen first short=1,second short=2. What will we see if we compile and execute our program on a machine that has the opposite ``endianess"?

A6: Of course, if the program was as printed in the book, it would not display the message first short=1,second short=2 as the pointer (p+1) points outside of where we stored the two short integers. So we corrected the program and after storing the two shorts in x, we re-point p to the beginning of x again. Now it performs correctly, and we can return to the original question.
The answer is -- no difference, we will see the same message. The fact that we are storing the values in x is coincidental, not essential to anything. We just store two shorts somewhere in memory and read them back. Of course, the storing and the reading is done according to the endianess, but it does not have any effect on the result.

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