Solution for Exercise 6-5

ex0605

#include <stdio.h>

int main()
{
    int blorf;

    blorf = 22;

    printf("The value of blorf is %d.\n",blorf);
    printf("The value of blorf plus 16 is %d.\n",blorf+16);
    printf("The value of blorf times itself is %d.\n",blorf*blorf);
    return(0);
}

Notes

* In Lines 10 and 11 demonstrate how you can calculate immediate values and display the results by using the printf() function.

* Even though a variable is used in a calculation, the expression's result is considered immediate because it's not stored anywhere. That's because:

* In C, you can use any variable to represent a value in an equation. In fact, most of your code will use variables in equations instead of immediate values.

* Manipulating the blorf variable in Lines 10 and 11 doesn't change its value. To prove it, add a fourth printf() statement that duplicates Line 9. The output proves that the value of blorf is not altered by its immediate manipulation in the printf() function.

Exercise ex0606 →