Homework Answers
Answer to First Question::
From main method modify_x() function is called 2 times.
If we check the function_x() function, we can see that it is not taking any function argument.
The function has a local variable x which is initialized to 10.
At the next statement (x+=20), it is basically doing x = x +20. i.e, the previous value of local variable x
is increased by 20.
So new value pof x become = 10+20 = 30
So, at the end of execution of first program, modify_x() will print 30.
Since the program will do the same thing again, it will again print 30.
Answer to Second Question:
In order to print differet values in the output, we can pass value of x as functon argument for modify_x().
So our new function definition will be:
void modify_x(int x){
x = x+20;
print("x is %d", x);
}
so if we are calling modify_x(20), which is followed by modify_x(30) ; then output will be like:
for first case , x is (20+20)=40
In second case:modify_x(30) will have output like 30+20 = 50
Add Answer to:
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