Thread: Compiler optimization: volatile

Hello,
I have some doubts about compiler optimizing C code. This is more related to the field of embedded system.

Code:

int a, b;

void experiment() {
    a = 8;
    b = a * 7;
    if(a == 8)
        printf("a equals 8!!!");
    else
        printf("A is NOT 8!!!");
}

In above code, inside the function experiment() value of a is not changing. So extreme optimization will change the code as follows

Code:

int a, b;

void experiment() {
a = 8;
b = a * 7;

printf("a equals 8!!!");
}

So in these cases it is necessary to declare variable a as volatile, since a could be modified by an external even ISR or multi threaded application.

This is clear for me. But what if the code is as follows

Code:

int a, b;

void experiment() {
    a = 8;
    b = a * 7;
    if(a == 8)
        printf("a equals 8!!!");
    else {
        printf("A is NOT 8!!!");
        a = 5;
    }
}

here a is changed to 5 inside else statement.

Q1. Will this avoid optimization?


Another example of using volatile is in loops.

Code:

void delay() {
    int x = 1000;
    while(x > 5)
        x--
}

I have seen in a tutorial that, if x in not declared volatile, compiler will optimize this.
Why????
Q2. Here value of x is modifying inside while loop. So what's the necessary of volatile here??

A1) No, it won't. It's the same situation as your previous example.

How come? here a is changing inside. So how could compiler ignore it.

You shouldn't use busy loops for delay anyway. What you need is to delay based on some clock value, which would be unpredictable for the compiler and thus not optimizable. Your system does have interrupts, right?

Like when counter value reaches til a certain value?

eg:

if TCINT is a counter

Code:

while(TCINT != 1000);