AVR: volatile variable resetting to zero - variables

Question

I have an interrupt service routine that contains the variable count and a variable state that changes when count reaches a certain value.
What I want my code to do is change and maintain state for a certain period of time determined by the value of count in the if statements in the ISR.
e.g. I want the variable state to equal 1 for 10 counts;
I want state to equal 0 for 5 counts.
Somewhat similar to changing the duty cycle of a PWM.
The problem I am having is that the variable state resets to zero around the end of the ISR or the end of the if statement, I'm not sure.
After searching for answers, I have found that it might be related to the gcc compiler's optimisation but I cannot find a fix for this problem besides declaring volatile variables, which I have already done.
Any help is appreciated thanks.
My code:
#include <avr/io.h>
#include <avr/interrupt.h>
volatile int count = 0;
volatile int state = 0;
int main(void)
{
cli();
DDRB |= (1 << PB2);
TCCR0B |= (1 << CS02)|(0 << CS01)|(1 << CS00);
TIMSK |= (1 << TOIE0);
sei();
while(1) {
...
}
}
ISR(TIM0_OVF_vect)
{
cli();
// user code here
count = count + 1;
if ((count > 5) && (state < 1) && !(PORTB & (1 << PB2))) {
state = 1;
count = 0;
}
else if ((count > 10) && (state > 0) && (PORTB & (1 << PB2))) {
state = 0;
count = 0;
}
sei();
}

Answer 1

First of all your cli() has no effect in your ISR. When an interrupt happens the I flag automatically cleared and restored at the end of ISR - so sei() is needless as well. However there is rational case when sei() inserted somewhere in the middle of an ISR service to let other ISR's to interrupt it. Though it is not beginner level to build such sophisticated ISR system.
Second: if you set up a conditional structure the elementary conditions should be chained by double operators. The evaluation of (count > 5) & (state < 1) cannot be predicted, correctly written it is (count > 5) && (state < 1). Obviously for 'or'-ing || is used.

Answer 2

The conditions of the if statements look suspicious if you are mimicking a PWM.
Why is the pin of port B being tested? The bodies of the if statements will only be entered if the pin (button) is pressed with the correct frequency.
Do you mean to set the pin on port B instead? Then you would want
count = count + 1;
if ((count > 5) && (state < 1)) {
PORTB &= ~(1 << PB2); // clears PB2
state = 1;
count = 0;
}
else if ((count > 10) && (state > 0)) {
PORTB |= 1 << PB2; // sets PB2
state = 0;
count = 0;
}