Thread: life time of variable ?

Hello Everyone

I am new to this forum. I am sorry if this question is already discussed on forum but I really don't understand clearly life time of variable.

Code:

#include <stdio.h>
int main(void)
 {  // start scope1 
   int i = 2;
  for (int  j = 0; j < 5; ++j)


      { //start scope2
         
         // j can be only accessible within scope 2
         
         printf("Hello \n");
      }// End scope2
      
  printf("%d", i);  // I can accessible within scope 1 and scope 2


  return 0;
}//  End scope1

I don't understand how variable create and destroy regrades memory ?

Code:

#include <stdio.h>
int main(void)
{  // start scope1 
  int i = 2;  // i begins at this ;
  for (int  j = 0;/* j begins at this ;*/ j < 5; ++j)
     { //start scope2
         // j can be only accessible within scope 2
         printf("Hello \n");
      }// End scope2
  printf("%d", i);  // I can accessible within scope 1 and scope 2

  return 0;
}//  End scope1

Variables begin their semantic existence at the ; following their declaration, and end at the closing brace of the scope block they're contained in.

> I don't understand how variable create and destroy regrades memory ?
Nor is there really any need to - at least in the short term.

The compiler is free to turn your code into this.

Code:

#include <stdio.h>
int main(void)
{
  int j;
  int i = 2;
  for (j = 0; j < 5; ++j)
    {
      printf("Hello \n");
    }
  printf("%d", i);  // I can accessible within scope 1 and scope 2
  return 0;
}

Now j has a physical lifetime which is the same as i.
But the compiler will enforce the rules to make sure you can only access j when you're allowed to.

Yes, if you're doing some really low level debugging, it might be useful to know that is the case.
But that knowledge is really only 'true' for that particular program compiled with one specific compiler.

You certainly can't assume it's true for all compilers, and it's definitely not something you would assume in the code by trying to write something sneaky to access variables outside of their declared scope.

> It means local variable is created when the function first starts, and stays alive until the function exits.

Not exactly. Local scope variables are not created when the function starts. They are created when "you" declare them. If you have a variable declared in the last line in your function, it'd be created when the compiler reaches that line, and not when your function first starts. And yes, it stays alive until the end of the function i.e. its lifetime is the scope of that function. This is typically the case for statically allocated variables. When you learn about dynamic allocation, you'll learn that the lifetime of a dynamically allocated variable is decided by you.

My bad, I meant to say stack allocated. Yeah, static variables have the lifetime same as that of the entire program. I'll edit my post above for the typo error.

Edit: I don't have edit permissions for messages so I can't edit the old message. What I meant was this:

"This is typically the case for stack allocated variables."

My bad, I meant to say stack allocated. Yeah, static variables have the lifetime same as that of the entire program. I'll edit my post above for the typo error.

Edit: I don't have edit permissions for messages so I can't edit the old message. What I meant was this:

"This is typically the case for stack allocated variables."

Ignore this. This is just me being stupid. @Hodor, you're completely right. With the exception of static objects, it would be right saying "typically the case for statically allocated variables". I was on the bus when I replied to your message and got "statically allocated" i.e. allocation at compile time mixed with static variables. It's a mess - what I was thinking - so you probably don't wanna know lol but thanks anyway for pointing that out.

Also, why can I not just be able to edit older messages...