CODEWITHSUNDEEP
cprogramming-languagespointers
Mohit
Mohit

Reputation: 539

Why address of a variable change after each execution in C?

int i=10;
printf("Address of i = %u",&i);

Output:
Address if i = 3220204848

Output on re-execution:
Address of i = 3216532594

I get a new address of i each time I execute the program. What does this signify?

Upvotes: 7

Views: 3146

Answers (4)

bhasker pratap singh
bhasker pratap singh

Reputation: 21

Disable ASLR using:

echo 0 | sudo tee /proc/sys/kernel/randomize_va_space

You will always see the same address.

Upvotes: 2

somnath Zalte
somnath Zalte

Reputation: 1

At the time of c program execution another processes are running.While executing a code again you will allocate new address previously allocated address will be allocate for another process.

Upvotes: -1

Jason B
Jason B

Reputation: 665

That's how operating systems work. When you declare a variable, you're asking the underlying systems to allocate a memory block (address) to store that data (or a pointer to another block if you're dealing with pointers, but here you've got a primitive, so it's just data being stored). The program doesn't care where the memory is, just that it exists, because it knows how to keep track of whatever it's given.

As the programmer, this really isn't that big of a deal unless you're doing some incredibly low-level work. The hardest part of this to really grasp, for most people, is that when you work with pointers, you can't equate things the same way you can primitives, because pointers consider their values (when using == as an equator) to be their memory addresses.

Upvotes: 0

Dean Harding
Dean Harding

Reputation: 72658

It signifies that your program is being loaded a different (virtual) address each time you run it. This is a feature called Address Space Layout Randomization (ASLR) and is a feature of most modern operating systems.

Upvotes: 11

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