Address of a variable

Question

Please help me to understand the output...(I understand that the variables in the same scope have same address) but here I'm not getting how this addresses are actually working out.

#include <stdio.h>

void f(int p)
{
printf("The address of p inside f() is %p\n", &p);
}

void g(int r)
{
printf("The address of r inside g() is %p\n", &r);
f(r);
}

int main()
{
int a = 55;

printf("The address of a inside main() is %p\n", &a);
f(a);
g(a);

return 0;
}

/*
The address of a inside main() is 0x7ffe12bb471c
The address of p inside f() is 0x7ffe12bb46fc
The address of r inside g() is 0x7ffe12bb46fc
The address of p inside f() is 0x7ffe12bb46dc
*/

https://code.sololearn.com/ca120a18a141

Answer 1

@Mirielle Thank you for this! But I'm in confusion...I need little more clarification.
My point is that :
If I'm creating a copy of the argument with different memory location then why f() local variable p and g() local variable r have same address? Aren't after the execution of f(), memory occupied by it's variable will get deallocated as the end of the scope of variable is reached? and when g() will execute it's variable should be get allocated to different memory location?