#include<stdio.h>
int main(void) {
int a[3] = {1,2,3};
printf("\n\t %u %u %u \t\n",a,&a,&a+1);
return 0;
}
Now i don't get why a and &a return the same value, what is the reasoning and the practical application behind it? Also what is the type of &a and could i also do &(&a) ?
Now i don't get why a and &a return the same value, what is the reasoning
a is the name of the array that decays to pointer to the first element of the array.
&a is nothing but the address of the array itself, although a and &a print the same value their types are different.
Also what is the type of &a?
Pointer to an array containing three ints , i.e int (*)[3]
could i also do &(&a) ?
No, address of operator requires its operand to be an lvalue. An array name is a non-modifiable lvalue so &a is legal but &(&a) is not.
Printing type of &a(C++ Only)
#include <typeinfo>
#include <iostream>
int main()
{
int a[]={1,2,3};
std::cout<< typeid(&a).name();
}
P.S:
Use %p format specifier for printing address(using incorrect format specifier in printf invokes Undefined Behavior)
T, the address of the first element has type ‘pointer to T’; the address of the whole array has type ‘pointer to array of n elements of type T’.
int i; you did not declare i as a pointer to int, yet &i is a pointer to int ;)