Why Do `t`, `&t[0]`, and a Pointer to `t` Have the Same Address?

Why t, &t[0], and t Converted to a Pointer Have the Same Address

An array is a contiguous block of memory containing elements of the same data type. When you declare an array, the compiler allocates a block of memory and creates pointers to each element within that block.

Case 1: t and &t[0]

When you use t on its own (without any subscripts), it undergoes an array-to-pointer conversion, which generates a pointer to the first element of the array. This behavior is equivalent to using &t[0], which explicitly returns the address of the first element. Hence, t and &t[0] have the same value.

Case 2: &t

However, when you apply the & operator to t, it doesn't trigger the array-to-pointer conversion. Instead, &t directly takes the address of t itself, resulting in a pointer to the entire array.

In memory, the first element of the array and the start of the array reside at the same location. Therefore, the pointers t, &t[0], and u (which is a pointer to t) all have the same address.

Output Explanation:

The given code demonstrates this behavior:

int t[10];
int * u = t;

cout << t << " " << &t << endl;
cout << u << " " << &u << endl;

Output:

0045FB88 0045FB88
0045FB88 0045FB7C

The output shows that t, &t[0], and u have the same value (0045FB88), confirming their equivalence as pointers.

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