How Do Array-to-Pointer Conversion and the Address-of Operator (&) Affect Array Addresses in C ?

Understanding Array Addresses and Pointer Conversion

In C , arrays and pointers are closely related, but understanding their relationship can be tricky. Let's explore the following code snippet to delve into the topic:

int t[10];

int * u = t;

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

The output you observe is:

0045FB88 0045FB88
0045FB88 0045FB7C

Deciphering the Output

The address of u (0045FB88) is understandable, as it points to the first element of the array t. However, the confusing part arises when analyzing the expressions involving t.

• t: It represents the address of the first element of the array, which is the same as &t[0].
• &t: Surprisingly, it also yields the same value (0045FB88) as t. But why?

Array-to-Pointer Conversion vs. Array Address

The key to understanding this behavior lies in how t is being used in the expressions.

• When t is used on its own, an array-to-pointer conversion automatically occurs. This conversion implicitly produces a pointer to the first element of the array.
• In contrast, when t is used as an argument to the & operator, no array-to-pointer conversion takes place. Instead, the & explicitly takes the address of the array itself.

Therefore, &t is not a pointer to the first element of the array but rather a pointer to the entire array.

Memory Locations

In memory, the first element of the array and the beginning of the array occupy the same location. This is why t, &t[0], and &t all have the same value.

In conclusion, the expressions involving t demonstrate the difference between array-to-pointer conversion and explicitly taking the address of an array using the & operator. Understanding this distinction is crucial for effectively working with arrays and pointers in C .

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