Why Does the Address of an Array Equal the Pointer to its First Element in C ?

Decoding the Array-to-Pointer Conversion: Why Address of Array Equals Pointer to First Element

In C , the concept of array-to-pointer conversion can lead to a seemingly peculiar observation: the address of an array is identical to the array converted to a pointer. To unravel this phenomenon, let's dissect the following code snippet:

int t[10];

int *u = t;

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

Output:

0045FB88 0045FB88
0045FB88 0045FB7C

While the output for the pointer u makes sense, understanding the equivalence of t, &t, and &t[0] might seem puzzling. Delving into the intricacies of array-to-pointer conversion sheds light on this behavior.

Array-to-Pointer Conversion

When using t on its own in an expression, an automatic conversion from the array type to a pointer type occurs, resulting in a pointer to the first element of the array. This conversion allows treating the array as a pointer.

Taking the Address of a Variable

When the & operator is used with t, no implicit array-to-pointer conversion takes place. Instead, the & operator explicitly derives the address of t itself, resulting in a pointer to the entire array.

Positional Equivalence

The first element of the array occupies the same memory location as the beginning of the array as a whole. This positional equivalence explains why the address of t (the array), &t (pointer to the array), and &t[0] (pointer to the first element) all have the same value.

In essence, the array-to-pointer conversion provides a means to access individual array elements via a pointer while &t grants access to the array as a single entity. This duality of representing an array both as a contiguous block and as an aggregation of individual elements is fundamental to effectively leveraging arrays in C code.

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