


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 <p>The output you observe is:</p><pre class="brush:php;toolbar:false">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 .
The above is the detailed content of How Do Array-to-Pointer Conversion and the Address-of Operator (&) Affect Array Addresses in C ?. For more information, please follow other related articles on the PHP Chinese website!

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