How Do Array-to-Pointer and Pointer-to-Array Conversions Affect Array Addressing in C/C ?

Pointers and Array Conversions

When working with arrays in C/C , understanding the relationship between array and pointer addressing is crucial. Consider the following code:

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

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

Output:

0045FB88 0045FB88
0045FB88 0045FB7C

Understanding the Behavior

The first line of output makes sense: u is a pointer to t, so both u and &u represent the address of the same memory location.

However, the second line raises questions: Why are t and &t[0] (or &t) equivalent? What does &t represent?

Array-to-Pointer Conversion

When t is used standalone in an expression, an implicit array-to-pointer conversion occurs. This conversion produces a pointer to the first element of the array, in this case t[0]. So, in the first line of output, t effectively becomes &t[0], which is why both t and &t[0] have the same value.

Pointer-to-Array Conversion

In contrast, when &t is directly used, no such conversion happens. Instead, &t explicitly calculates the address of t, which is a pointer to the entire array.

Memory Address

The first element of an array and the start of the array occupy the same memory location. Consequently, the pointers t (which is equivalent to &t[0] due to array-to-pointer conversion) and &t (which points to the start of the array) have the same value.

Conclusion

Understanding the distinction between array-to-pointer conversion and pointer-to-array conversion is key to comprehending the different ways of addressing arrays and pointers in C/C . This understanding is essential for efficient memory management and data manipulation.

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