adress of operator on array

[joseplusplus]

Good day everyone !

I believe I'm confused with the use of the address-of operator in two distinct cases, Let me clarify:

int num = 10;
int *p = #

In the code above I must use the address-of operator to get the address of num and assign it to p, Pretty Logical to me.

But in this case:

int x[10]; 
int *p = x;

The address-of operator in not required, moreover it is NOT allowed by the compiler. Could somebody please explain the difference to me?, especially why the & operator is not needed in the last case

Thank very much,

Jose

[Disch]

Arrays can be implicitly cast to a pointer to the first element of the array. It's allowed for convenience:

int x[10];

int* p = x;  // <- this is just short-hand for...
int* p = &x[0];  // <- this 

You can use the & operator on array names, but the type is different. You get a pointer to the array itself, not to the first element in the array.

This is easier to visualize by typedefing the array:

typedef int int5[5];  // make a new type 'int5'

int foo[5];  // <- now this...
int5 foo;  // <- is the same as this

int* bad = &foo;  // this fails because 'foo' is an int5, not an int
int5* good = &foo;  // this works 

[naivnomore]

In your second case, you do not need the address of operator because the array name automatically decays to a pointer to the first element of your array. In other words, your pointer p is pointing to x[0].

However, if you use the address of operator with the array name, it will give you the address of the array containing the 10 integer elements and NOT to the first element of the array. Therefore, you cannot assign that address to a pointer to a single integer element. This is why your compiler is throwing an error. The correct assignment in this case would be as follows.

int (*p)[10] = &x;

However, in both cases, p will display the same address. You will see the difference only when you perform pointer arithmetic. For instance, when you increment p, p will increment by an element in the first case where as it will increment by 10 elements in the second case.

[joseplusplus]

Thank you guys,

your explanations have solved my doubts. I kind of had the idea, but it's hard to cram so much information as I'm doing this days in an attempt to learn the language as fast as I can.

Keep on the good job,

Best Regards,

Jose