the meaning of '&' in C when using with array

Question

I am learning pointer right now, and '&' operator is messing my mind in terms of datatype, especially when it's used with array.

I see that '&' operator is giving me the first memory address of whatever it is used with.

But I can't understand why this is changing the datatype when I use it with array's name. Is this just the way it is? Or are there any reasons that I don't know yet?

so, my question is,

why do the datatype of array1 and array 2 is different even though they are just the name of array?
why does the datatype changes just because I added '&'?

int array1[5];

int *p1 = array1;    //why does this works
int *p2 = &array1;   //but not this one?

the gcc says that one on the top right is 'int*' but one with '&' is 'int(*)[5]'.

int array2[5][5];

int (*q1)[5] = array2;    //why does this wokrs
int (*q2)[5] = &array2;   //but not this one?

the gcc says that one on the top right is 'int( * )[5]' but one with '&' is 'int( * )[5][5]'.

Does this answer your question? Why can't I treat an array like a pointer in C? — gspr
– gspr, Commented May 15, 2021 at 10:52
The unary & operator gives the address of its operand. The data type of the result is pointer to whatever type that operand has. So if you have int a, then &a has type int *. If you have int b[10] then &b has type int (*)[10]. — Tom Karzes
– Tom Karzes, Commented May 15, 2021 at 10:53
Actually once you're creating an array, the first member of this array becomes a pointer which points to the array1in your example int array1[5];. So if you do this int *p2 = &array1;you will be taking an address of a pointer since array1is actually a pointer.. — Ali Ziya ÇEVİK
– Ali Ziya ÇEVİK, Commented May 15, 2021 at 10:59

Greg · Accepted Answer · 2021-05-15 11:02:52Z

1

In C, array name gets converted to a pointer to the first element of the array. So you do not need to use '&' to get the address. When you do this, what you are getting is a pointer to an array.

Comments

hsuecu · Accepted Answer · 2021-05-15 10:59:56Z

0

Pointers in c have very deep implementation. Pointers , as name suggest pointes to a data location. Pointers in c have a datatype which tells about the data type of data.

int * ptr;

this is pointer daclaration.

int i = 30; ptr = &i ; // this returns address of variable t

in terms of array both arr, &arr return address of array

int (*p)[5] = &arr or arr , both works , otherwise send the error message with question.

answered May 15, 2021 at 10:59

hsuecu

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