CODEWITHSUNDEEP
cpointers
Expert Novice
Expert Novice

Reputation: 1963

Why does this declaration not work

In int (*x)[10]; x is a pointer to an array of 10 ints

So why does this code not compile:

int arr[3] ;

int (*p)[3] =arr;

But this works:

int  arr[3];

int (*p)[3] =&arr;

Upvotes: 5

Views: 117

Answers (2)

Michael Burr
Michael Burr

Reputation: 340178

arr is an expression that evaluates to an int* (this is the famous 'arrays decay to pointer' feature).

&arr is an expression that evaluates to a int (*)[3].

Array names 'decay' to pointers to the first element of the array in all expressions except when they are operands to the sizeof or & operators. For those two operations, array names retain their 'arrayness' (C99 6.3.2.1/3 "Lvalues, arrays, and function designators").

Upvotes: 10

Maxim Egorushkin
Maxim Egorushkin

Reputation: 136208

It doesn't work for exactly the same reason as:

int i;
int* pi = i; // error: no conversion from int to int*

Upvotes: 0

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