What's the differences between `int[1]` and `int()[1]`?

Question

using T1 = int*[1];
using T2 = int(*)[1];

T1 t1;
T2 t2;

t1[0] = 0;   // ok
t2[0] = 0;   // error : array type 'int [1]' is not assignable
t2    = t1;  // error : array type 'int [1]' is not assignable
t2    = &t1; // error : array type 'int [1]' is not assignable
t2    = 0;   // ok

Why is t2[0](/t1) not assignable?

What's the differences between int*[1] and int(*)[1]?

Update:

int n[1];
t2 = &n; // ok

score 2 · Accepted Answer · answered Mar 17 '18 at 10:37

2

int*[1] is an array of length 1, whose element is int*.

int(*)[1] is a pointer, which points to array int[1]. Thus t2[0] is an array int[1], which is not assignable.

answered Mar 17 '18 at 10:37

llllllllll

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@szxwpmj It's obvious, what do you expect ? – llllllllll Mar 17 '18 at 10:46
@szxwpmj why would it be? `t1` is array of pointers to ints not array of ints – Killzone Kid Mar 17 '18 at 10:46

What's the differences between `int*[1]` and `int(*)[1]`?

1 Answers1

What's the differences between `int[1]` and `int()[1]`?