CODEWITHSUNDEEP
c++pointers
dragosb
dragosb

Reputation: 627

Pointer assignment different behaviour

Hi I tried the following code: 

int *a=&matrice[4][0];
cout<<*a<<"\n\n";

int *b;
b = &matrice[4,4];

Where matrice is defined like this: int matrice[5][5]; and I filled it with some normal int values.

The first assignment works but the second doesn't with the error :

a value of type "int (*)[5]" cannot be assigned to an entity of type "int * " .

I expected both of them to give me an error at compile time.

Could you explain why there is this different behavior ?

Upvotes: 2

Views: 155

Answers (3)

haccks
haccks

Reputation: 105992

&matrice[4][0] is of type int * and you are assigning it to a which is also of type int *. Nothing wrong in this. You will not get any type of error.

matrice[4,4] is not representing a 2D matrix rather it evaluates to matrice[4] because of the effect of the , operator and &matrice[4]is of type int(*)[5]. Try this; 

 b = &matrice[4][4];

Upvotes: 2

unxnut
unxnut

Reputation: 8839

In the first case, you are assigning a pointer address to a. matrice is internally defined as a pointer to an array of pointers and you are getting the address of one such vector into a.

The second one is using the comma operator inside the index reference. That is incorrect as well.

Upvotes: 0

Delan Azabani
Delan Azabani

Reputation: 81384

The expression

&matrice[4,4]

does not do what you have expected, because the expression

4,4

evaluates to 4. This is an example of the comma operator, which evaluates both operands and returns the second operand.

Upvotes: 7

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