Syntax difference between pointer to int array and pointer to int

Question

Let us consider the below code:

int y = 20;
int *pointer_of_y = y; //Intentionally missed &
printf("%p\n",pointer_of_y);
printf("%p\n",&y);
printf("%d\n",*pointer_of_y);

Output:

0x14 //Value of 20 in hex
0x7fff56747af8 //Address of Y Variable
Segmentation fault: 11 //Value at pointer: Due to missing &

The above code does not execute because I have missed & in pointer initialisation.

Let us consider the same code if it is initialised as array.

int x[]= {10,20,30,40};
int *pointer_of_x = x;
printf("%p\n",pointer_of_x);
printf("%p\n",&x);
printf("%d\n",*pointer_of_x);

The output :

0x7fff5b5eaad0  //Pointer value
0x7fff5b5eaad0  //Address of x
10 //Value at address pointed

How does the second example work even though I have missed the & in the pointer initialisation. Why does being a array make a difference?

ameyCU · Accepted Answer · 2015-09-20 13:40:19Z

4

When the name of an array is used in an expression context other than as the operand to sizeof or & it decays to a pointer to its first element.

So here -

int x[]= {10,20,30,40};

x will give address of first element of array and has type int *. Therefore this works -

int *pointer_of_x = x;   //works

answered Sep 20, 2015 at 13:40

ameyCU

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5 Comments

alk Over a year ago

I'd remove "the name of", as the name is not relevant in this context.

user2371524 Over a year ago

@alk I think it is relevant, but more accurate would be the identifier of

alk Over a year ago

@FelixPalmen: In C names or identifiers are for the parser only. "Expression evaluating to an array" might do.

user2371524 Over a year ago

@alk so? doesn't matter at the syntax level, and that's what was asked.

alk Over a year ago

@FelixPalmen: The question is not about syntax (syntax error: int i = 1; int * pi := i;). It is about semantics (semantical "error": int i = 1; int * pi = i;)

alk · Accepted Answer · 2015-09-20 13:39:51Z

2

If an array of type T is assigned to a pointer of type T* then the array decays to the address of its 1^st element.

answered Sep 20, 2015 at 13:39

alk

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