CODEWITHSUNDEEP
csizeof
onezero
onezero

Reputation: 23

What does sizeof (int) * p semantically mean?

What does sizeof (int) * p semantically mean? Is it:
1. sizeof( (int) *p )
or
2. ( sizeof(int) ) * p
and what rule makes the expression to be evaluated this way?

Upvotes: 2

Views: 1619

Answers (2)

John Bode
John Bode

Reputation: 123458

sizeof is a unary operator, which has a higher precedence than binary *, so the expression sizeof (int) * p is parsed as (sizeof (int)) * p. Here's the parse tree:

             *
           /   \
       sizeof   p
         |
       (int)

Edit

From onezero's comment:

but can't the expression be evaluated like sizeof( (int) *p ) ,as sizeof operator, type-cast operator and *(dereference) operator have same precedence and associates from right to left?

Here's the relevant syntax (from the ballot draft of the C2011 standard):

(6.5.3) unary-expression:
    postfix-expression
    ++ unary-expression
    -- unary-expression
    unary-operator cast-expression
    sizeof unary-expression
    sizeof ( type-name )
    alignof ( type-name )

As you can see, the (int) is interpreted as part of the sizeof expression, full stop. It is not interpreted as part of a cast-expression. The syntax doesn't allow sizeof to be followed directly by a cast-expression; there must be a unary-operator (one of &, *, +, -, ~, !) between them.

Upvotes: 4

SpacedMonkey
SpacedMonkey

Reputation: 2773

Doesn't it depend what p is?

void main()
{
   int p = 3;

   printf("%d", sizeof(int) *p);
}

Will output 12.

void main()
{
   char *p = "a";

   printf("%d", sizeof((int) *p));
}

Will output 4.

Upvotes: -1

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