why does *ptr++ act like *(ptr++) and not (*ptr)++? [duplicate]

Answer 1

From wikipedia:

For the ISO C 1999 standard, section 6.5.6 note 71 states that the C grammar provided by the specification defines the precedence of the C operators

This means that the operator precedence is governed by C standard.

Answer 2

The differences seem confusing. Is this, perhaps, compiler dependent?

No, the precedence of operators is defined in the c lang spec. And and so *prt++ is always deferencing the pointer before the post-increment occurs.

Answer 3

It has to do with operator precedence. The * operator has a lower precedence than ++ so it occurs last.

Here's a Wikipedia chart that lists all the operators: http://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B#Operator_precedence

You can see in the chart that postfix ++ has a precedence of 2 while * dereference has a precedence of 3. (The numbers are slightly backwards, as lower numbers have higher precedence).

Answer 4

Because of precedence (that's just how C works).

C FAQ on the * exact * subject

The postfix ++ and -- operators essentially have higher precedence than the prefix unary operators. Therefore, *p++ is equivalent to *(p++);

Answer 5

Operator precedence. The ++ operator "binds more tightly" than the * operator.

Here's the table, in order of precedence. http://isthe.com/chongo/tech/comp/c/c-precedence.html

This is not compiler dependent. It will always behave this way.

Answer 6

because of operator precedence

the postfix ++ has a higher precedence than the * operator. It's not compiler dependent.

*ptr += 1 will increase the value pointed to by ptr by one (or call the appropriate overloaded operator) *ptr % 8 will calculate the remainder of the value pointed to by ptr divided by 8 *ptr / 8 will calculate the division of the value pointed to by ptr and 8