a = &b vs *a = &b — pointer assignment

Question

I have a pointer and a variable:

int *a;
int b;

Is there any difference between assignments

a = &b;

and

*a = &b;

and what are they called (like pointer declaration or something) ?

Answer 1

Types matter.

• In case of a=&b, the assignment is valid. You're assigning the address of an integer (type: int *), to another variable of type int *, so this is legit.

• In case of *a=&b, this is a constraint violation (for assignement operator, see chapter §6.5.16.1/p1, Constraints, for Simple assignment) and thus not a valid C syntax, thus not required to be compiled by any conforming compiler. To make it a valid C syntax, we need to enforce a typecast, something like

  *a= (int) &b;
  

  would make it a syntactically valid C statement which meets the required constraint.

  Even, after that, the result is implementation defined.^#note Here, you're basically trying to assign the the address of an integer (type: int *) to another variable of type int (*a is of type int). Conversion from a pointer to integer is implementation defined behaviour.

  Quoting C11, chapter §6.3.2.3, Pointers

  Any pointer type may be converted to an integer type. Except as previously specified, the result is implementation-defined. If the result cannot be represented in the integer type, the behavior is undefined. [....]

[....] And what are they called?

They both are assignment statements.

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Note:

Considering a points to a valid memory location already. Otherewise, dererefencing an invalid pointer invokes undefined behavior on it's own.

Answer 2

Pay attention to types on the left and on the right of =.

&b is int *, a is also int * but *a is int.

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It's a bit confusing that * has different meanings:

int *a; — here * means that a will be a pointer;

*a = ...; — here * means that we change not address stored in a but value which is located at the address.

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So a = &b means "write the address of b to a",

but *a = &b means "write the address of b to *a, i.e. to the address which is stored in a".

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Let's suggest that we have this situation:

   Address  Value
 a 0x0001   0x0004
 b 0x0002   70
   0x0003   80
   0x0004   90

At the moment a is 0x0004 and *a is 90.

If you do a = &b, a will be 0x0002 and *a will be 70.

But if you do *a = &b, a will not change, but *a, i.e. value at the address 0x0004, will change to 0x0002.