Why does sizeof(x)++ compile? [duplicate]

Question

I have run into the following code snippet:

int a = 3; printf("%d", sizeof(a)++); 

Apparently this will compile with GCC 9.3.0 and -std=c99. While this does not compile:

printf("%d", sizeof(3)++); 

GCC prints an error

error: lvalue required as increment operand

Before I have compiled the first snippet I would have expected such an error.

The operand of the postfix ++ operator shall be an lvalue as of C99 standard

The operand of the postfix increment or decrement operator shall have qualified or unqualified real or pointer type and shall be a modifiable lvalue.

About the return value of the sizeof Operator (as expected):

The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

The value of the result is implementation-defined, and its type (an unsigned integer type) is size_t, defined in (and other headers).

How is it possible the sizeof(a)++ compiles? Is this undefined behavior or am I missing something?

Answer 1

The sizeof is an operator, not a function, and as any operator it has a precedence, which is in C lower than of the ++ operator. Therefore the construct sizeof(a)++ is equivalent to sizeof a++ which is in turn equivalent to sizeof (a++). Here we have postincrement on a which is an lvalue so it is perfectly legal. If instead of a you have 3 which is not lvalue, the compilation will fail.