Why does sizeof(x)++ compile?

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?

Maybe the parentheses are confusing you. You can also write it like this: `sizeof a++`. — Fiddling Bits, May 07 '20 at 17:18
It's the same as `sizeof (a++)`. `sizeof` is an operator and not a function. And it is lower precedence than `++` — Eugene Sh., May 07 '20 at 17:20
Essentially the same as https://stackoverflow.com/questions/46653406/why-does-sizeofmy-arr0-compile-and-equal-sizeofmy-arr0 — Guy Incognito, May 07 '20 at 17:28

score 45 · Accepted Answer · answered May 07 '20 at 17:28

45

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.

answered May 07 '20 at 17:28

Eugene Sh.

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I swear... I lived a happier life before visiting this answer! – Cort Ammon May 08 '20 at 02:47

Why does sizeof(x)++ compile?

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