Increment operator is not invoked at sizeof(++n) expression [duplicate]

Question

In C or C++, increment and decrement operator (++n, --n) are not performed when it is in a sizeof() operator.

int n = 100; int size_int = sizeof(++n); std::cout<<n; 

I have written this code and run the program. Of course, I think 101 will be showed for me. But, n was not 101, it was 100.

Why is that?

Answer 1

In C++, the expressions inside sizeof is not evaluated, except for C99's VLA ^{as mentioned in comments, since this was earlier tagged for C too}

The sizeof operator is calculated at compile time.

Only the type of the expression (that is calculated at compile time) is then used by sizeof.

From C++ Standard § 5.3.3 Sizeof

The sizeof operator yields the number of bytes in the object representation of its operand. The operand is either an expression, which is an unevaluated operand (Clause 5), or a parenthesized type-id.

^{Some C++ compilers provide VLAs as an extension as commented below.}

Answer 2

In C the operand of sizeof is not evaluated at except for variable length arrays:

6.5.3.4. p2:

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

If you put n++ into a variable length array, for example:

int n = 1 ;  sizeof( int(*)[n++] ) ; 

then it is unspecified if the operand is evaluated.

6.7.6.2. p5

If the size is an expression that is not an integer constant expression: if it occurs in a declaration at function prototype scope, it is treated as if it were replaced by *; otherwise, each time it is evaluated it shall have a value greater than zero. The size of each instance of a variable length array type does not change during its lifetime. Where a size expression is part of the operand of a sizeof operator and changing the value of the size expression would not affect the result of the operator, it is unspecified whether or not the size expression is evaluated.