A simple question: is variably-modified type a VLA (variable length array) only?

C11, 6.10.8.3 Conditional feature macros, 1 (emphasis added):

__STDC_NO_VLA__ The integer constant 1, intended to indicate that the implementation does not support variable length arrays or variably modified types.

Does it mean that there is a variably-modified type, other than VLA? Any examples?

What is the relationship between "variably modified type" and "variable length array"?

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Extra: the definition of "variable length array" depends on the definition of "known constant size":

If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type; otherwise, the array type is a variable length array type.

However, the definition of "known constant size" depends on the definition of "variable length array":

A type has known constant size if the type is not incomplete and is not a variable length array type.

A bit confused.

Related DR: http://www.open-std.org/jtc1/sc22/wg14/www/docs/dr_312.htm.

CodePudding user response：

Does it mean that there is a variably-modified type, other than VLA? Any examples?

According to the C Standard (6.7.6 Declarators)

3 A full declarator is a declarator that is not part of another declarator. If, in the nested sequence of declarators in a full declarator, there is a declarator specifying a variable length array type, the type specified by the full declarator is said to be variably modified. Furthermore, any type derived by declarator type derivation from a variably modified type is itself variably modified.

Here is a demonstration program

#include <stdio.h>

int main( void )
{
    for ( size_t n = 1; n < 10; n   )
    {
        typedef int ( *Ptr )[n];
        Ptr p;

        printf( "sizeof( *p ) = %zu\n", sizeof( *p ) );
    }
}

The program output is

sizeof( *p ) = 4
sizeof( *p ) = 8
sizeof( *p ) = 12
sizeof( *p ) = 16
sizeof( *p ) = 20
sizeof( *p ) = 24
sizeof( *p ) = 28
sizeof( *p ) = 32
sizeof( *p ) = 36

In this program the pointer type Ptr defined like int( * )[n] is a variably modified type.

This quote

A type has known constant size if the type is not incomplete and is not a variable length array type.

means that the sizeof operator for such types is evaluated at compile-time opposite to the evaluation at run-time for variable length array types and the size of such a type is not changed during the program execution.

The quote has a different meaning relative to the quote below

If the size is an integer constant expression and the element type has a known constant size, the array type is not a variable length array type; otherwise, the array type is a variable length array type.

that says about how distinguish a declaration of a variable length array from a declaration of a non-variable length array.

CodePudding user response：

C 2018 6.7.6 3 says:

If, in the nested sequence of declarators in a full declarator, there is a declarator specifying a variable length array type, the type specified by the full declarator is said to be variably modified. Furthermore, any type derived by declarator type derivation from a variably modified type is itself variably modified.

Therefore int (*)[n], for some non-constant n, is a variably modified type even though it is a pointer. Further, int [3][n] is a variably modified type.