The first quote from the C Standard is incorrect. It sounds like
— An adjusted parameter type in a function definition is not a
complete object type (6.9.1)
That is you omitted the word complete.
For example in a function declaration that is not at the same type its definition you may specify an incomplete object type like
void f( size_t, size_t, int [][*] );
In this function declaration the declaration of the third parameter is not a complete object type because the size of the array elements is unknown.
Here is a demonstrative program
#include <stdio.h>
void f( size_t, size_t, int [][*] );
void f( size_t m, size_t n, int a[][n] )
{
for ( size_t i = 0; i < m; i++ )
{
for ( size_t j = 0; j < n; j++ )
{
a[i][j] = n * i + j;
}
}
}
void g( size_t, size_t, int [][*] );
void g( size_t m, size_t n, int a[][n] )
{
for ( size_t i = 0; i < m; i++ )
{
for ( size_t j = 0; j < n; j++ )
{
printf( "%d ", a[i][j] );
}
putchar( '\n' );
}
}
int main(void)
{
size_t m = 2, n = 3;
int a[m][n];
f( m, n, a );
g( m, n, a );
return 0;
}
Its output is
0 1 2
3 4 5
Here in the program these two function declarations
void f( size_t, size_t, int [][*] );
and
void g( size_t, size_t, int [][*] );
have a parameter declaration with an incomplete object type.
You may not use such a declaration that at the same type is its definition like for example
void f( size_t m, size_t n, int a[][*] )
{
// ...
}
Because the compiler is unable to determine the pointer type after adjusting the third parameter to pointer. That is the pointer will have an incomplete object type int ( * )[].
C99on certain work for the reason it has been proved fit for the purpose by years of development in a highly regulated industry.) Because this question is careful to point out it is specific toC99, it is completely viable, and it makes perfect sense in that context. – Accusatory