How to deconstruct complex C/C++ statements/declarations?

April 30, 2023

Take the following snippet as an example.

char* const (*(* const bar)[5])(int)

Cannot seem to make sense of it or more so, cannot identify the initial point from where to begin making sense of it.

>Solution :

Let’s consider the declaration step by steap

char* const (*(* const bar)[5])(int);

This part

* const bar

declares a constant pointer with the name bar that points to to an array of 5 elements of a pointer type

*(* const bar)[5]

that is pointer to the function type

char* const (int)

Here is a demonstration program.

#include <stdio.h>

char * const f1( int n )
{
    printf( "f1 called with n = %d\n", n );

    return NULL;
}

char *const f2( int n )
{
    printf( "f2 called with n = %d\n", n );

    return NULL;
}

char *const f3( int n )
{
    printf( "f3 called with n = %d\n", n );

    return NULL;
}

char *const f4( int n )
{
    printf( "f4 called with n = %d\n", n );

    return NULL;
}

char *const f5( int n )
{
    printf( "f5 called with n = %d\n", n );

    return NULL;
}

int main( void )
{
    char *const ( *a[5] )( int ) = { f1, f2, f3, f4, f5 };

    for (int i = 0; i < 5; i++)
    {
        a[i]( i );
    }

    putchar( '\n' );

    char *const ( *( *const bar )[5] )( int ) = &a;

    for (int i = 0; i < 5; i++)
    {
        ( *bar )[i]( i );
    }

}

The program output is

f1 called with n = 0
f2 called with n = 1
f3 called with n = 2
f4 called with n = 3
f5 called with n = 4

f1 called with n = 0
f2 called with n = 1
f3 called with n = 2
f4 called with n = 3
f5 called with n = 4

Using typedef declarations could make the declaration of the pointer much more readable.