Question

Look at the strange line from function f:

typedef char thing[1];

void f(thing t){
    thing *p;

    /* Strange line. Changing t to  &t is wrong. 
       C++ require a (thing *) cast.
     */
    p = t;

    *(*p) = 'C';
}

int main(int argc, char* argv[])
{
    thing g;
    thing *h;

    /* Works as expected. */
    h = &g;  

    g[0] = 'A';
    *(*h) = 'B';
    f(g);

    /* g[0] is now  'C' */

    return 0;
}

Changing f to

void f(thing t){
    char **p; 
    p = &t; /* note the & */
    *(*p) = 'C';
}

also works.

I was expecting p=&t to be the right instruction, not p=t.
After all t is a thing and &t must be a thing*.

What happened in that strange line?

Answer 1

Just an alternative solution:

You can prevent the array-to-pointer decay by passing the array by reference. This should work as you want:

void f(thing& t){
    thing *p;

    p=&t;

    *(*p)='C';
}