Defining pointers to array of structs

Question

I came across this code:

typedef struct {
    char *name;
    char *value;
} SPAM;

typedef struct {
    char *name;
    int num_spams;
    SPAM *spams;
} EGG;

SPAM my_spams[2] = {
    { "name1", "value1" },
    { "name2", "value2" },
};

EGG my_eggs[1] = {
    { "first egg", 1, my_spams }
};

EXPORT(int) getSPAMANDEGGS(EGG **eggs)
{
    *eggs = my_eggs;
    return 1;
}

In this declaration shouldn't the definition of EGG struct have SPAM **spams; as the definition of the spams member since we store an array of SPAMs there later?

Arrays decay to pointers. So SPAM *spams can point to the first element of an array of SPAM. — Barmar
– Barmar, Commented May 10, 2018 at 0:42
Just like you can write char foo[] = "abc"; char *foo_ptr = foo; — Barmar
– Barmar, Commented May 10, 2018 at 0:43
{ "first egg", 1, my_spams } should be { "first egg", 2, my_spams }, because num_spams should correspond to the number of elements in my_spams array, which is 2. — Sergey Kalinichenko
– Sergey Kalinichenko, Commented May 10, 2018 at 0:48

Sergey Kalinichenko · Accepted Answer · 2018-05-10 00:46:01Z

1

No, it should not. An array of SPAM values is expressed through a single pointer, so SPAM *spams is correct.

The reason the parameter of the getSPAMANDEGGS function needs two asterisks is that it sets a pointer passed to it from the outside:

EGG *eggArray; // First asterisk is due to *
int eggArraySize = getSPAMANDEGGS(&eggArray); // Second asterisk is due to &

Without the & operator getSPAMANDEGGS would be unable to set a new value to eggArray pointer.

answered May 10, 2018 at 0:46

Sergey Kalinichenko

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