Thread: Passing a struct

1. You don't need the [1] when declaring your name field of the struct.
2. You would be better off declaring an array of team structs perhaps called teams to work with.
3. Your initialization of your team structs is a bit off.

Code:

const int NUM_TEAMS = 12;

typedef struct
{
    char name[20];
    int p, w, d, l, f, a, t;
} team;

...

// Example initialization...
team teams[NUM_TEAMS] = { {"Packers",1,0,0,0,0,0,0}, {"Steelers",2,0,0,0,0,0,0},
                          ... etc ...
                          {"Ravens",11,0,0,0,0,0,0}, {"Cowboys",12,0,0,0,0,0,0} };

Your function that modifies this array should accept a pointer to type team:

Code:

// This sample loops through all the teams and does something
void func(team* teams)
{
    int loop;
    for( loop = 0; loop < NUM_TEAMS; ++loop )
    {
        // Inside this loop access the individual teams data by using teams[loop].whatever
        // where whatever is the data member you want to check/access/modify
    }
}

You call the function like so:

Code:

func(teams);

Originally Posted by Drainy

How do I declare a prototype a function like that?

Code:

void func(team*);

Originally Posted by Drainy

and

Code:

void enterteams(team* teams)

Just to verify I am understanding pointers correctly, that is basically saying that I am passing in the address to a team struct definition inside of the teams array?

Is that supposed to be a prototype or a part of the actual function definition? It merely states that teams is a pointer to a memory location that could contain one or more team structs.

When you actually call the function like so:

Code:

team teams[NUM_TEAMS];
...
enterteams(teams);

That is saying that we are passing the address (a pointer) to the first element (a team struct) of the array teams to the function. Array names used in this manner are taken to mean the address of the first element.