struct args中的函数指针？

Question

foo is a struct with 5 scalar variables ( A , B , C , D , F ) and one array ( E ). What is confusing me is what f[0] , f[1] , and f[2] are in this context and what is happening here.

int     
bar(struct foo *f)
{
    f[1].C = f[0].B > f[2].C;
    f[0].E[-1] = f[0].D;
    f[0].A = f[1].C;
}

Are f[0] , f[1] , and f[2] individual structures with member variables? Can someone please explain? Thanks.

Answer 1

Are f[0] , f[1] , and f[2] individual structures with member variables?

Yes. f is a pointer to an array of struct foo instances f[0] is the first such member of that array, f[1] is the second member, etc. You might call it like this:

struct foo fArray[3];
// ... Initialize fArray[0], fArray[1], fArray[2] etc. ...
bar(fArray);

Answer 2

What you are doing is, you are passing a reference (pointer) to an array of struct foo to the function bar .

You must somewhere have a code that is similar to following:

struct foo  myFoos[10]; // an array with 10 elements of struct foo
struct foo *mallocedFoos;
// here goes some code to initialize the elements of the array
bar(&myFoos[0]);          // pass a reference to (address of/pointer to) the array

// or something like this is happening
mallocedFoos = malloc(sizeof(struct foo) * 10);
// here goes some code to initialize allocated memory
bar(mallocedFoos);        // pass the 'struct foo *' to the function

To understand the concept better, see this example.

Answer 3

Yes, in this context, f[0] , f[1] , etc., are elements of an array of type struct foo .

The more interesting thing to me is this line:

f[0].E[-1] = f[0].D;

I didn't realize negative indexes were allowed, but this question explains that array indexing is just pointer math, so it's an obfuscated way of saying:

f[0].D = f[0].D;

Which is basically useless as far as I know.

Also interesting:

f[0].C = f[0].B > f[2].C;

This would set f[0].C to a boolean, which is not usually compared with a > operator, so it's possible that different C members are used for different functions.

I feel that your confusion is warranted, given the strange nature of this function.

Answer 4

In this case f is an array of structures Similar to
struct node = { int A; //A to D and F are scalar variables int B; int C; int D; int E[10]; //E is an array of integers int F; } struct node = { int A; //A to D and F are scalar variables int B; int C; int D; int E[10]; //E is an array of integers int F; } struct node f[10]; //f is an array of structs struct node f[10]; //f is an array of structs

For more details you can also refer How do you make an array of structs in C?