c++ Initializing a struct with an array as a member

Drew Shafer picture Drew Shafer · Apr 16, 2010 · Viewed 75.4k times · Source

Edited again because it originally wasn't clear that I'm trying to initialize the arrays at compile time, not at run time...


I've got the following reduced testcase:

typedef struct TestStruct
{
    int length;
    int values[];
};

TestStruct t = {3, {0, 1, 2}};
TestStruct t2 = {4, {0, 1, 2, 3}};

int main()
{
    return(0);
}

This works with Visual C++, but doesn't compile with g++ under linux. Can anyone help me make this specific kind of initializer portable?

Additional details: the actual structure I'm working with has several other int values, and the array can range in length from a single entry to over 1800 entries.

EDIT: I think (but am not sure) that this is not a VLA issue. To clarify, I'm trying to get the compiler to do the work for me at compile-time. The length of the array at run-time is constant. Apologies if I'm wrong; I'm primarily a c#/Perl/Ruby programmer who is stuck maintaining this legacy app...

Any help much appreciated. Thanks!

Answer

Evan Teran picture Evan Teran · Apr 16, 2010

c++ doesn't have the same flexible array member as last element as c99. You should use a std::vector if you don't know how many elements or you should specify how many if you do.

EDIT: You have said in your edit that the array is a runtime constant, so specify the size and it should work fine. g++ has no problem with the following code:

struct TestStruct { // note typedef is not needed */
    int length;
    int values[3]; // specified the size
};

TestStruct t = {3, {0, 1, 2}};

int main() {
    // main implicitly returns 0 if none specified
}

EDIT: to address your comment, you could use templates like this:

template <int N>
struct TestStruct {
    int length;
    int values[N];
};

TestStruct<3> t3 = {3, {0, 1, 2}};
TestStruct<2> t2 = {2, {0, 1}};

int main() {}

The only problem is that there is no easy way to put both t2 and t3 in a container (like a list/vector/stack/queue/etc because they have different sizes. If you want that, you should use std::vector. Also, if you are doing that, then it isn't necessary to store the size (it is associated with the type). So you could do this instead:

template <int N>
struct TestStruct {
    static const int length = N;
    int values[N];
};

TestStruct<3> t3 = {{0, 1, 2}};
TestStruct<2> t2 = {{0, 1}};

int main() {}

But once again, you cannot put t2 and t3 in a "collection" together easily.

EDIT: All in all, it sounds like you (unless you store more data than just some numbers and the size) don't need a struct at all, and can't just use a plain old vector.

typedef std::vector<int> TestStruct;


int t2_init[] = { 0, 1, 2 };
TestStruct t3(t3_init, t3_init + 3);

int t2_init[] = { 0, 1 };
TestStruct t2(t2_init, t2_init + 2);

int main() {}

Which would allow you to have both t2 and t3 in a collection together. Unfortunately std::vector doesn't (yet) have array style initializer syntax, so i've used a shortcut. But it's simple enough to write a function to populate the vectors in a nice fashion.

EDIT: OK, so you don't need a collection, but you need to pass it to a function, you can use templates for that to preserve type safety!

template <int N>
struct TestStruct {
    static const int length = N;
    int values[N];
};

TestStruct<3> t3 = {{0, 1, 2}};
TestStruct<2> t2 = {{0, 1}};

template <int N>
void func(const TestStruct<N> &ts) { /* you could make it non-const if you need it to modify the ts */
    for(int i = 0; i < N; ++i) { /* we could also use ts.length instead of N here */
        std::cout << ts.values[i] << std::endl;
    }
}

// this will work too...
template <class T>
void func2(const T &ts) { 
    for(int i = 0; i < ts.length; ++i) {
        std::cout << ts.values[i] << std::endl;
    }
}

int main() {
    func(t2);
    func(t3);
    func2(t2);
}