C++: how to assign static array to a pointer inside a struct

Question

I have C-code that I need to compile to C++ and need to minimally change it.

In C, the following works

typedef struct _type_t
{
    int a;
    int b;
    int c[];
}type_t;

type_t var = {1,2,{1,2,3}};

But in C++11, it gives the error

error: too many initializers for int [0]

But I cannot give type_t.c a constant size because it needs to work for any size array.

So I need to change the struct to

typedef struct _type_t
    {
        int a;
        int b;
        int *c;
    }type_t;

But then I need to change

type_t var = {1,2,{1,2,3}};

to something else because current code gives the error

error: braces around scalar initializer for type int*

Casting 3rd element to(int[]) gives error

error: taking address of temporary array

This is from micropython, parse.c:

#define DEF_RULE_NC(rule, kind, ...) static const rule_t rule_##rule = { RULE_##rule, kind, #rule, { __VA_ARGS__ } };

How do I initialize the array and assign it to type_t.c ?

gcc warns with your C code: "warning: initialization of a flexible array member [-Wpedantic]" ( gcc -Wall -pedantic -std=c11 ) — user2672107
– user2672107, Commented Nov 2, 2017 at 13:15
You need to know the size of c to do this. For example, it would have worked if int c[]; was int c[3]; — DimChtz
– DimChtz, Commented Nov 2, 2017 at 13:17
@Adrian - No, it doesn't work in C. It works under your previous compiler, which offered an extension. The C language specification explicitly forbids it. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Nov 2, 2017 at 13:25
@StoryTeller I guess you're right: it's not allowed by C standard but it works in gcc. — Bob
– Bob, Commented Nov 2, 2017 at 13:31

Ron · Accepted Answer · 2017-11-02 13:15:26Z

7

Use std::vector and aggregate initialization:

struct type_t
{
    int a;
    int b;
    std::vector<int> c;
};

int main() {
    type_t var = { 1, 2, { 1, 2, 3 } };
}

edited Nov 2, 2017 at 13:15

answered Nov 2, 2017 at 13:14

Ron

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6 Comments

Bob Over a year ago

That's not an option. I need to modify it as little as possible i.e. not introduce use of stl.

Ron Over a year ago

Sure it's an option. It's the preferred option.

Bob Over a year ago

then I have to modify all the code that uses type_t

Algirdas Preidžius Over a year ago

@Adrian std::vector has operator[] overloaded, so you can, still, access the elements stored in a std::vector, as you would access array elements.

Zinki Over a year ago

@Adrian then you'll need to accept that C++ is not C and there are things you will have to do differently than you did in C.

|

user7860670 · Accepted Answer · 2017-11-02 13:18:24Z

6

In order to do this you need to make your array really static:

typedef struct _type_t
{
    int   a;
    int   b;
    int * c;
}type_t;

int items[3] = {1,2,3};
type_t var = {1,2, static_cast< int * >(items)};

answered Nov 2, 2017 at 13:18

user7860670

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6 Comments

Bob Over a year ago

Since this works, why isn't it possible to declare an array inside the {}?

StoryTeller - Unslander Monica Over a year ago

@Adrian - Because C++ doesn't allow. You can't force the language to be something it's not.

Jeffrey Over a year ago

This solution is much more in-line with the task of converting existing C code to C++ than the highest-voted one.

Jeffrey Over a year ago

@StoryTeller this leaves the memory footprint of the object indentical to what it was before. There's a chance that somewhere lurks an access to _type_t.c typcasted to int*. Or a memcpy of _type_t assuming a specific size. Or complexity in addind std::vector. I've converted legacy codebase before and typically, the less of change the better. Even if you end-up up with hackish stuff like that.

StoryTeller - Unslander Monica Over a year ago

@Jeffrey - To be perfectly honest, the way I see this, the proper conversion is to keep compiling this as C with extensions. Just wrap it tightly and snugly in a standard compliant hazmat suit of code, and leave it be.

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DeepakKg · Accepted Answer · 2017-11-02 13:37:38Z

1

C++ doesn't allow this statement expression. You may use lightweight std::initialiser_list to achieve the objective

#include <initializer_list>
typedef struct _type_t
{
    int a;
    int b;
    std::initializer_list<int> c;
}type_t;

 type_t var = {1,2,{1,2,3,4}};

answered Nov 2, 2017 at 13:37

DeepakKg

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5 Comments

StoryTeller - Unslander Monica Over a year ago

initializer_list doesn't offer operator[]. The OP won't be able to use it as a drop in replacement.

DeepakKg Over a year ago

Story Teller : Yes it doesn't provide subscript and one has to use the iterator to traverse through it. But its very lightweight as compared to vector and depending upon one's requirement they may choose to use it

StoryTeller - Unslander Monica Over a year ago

It's too lightweight I'm afraid. For one there's object lifetime considerations to think about. Is the member c going to prolong the life of the temporary array created from {1, 2, 3, 4}? I don't think it will. Since the OP also wants a drop in replacement, it doesn't go towards helping here.

Massimiliano Janes Over a year ago

it's also not lightweight enough, there's no guarantee that such a type_t remains POD; so even if it were correct, you'd gain very little over std::vector

DeepakKg Over a year ago

I guess the requirements of the function / program should decided what to choose. Its an option, may not be the best for every situation, but memory and complexity wise its definitely better than the vector because there is no requirement of growing it dynamically (As per the code above)

Massimiliano Janes · Accepted Answer · 2017-11-02 14:13:30Z

0

yet another alternative might be

    ...
    int * c;
}type_t;

type_t var = {1,2, []{ static int x[] = {1,2,3}; return x; }() };

PS: yes, I know it's crazy, but it's somewhat more respectful of OP requirements ( it can be incorporated in a dropin replacement macro as given )

edited Nov 2, 2017 at 14:13

answered Nov 2, 2017 at 13:52

Massimiliano Janes

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