C ++：如何将静态数组赋给结构内的指针

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 ?

Answer 1

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 } };
}

Answer 2

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)};

Answer 3

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}};

Answer 4

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 )

Answer 5

In C++ arrays are stored in stack memory so you must have to specify array size on the time of declaring it. If you want to declare array with arbitrary size use vector or do this

struct type_t
{int a;
 int b;
 int *c = new int[size];
}var{1,2,{1.,..}};

Here size is any arbitrary integer and {1.,..} are values of elements of dynamic array seperated by comma.