Dynamic string array allocation error

Question

I am trying to allocate a dynamic string array in the following way and I get an error:

struct Test
{
    std::string producer[];
    std::string golden[];
};

Test test1 =
{
        {"producer1", "producer2"} ,
        {"golden1"  , "golden2"}
};

The error i get is that there are too many initializers for std::string[0], but if I get off the array part it works:

struct Test
{
    std::string producer;
    std::string golden;
};

Test test1 =
{
        "producer1" ,
        "golden1"
};

Thanks in advance!

std::string producer[]; You cannot declare an array without saying how many elements is should contain. — Bo Persson
– Bo Persson, Commented Apr 2, 2012 at 7:24
@Bo you can, but you cannot define it, you have to do it dynamically. — oh whatever
– oh whatever, Commented Apr 2, 2012 at 7:25
@littleadv Yes, but a non-static member declaration is a definition, and requires the size. And you can't dynamically allocate any thing that has been declared. To dynamically allocate a string[], you'd have to define the object as a string*, not a string[]. The only places where and actual string[] would be legal is as an extern, or as a reference parameter to a function (but that's a bit tricky, since you cannot then pass the function an array whose size is known). — James Kanze
– James Kanze, Commented Apr 2, 2012 at 7:52
@James, i won't explain it in the comment, but you're wrong. Note that he's defining a type here. — oh whatever
– oh whatever, Commented Apr 2, 2012 at 8:00
@littleadv Exactly. He's defining a class, and non-static data member declarations in a class are defintions, and require the size, see §9.2/10 ("Non-static data members shall not have incomplete types.") — James Kanze
– James Kanze, Commented Apr 2, 2012 at 8:26

Community · Accepted Answer · 2017-05-23 10:34:38Z

3

You cannot initialize zero-sized arrays this way, because you have to dynamically allocate the memory. You can only do what you do if you specify the dimensions in the type definition.

See my answer to a similar question here.

edited May 23, 2017 at 10:34

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answered Apr 2, 2012 at 7:23

oh whatever

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Comments

user1149224 · Accepted Answer · 2012-04-02 07:51:05Z

You can use C++11 uniform initialization:

struct Test
{
    vector<string> producer;
    vector<string> golden;  
};

Test test1 =
{
    {"producer1", "producer2"},
    {"golden1", "golden2"}
};

The following sample code compiles fine with g++:

#include <iostream>
#include <string>
#include <vector>

using namespace std;

struct Test
{
    vector<string> producer;
    vector<string> golden;  
};


ostream & operator<<(ostream & os, const Test & test)
{
    os << "------------------------" << endl;
    os << "producer={" << endl;
    for (auto& p : test.producer)
    {
        os << '\t' << p << endl;
    }
    os << "}" << endl;

    os << "golden={" << endl;
    for (auto& p : test.golden)
    {
        os << '\t' << p << endl;
    }
    os << "}";
    os << "------------------------" << endl;

    return os;   
}



int main()
{
    Test test1 =
    {
         {"producer1", "producer2"},
         {"golden1", "golden2"}
    };

    cout << test1 << endl;


    return 0;
}

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