Struct array typedef in struct shows incomplete data

Question

I'm having trouble with a piece of code where a typedef array is created of a struct. That typedef is then used in another struct.
When receiving the typedef in a function and initialising the struct with the typedef in it, I only get data of the first element in the array.

Below I have a simplified example of what I'm getting at the moment.

struct simple_struct {
    double a;
};

typedef struct simple_struct arr_typedef[2];

struct struct_with_typedef {
    const arr_typedef typedef_arr;
};

void foo(const arr_typedef arg_s) {
    struct struct_with_typedef test = {
        *arg_s
    };

    int i;
    for (i = 0; i < 2; i++) {
        printf("value: %f \n", test.typedef_arr[i].a);
    }
}

int main(int argc, char* argv[]) {
    arr_typedef d_test = { {1}, {2} };
    foo(d_test);

    return 1;
}

When compiled, using gcc 4.4, and run I see the following output:

~/public > ./test                  
value: 1.000000 
value: 0.000000

Would someone be able to explain why the value of the second item isn't available?
If I leave out the dereference I get the following error whilst compiling, which I also don't get:

test.c:82: error: incompatible types when initializing type 'double' using type 'const struct simple_struct *'

I have removed the typedef and the same result persists. Sort of understand that indeed that one initialiser value is given. But there is only one member in the struct, or are they expanded in the background?
If so, how would you initialise both values?

struct simple_struct {
    double a;
};

struct struct_with_arr {
    const struct simple_struct struct_arr[2];
};

void foo(const struct simple_struct arg_s[2]) {
    struct struct_with_arr test = {{*arg_s}};

    int i;
    for (i = 0; i < 2; i++) {
        printf("value: %f \n", test.struct_arr[i].a);
    }
}

int main(int argc, char* argv[]) {
    struct simple_struct d_test[2] = { {1}, {2} };
    foo(d_test);

    return 1;
}

Of course you only get the 1st element, that's what *arg_s does. That being said, hiding an array behind a typedef is very bad practice, as bad as hiding pointers behind typedefs. The reason you can't find this bug, is in fact because you hid the array behind a typedef. — Lundin
– Lundin, Commented Mar 16, 2018 at 10:48
You only provide 1 initializer value: *arg_s. If you are confused, fine! Don't do such typedef's! — Gerhardh
– Gerhardh, Commented Mar 16, 2018 at 10:49
Thanks, I have updated the question with some more info base on your comments — Alex_na
– Alex_na, Commented Mar 16, 2018 at 11:07

Vlad from Moscow · Accepted Answer · 2018-03-16 11:31:35Z

4

In this declaration

struct struct_with_typedef test = {
    *arg_s
};

there is used an object of the type struct simple_struct to initialize an array of the type const arr_typedef.

Firstly you need to enclose the initializer in braces and add an initializer for the second element of the array if you want to do so.

A correct way to initialize the array is the following

#include <stdio.h>

struct simple_struct 
{
    double a;
};

typedef struct simple_struct arr_typedef[2];

struct struct_with_typedef 
{
    const arr_typedef typedef_arr;
};

void foo( const arr_typedef arg_s ) 
{
    struct struct_with_typedef test = 
    {
        { arg_s[0], arg_s[1] }
    };

    for ( int i = 0; i < 2; i++ ) 
    {
        printf("value: %f \n", test.typedef_arr[i].a);
    }
}

int main( void ) 
{
    arr_typedef d_test = { {1}, {2} };
    foo(d_test);

    return 0;
}

The program output is

value: 1.000000 
value: 2.000000

You can write the initializers also the following way

struct struct_with_typedef test = 
{
    { *arg_s, *( arg_s + 1 ) }
};

Take into account that an array designator used as an initializer expression is implicitly converted to pointer to its first element. Dereferencing the pointer you get the first element itself.

edited Mar 16, 2018 at 11:31

answered Mar 16, 2018 at 11:21

Vlad from Moscow

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1 Comment

Alex_na Over a year ago

Thank you! Now I understand it

M.M · Accepted Answer · 2018-03-16 11:30:35Z

1

There is no syntax in C for initializing an array with a single initializer representing another array to copy values from.

To fix your code you could do one of these options:

List the members: struct struct_with_arr test = {{arg_s[0], arg_s[1]}};
Change the function to accept struct struct_with_arr as the parameter type, in which case you can use struct struct_with_arr test = arg;
Copy without an initializer: struct struct_with_arr test; memcpy(&test.struct_arr, arg_s, sizeof test.struct_arr);. (Actually you can't use this option since you have defined the struct element as const ... not really a great idea in the first place in my opinion)

For completeness I will mention the code:

struct struct_with_arr foo = *(struct struct_with_arr *)arg_s;

This is one of those things that is technically undefined behaviour (if the argument source was not actually a struct_with_arr) but is likely to work on any actual compiler that exists. My advice would be to not do this.

edited Mar 16, 2018 at 11:30

answered Mar 16, 2018 at 11:24

M.M

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1 Comment

Alex_na Over a year ago

Thanks for your answer! Sadly we can't change anything of the const or the function prototype, as these our outside our control. So expanding them is the way to go

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