Is there a way to modify this C++ struct assignment block to work in straight C

Question

The following code resides in a device that will issue a deviceId (LXdeviceInfo) when enumerated through an IrDA sockets connection. This is only important to explain why I would like to keep the data types as similar as possible, but be able to compile using ansi C

With #includes of windows.h and af_irda.h, the following code compiles with no errors in a C++ compiler, but breaks in a C compiler just below the struct assignment (see ERROR here). Ideally I would like to initialize the struct member 'ID' to be an array of characters while keeping it typed just as it is in the original code so I can test the value of LXdeviceInfo just as it would appear when querying it from a call to the device from a PC socket connection.

Is there some way to modify this assignment block to work in straight C?

#include <windows.h>
#include <af_irda.h>

#define IR_HINT_COMPUTER  0x04
#define IR_HINT_EXT       0x80
#define IR_HINT_OBEX      0x20
#define IR_HINT_IRCOMM    0x04
#define IR_CHAR_ASCII       0
#define PROD_FAMILY_NAME ("product name goes here")

#define uint8_t unsigned char

const struct {
    uint8_t hint1;
    uint8_t hint2;
    uint8_t charset;
    uint8_t ID[sizeof(PROD_FAMILY_NAME)];
} devInfoStorage = 
{
    IR_HINT_COMPUTER | IR_HINT_EXT,   // hint1
    IR_HINT_OBEX | IR_HINT_IRCOMM,    // hint2
    IR_CHAR_ASCII,                    // charset
    PROD_FAMILY_NAME                  // Prod ID string
}; // ERROR here: Innvalid initialization type: found 'pointer to char' expected 'unsigned char'

const uint8_t *LXdeviceInfo = (uint8_t *) &devInfoStorage;

/* The size of the device info */
const uint8_t LXdeviceInfoLen = sizeof(devInfoStorage);



void main(void)
{

    #define DEVICE_LIST_LEN    10

    unsigned char DevListBuff[sizeof (DEVICELIST) -
                              sizeof (IRDA_DEVICE_INFO) +
                              (sizeof (IRDA_DEVICE_INFO) * DEVICE_LIST_LEN)];

    int DevListLen = sizeof (DevListBuff);
    PDEVICELIST pDevList;

    pDevList = (PDEVICELIST) & DevListBuff;
         //code continues.

}

Answer 1

Remove the parentheses around the string literal. The parentheses make the macro expand into an expression that will decay into a pointer type, which makes it not compile in your C compiler. A pointer type cannot be used to initialize an array. Without the parentheses, the string literal is used to initialize the array.

#define PROD_FAMILY_NAME "product name goes here"

The C standard states that a parenthesized expression takes on the same type as the unparenthesized expression, C.99 §6.5.1 ¶5:

A parenthesized expression is a primary expression. Its type and value are identical to those of the unparenthesized expression. It is an lvalue, a function designator, or a void expression if the unparenthesized expression is, respectively, an lvalue, a function designator, or a void expression.

However, while a string literal is an expression, the converse is not true. Specifically, a string literal in of itself is not a type, but a defined entity. The initialization of arrays makes a specific allowance for a string literal, C.99 §6.7.8 ¶14:

An array of character type may be initialized by a character string literal, optionally enclosed in braces.

The other allowed initializers for an array are described in C.99 §6.7.8 ¶16:

Otherwise, the initializer for an object that has aggregate or union type shall be a brace enclosed list of initializers for the elements or named members.

A parenthesized expression is not a string literal, nor a brace enclosed list of initializers.