C initialize const struct member with existing const variable

Question

I'm using default C under gcc.

My code:

typedef struct _OpcodeEntry OpcodeEntry;

//

struct _OpcodeEntry
{
    unsigned char uOpcode;
    OpcodeMetadata pMetadata;
};

//

const OpcodeMetadata omCopyBytes1 = { 1, 1, 0, 0, 0, 0, &CopyBytes };

const OpcodeEntry pOpcodeTable[] =
{
    { 0x0, omCopyBytes1 },
};

Errors:

error: initializer element is not constant
error: (near initialization for 'pOpcodeTable[0].pMetadata')

If I change omCopyBytes1 to what it's actually set to in the above line, the code compiles fine. What am I doing wrong?

Sergey Kalinichenko · Accepted Answer · 2013-03-26 01:23:45Z

7

You cannot use omCopyBytes1 to initialize a member of pOpcodeTable[] array, because omCopyBytes1 is a variable that is run-time constant, not a compile-time constant. Aggregate initializers in C must be compile-time constants, that's why the code from your post does not compile.

As a variable, omCopyBytes1 has its own place in memory, which is initialized to an array of items. You can use such variable by a pointer, like this:

struct _OpcodeEntry {
    unsigned char uOpcode;
    const OpcodeMetadata *pMetadata;
};
...
const OpcodeEntry pOpcodeTable[] = {
    { 0x0, &omCopyBytes1 }, // This should work
};

Alternatively, you can make it a preprocessor constant:

#define omCopyBytes1 { 1, 1, 0, 0, 0, 0, &CopyBytes }

If defined in this way, the omCopyBytes1 would no longer be a variable: it would be a preprocessor definition that vanishes before the compiler is done. I would recommend against the preprocessor method, but it's there in case you must do it.

answered Mar 26, 2013 at 1:23

Sergey Kalinichenko

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1 Comment

jamesdlin Over a year ago

Are constant expressions really required to initialize all aggregates in C99? I don't see anything in section 6.7.8 of the C99 standard saying so (other than for objects of static storage duration).

R.. GitHub STOP HELPING ICE · Accepted Answer · 2013-03-26 13:29:46Z

0

In C, initializers for objects of static storage duration must be constant expressions. A const-qualified variable is not a constant expression.

edited Mar 26, 2013 at 13:29

answered Mar 26, 2013 at 1:23

R.. GitHub STOP HELPING ICE

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1 Comment

jamesdlin Over a year ago

Surely you don't mean all initializers must be constant expressions, unless you're claiming that something like int x = rand(); is illegal.

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