How to initialize a dynamic array of pointers to NULL in C?

Question

I am fairly new to C and I don't understand why the following two statements do not create the same result:

char *fields[14] = {NULL};

const int num_fields = 14;
char *fields[num_fields] = {NULL};

Option 1 works, but option 2 does not. It says "variable-sized object may not be initialized" and it gives a warning "warning: excess elements in array initializer". I use gcc 4.2.1 on OSX.

Thanks for sharing your thoughts!

I think this is probably because a const int isn't necessarily a constant value, crazy as that may seem. I believe you can obtain a pointer to the const and modify the underlying value. — Lazarus
– Lazarus, Commented Jan 31, 2012 at 14:29

cnicutar · Accepted Answer · 2012-01-31 14:32:10Z

5

The second object is called a VLA (Variable Length Array), well defined by C99. To achieve what you want you can use this:

for (i = 0; i < num_fields; i++)
    fields[i] = NULL;

The gist of the issue is that const int num_fields is very different from 14, it's not a constant, it's read-only.

edited Jan 31, 2012 at 14:32

answered Jan 31, 2012 at 14:26

cnicutar

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4 Comments

Armen Tsirunyan Over a year ago

+1 but technically it's not initialization, but assignment. I guess it would be correct to say that VLAs cannot be initialized per se.

cnicutar Over a year ago

@ArmenTsirunyan I changed the wording.

DrDee Over a year ago

Thank you cnicutar for your explanation!

Dani Cricco Over a year ago

In a VLA (Variable Length Array) the compiler automatically handles the storage management. Since the size won´t be known until runtime, you can´t use the initializer. As everybody said. You need a for loop.

codestriker · Accepted Answer · 2012-01-31 17:23:07Z

3

Even if you define num_fields with const keyword, compiler interprets it as variable only. you can have alternative for this by defining following macro:

#define num_fields 14

char *fields[num_fields] = {NULL};

answered Jan 31, 2012 at 17:23

codestriker

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Comments

Blagovest Buyukliev · Accepted Answer · 2012-01-31 14:28:45Z

1

Although num_fields has a const qualifier, it is still considered a variable by the compiler.

Therefore, you are attempting to declare a variable-sized array, and initialisers (the {NULL} part) cannot be used in conjunction with them.

answered Jan 31, 2012 at 14:28

Blagovest Buyukliev

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1 Comment

DrDee Over a year ago

Thank you Blagovest Buyukliev for your response.

MikeW · Accepted Answer · 2017-07-12 15:47:40Z

0

Your construction works in C++, where a const int will be treated as a compile-time constant ("constant expression"), and hence available for use as a compile-time array size.

(This aspect was one of B. Stroustrup's design goals for C++, to eliminate the need for compile-time macros if possible)

However in C, your definition of "num_fields" effectively declares a read-only memory location with your preset value, and hence is not under C rules a "constant expression" valid at compile time, and hence may not be used as an array size at the outermost 'program' scope.

edited Jul 12, 2017 at 15:47

answered Jul 12, 2017 at 15:40

MikeW

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