How do I #define an array in C?

Question

Is the following syntax to define an array valid syntax in C?

#define array[] { \
for (j=0; j<N_frequencysteps;j++)  \
{ \
array[j] = (function); \
} \
}

If not, how do I define an array in C?

@user442836 - did you try compiling it? The quick answer is no, that isn't valid syntax. — Carl Norum
– Carl Norum, Commented Sep 8, 2010 at 21:08
Can you describe what you are calling an array? Maybe in n comparison to another language? — Rod
– Rod, Commented Sep 8, 2010 at 21:13
For an overview of defining, initializing, and using arrays, see exforsys.com/tutorials/c-language/c-arrays.html. — bta
– bta, Commented Sep 8, 2010 at 21:26

Jonathan Leffler · Accepted Answer · 2010-09-08 21:28:03Z

It depends on how you define 'valid syntax'.

The C pre-processor will accept it as an object-like macro.
The C compiler will not accept the output as legitimate C.

Consider an invocation:

array[23];

The C compiler sees:

[] { for (j=0; j<N_frequencysteps;j++) { array[j] = (function); } } [23];

That is gibberish (even with the 'enter code here' removed).

How can you define an array in C?

enum { ARRAYSIZE = 23 };
int array[ARRAYSIZE];
int j;

for (j = 0; j < ARRAYSIZE; j++)
    array[j] = 0;

You could also use an initializer, but you might get compiler warnings unless you are thorough:

int array[ARRAYSIZE] = { 0 };

One more possibility: if you want to define an array and initialize it with calls to a specific function, then you could, I suppose, try:

#define ARRAY(type, name, size, initializer) type name[size]; \
          for (int j = 0; j < size; j++) name[j] = initializer

This could be used as:

ARRAY(int, array, 23, j % 9);

(Note that this initializer depends on an implementation detail of the macro - not a good idea. But I don't think I'd ever use such a macro, anyway.) It depends on you being in C99 mode if you have more than one of these in a particular block of code (since it would then mix code with declarations) - and also because the declaration in the for loop is only supported in C++ and C99, not in C90.

+1. My preprocessor says warning: missing whitespace after the macro name, which at least is a hint that it's a bad idea.
@Carl: the whole thing is horrid - and yes, it is customary (but not mandatory) to put a space after the name of an object-like macro such as this array macro. And the space then emphasizes how horrid it is.

Himanshu · Accepted Answer · 2010-09-08 21:18:59Z

2

This looks really wrong. Why not define array in a normal way?

int array[SIZE];
for (j=0; j<N_frequencysteps;j++)
{  
   array[j] = (function);
}

answered Sep 8, 2010 at 21:18

Himanshu

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2 Comments

Jonathan Leffler Over a year ago

The '(function)' notation could only be a way of assigning a pointer to function to an array of pointers to functions, assuming 'function' is in fact the name of a function. And presumably the N_frequencysteps should be SIZE or vice versa?

Himanshu Over a year ago

Yes, SIZE should be larger than or equal to N_frequencysteps

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