How can I specify an array at runtime?

Question

What I'm Trying To Do

Basically, I've got several possible arrays that I define with macros:

#define ARRAY_ONE  {0, 2, 7, 8}
#define ARRAY_TWO  {3, 6, 9, 2}
#define ARRAY_THREE  {3, 6, 4, 5}
//etc...

At runtime, I have a C-Array that gets used in a lot of places in a certain class. I want this array to use one of the #define values, i.e:

int components[4];

if (caseOne)
{
    components = ARRAY_ONE;
}
else if (caseTwo)
{
    components = ARRAY_TWO;
}
else if (caseThree)
{
    //etc...
}

-

The Problem

However, the above code does not work. Instead, I get a weird error

Expected expression before '[' token

Would anyone mind explaining what's going on, and how I could achieve what I'm attempting to? Any help would be much appreciated - Thanks!

Tim Dean · Accepted Answer · 2011-08-15 04:29:40Z

4

I don't think that C arrays can be initialized using the curly-brace syntax after they've been declared. You can only do that when initializing them while declaring them.

Try adjusting the previously posted answer with:

const int ARRAY_ONE[] = {0, 2, 7, 8};
const int ARRAY_TWO[] = {3, 6, 9, 2};
const int ARRAY_THREE[] = {3, 6, 4, 5};

int *components;
if (case1) {
    components = ARRAY_ONE;
} else if (case2) {
    components = ARRAY_TWO;
} else if (case3) {
    components = ARRAY_THREE;
}

answered Aug 15, 2011 at 4:29

Tim Dean

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3 Comments

Tommy Over a year ago

That's entirely true. The most obvious solution that I can think of is to use memcpy to copy the required array into components.

Jordan Smith Over a year ago

Ah! I see what the previous answer was trying to say now, and using pointers as above seems to be a good solution in my case. Thanks!

Chuck Over a year ago

More precisely, variables in C can't be initialized except at the time they're declared. After that, you're simply assigning them, and you can't assign a C array.

Tom Dalling · Accepted Answer · 2011-08-15 04:14:28Z

0

I can't really work out what the error is. I suspect it might be coming from some code you haven't posted. Does it say the error is on the int components[4]; line?

Would this do? I uses constants instead of defines.

const int ARRAY_ONE[] = {0, 2, 7, 8};
const int ARRAY_TWO[] = {3, 6, 9, 2};
const int ARRAY_THREE[] = {3, 6, 4, 5};

int* components = ARRAY_ONE;

int whatever = components[2];

answered Aug 15, 2011 at 4:14

Tom Dalling

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1 Comment

Jordan Smith Over a year ago

Thanks for the reply! The error is on the components = ARRAY lines. Your method doesn't seem to account for the fact that components could be either Array one, two, or three - this is where I'm having trouble.

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

try this:

int ARRAY_ONE[] = {0,2,7,8};

int ARRAY_TWO [] = {3,6,9,2};

int ARRAY_THREE[] = {3,6,4,5};

int components[4];

int count =sizeof(components)/4 //this will get array length, or you can just put array lenght ;

if (case1)
   for (int i =0; i< count; i++)
      components[i] = ARRAY_ONE[i];
else if (case2)
   for (int i =0; i< count; i++)
      components[i] = ARRAY_TWO[i];

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Aug 15, 2011 at 4:34

HelmiB

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