How to define different struct to single variable

Question

I am declaring one variable based on the condition:

if(abc == 2)
{
price* x;
    x = (price *)malloc(2 * sizeof(price*));
}
else if(abc == 3)
{
store* x;
    x = (store *)malloc(2 * sizeof(store*));
}

// x is getting used in other function
xyz(&x);

while compile, it is throwing error: error: 'x' was not declared in this scope. I understand that, since x is not defined in the function scope, it is throwing error.

I tried to declare void* x, but that also didn't work. Is there any way we can achieve this?

Why didn't void * work? It will work if you do it correctly. What problems you face when using void *? — P.P
– P.P, Commented Aug 3, 2015 at 9:51
Thanks, But it is throwing error if I am using x like this: xyz(&x[0]); error: pointer of type 'void ' used in arithmetic , error: 'void' is not a pointer-to-object type — user1138143
– user1138143, Commented Aug 3, 2015 at 9:57
As the error says void * is an incomplete type. Pass xyz(x) instead of indexing it. I am wondering what xyz() does with x without knowing the type of x. — P.P
– P.P, Commented Aug 3, 2015 at 9:59
above code , is a sample piece of code. I didn't posted the actual code. In xyz, we need to pass the parameter like &x[0] . As of now, I have two different function for each struct , which is working fine. I am trying to refactor the code, by adding this if condition. — user1138143
– user1138143, Commented Aug 3, 2015 at 10:03
xyz needs to cast it back to price* or store* before it can index it. And it needs some way to know which one it points to. — Barmar
– Barmar, Commented Aug 3, 2015 at 10:04

Daniele · Accepted Answer · 2015-08-03 09:50:41Z

3

You have to declare x before the if. The problem is the scope of the variable not the type!

answered Aug 3, 2015 at 9:50

Daniele

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Comments

Jarod42 · Accepted Answer · 2015-08-03 09:50:50Z

1

It seems you want

void* x = 0;

if(abc == 2)
{
    x = (price *)malloc(2 * sizeof(price*));
}
else if(abc == 3)
{
    x = (store *)malloc(2 * sizeof(store*));
}

btw, size or type seems suspicious in malloc line.

I think you want

x = (price **)malloc(2 * sizeof(price*));

or

x = (price *)malloc(2 * sizeof(price));

answered Aug 3, 2015 at 9:50

Jarod42

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2 Comments

user1138143 Over a year ago

Thanks, But it is throwing error if I am using x like this: xyz(&x[0]); error: pointer of type 'void ' used in arithmetic , error: 'void' is not a pointer-to-object type

Jarod42 Over a year ago

@user1138143: What is xyz ? Is xyz(x) sufficient ?

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