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Static array in C


Static array in C

By : Algenis
Date : November 22 2020, 03:03 PM
this one helps. What you are using is variable length array. Which is supported by C99 and latter. But note that VLA has automatic storage duration unlike dynamic memory allocated by malloc family functions.
Also note that compile time allocation is not equivalent to static array. static array and static allocation are different.
code :


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array declared with static keyword: what is static, the pointer or the whole array?

array declared with static keyword: what is static, the pointer or the whole array?


By : user2626881
Date : March 29 2020, 07:55 AM
wish of those help From the ISO C99 standard (section 6.2.1, "Scopes of identifiers"):
How do I use a static array element as an index to a static array of objects that are of different template instantiatio

How do I use a static array element as an index to a static array of objects that are of different template instantiatio


By : user3716228
Date : March 29 2020, 07:55 AM
it fixes the issue For some reason using constexpr instead of const in the following line works for me. I don't know why yet. See a question I asked on the subject.
code :
// static const size_t ARRAYSIZES[2] =   // Does not work
static constexpr size_t ARRAYSIZES[2] =  // Works
{
    1,
    2
};
static DataParent DataTable[ 2 ] =
{
    DataChild< 1 >(&size1Array),
    DataChild< 2 >(&size2Array)            
};
static DataParent* DataTable[ 2 ] =
{
    new DataChild< 1 >(&size1Array),
    new DataChild< 2 >(&size2Array)            
};
static std::unique_ptr<DataParent> DataTable[ 2 ] =
{
    new DataChild< 1 >(&size1Array),
    new DataChild< 2 >(&size2Array)            
};
static constexpr size_t ARRAYSIZES[2] =  { ... };
static const size_t ARRAYSIZES[2] =  { ... };
double read_double()
{
   double v;
   std::cin >> v;
   return v;
}

const double v1 = read_double();
constexpr double v2 = read_double();
C# Static arrays - asigning values of static array to other static array

C# Static arrays - asigning values of static array to other static array


By : vinothkumar
Date : March 29 2020, 07:55 AM
This might help you When you assign a2=a1 you do a shallow copy. A reference (somehow equivalent to the old pointer) to a1 is copied to a2. They now point to same data and changing one changes the other. To create a deep copy, use the Clone() function, like this:
code :
a2 = a1.Clone() as double[];
Cannot return a static reference to a temporary array even when the array contains a value that also has the static life

Cannot return a static reference to a temporary array even when the array contains a value that also has the static life


By : Stefan Caunter
Date : March 29 2020, 07:55 AM
help you fix your problem The RFC that added the automatic promotion of references to values to 'static states:
code :
const fn bar() -> &'static StructType {
    &StructType("asdf")
}

const fn combo() -> &'static [&'static StructType; 1] {
    &[Self::bar()]
}
fn wombo() -> &'static [&'static StructType] {
    Self::combo()
}
fn example<T>() {
    static NO_CAN_DO: T = unimplemented!();
}
error[E0401]: can't use type parameters from outer function
 --> src/lib.rs:2:23
  |
1 | fn example<T>() {
  |    ------- - type variable from outer function
  |    |
  |    try adding a local type parameter in this method instead
2 |     static NO_CAN_DO: T = unimplemented!();
  |                       ^ use of type variable from outer function
When assigning a static array to a non-static array in C#, is it done by reference or value?

When assigning a static array to a non-static array in C#, is it done by reference or value?


By : Van Do
Date : March 29 2020, 07:55 AM
should help you out The code in LoadArray doesn’t change any arrays. It assigns what the methods return, which apparently are new ones and you assign them to the variables. Therefore if you store a reference to a previous array somewhere it will still retain its value.
It would be different if the arrays were already created and you changed the values in them. Then what you expected to see would happen.
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