[Haskell-cafe] Specify array or list size?

Thomas Davie tom.davie at gmail.com
Sat May 7 15:20:15 EDT 2005

On May 7, 2005, at 8:07 PM, Marcin 'Qrczak' Kowalczyk wrote:

> Thomas Davie <tom.davie at gmail.com> writes:
>> I'm not familiar with your C++ example (not being familiar with C++),
>> but I think that it's a bit of a stretch of the imagination to say
>> that C "introduces a variable of type "array of 50 ints"", the fact
>> that this is now an array of 50 integers is never checked at any
>> point in the compilation or run, and I'm not sure it can be even if
>> K&R had wanted to.
> The size is taken into account when such array type is an element of
> another array, and by sizeof.
> int (*p)[50]; /* p may legally point only to arrays of 50 ints each */
> ++p; /* p is assumed to point into an array, and is moved by one
>         element, i.e. by 50 ints */
I'm not sure what you're trying to prove by saying that... There is  
still no type information that says that the contents of p are an  
array of 50 elements... I can still attempt to access element 51 and  
get a runtime memory error.  The type of p is still int**, not  
"pointer to array of 50 ints"

>> As an example:
>> int bobsArray[5];
>> bobsArray[6] = 23;
>> is not badly typed - it is merely a badly broken program.
> Because the array size is not taken into account by indexing. But it's
> a part of the type. These issues are independent, for example in C#
> both are the opposite.
I don't think it is part of the type... Does the compiler ever know  
any more about the type of bobsArray other than it's a pointer to an  
integer?  I think that the above code can be directly translated to:

int *bobsArray;
bobsArray = (int *)malloc(5 * sizeof(int));
bobsArray[6] = 23

Which stores exactly the same about on type information.


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