address representation in ada
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Published on 2010-04-27T09:31:54Z
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2010/04/27
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Hi all,
I have pasted a code below which is in ada language.I need some clarification on some implementations.
C : character;
Char : character; type Myarr_Type is array (character range 'A'..'K') of character; Myarr : Myarr_Type := ('A','B','C','D','E','F','G','H','I','J','K');
Next_Address := Myarr'address
Last_Address := Next_Address + Storage_Offset'(40);
my_func(int P1,int P2)
{
return P2 + Storage_Offset'(4);
}
Last_Address := Next_Address + Storage_Offset'(4);
Now my doubt is 1)what does P2 + Storage_Offset'(4) actually mean.Does that mean that its returning the address of the next element in the array which is 'B'.Storage_Offset'(4) in ada --does this mean 4 bits or 4 bytes of memory. 2) If i assume that Last_Address points to last element of the array which is 'K',how does the arithmentic Storage_Offset'(40) satisfies the actual implementation?
Please get back to me if u need any more clarifications.
Thanks
Maddy
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