E2099 Overflow in conversion or arithmetic operation - delphi

I want to compare an int64 with a variable like this:
const GB=1073741824;
if DiskFile.Size< 1*GB then
It works with 1 but not with 3:
if DiskFile.Size< 3*GB then
This post (Strange Delphi integer multiplication behavior) explains why. I agree with that explanation. The result of 2*GB cannot fit in 'integer'. What I don't understand is why the compiler chooses integer instead the int64? As in the case of:
if DiskFile.Size< 3073741824 then <--------- almost 3GB
that works.
There is any way to write the last line of code in the 3*GB style (using constants) BUT without defining a new constant for 1GB, 2GB, 3GB, 4GB, etc ?

The first thing to be clear of here is that the integer overflow occurs in the compiler. The compiler has to evaluate your expression because it is a constant expression and they are evaluated by the compiler.
The documentation is a little sparse (and I am being kind here) on how the compiler treats your expression. We can infer, at least empirically, that the compiler attempts to perform 3*GB in a signed integer context. That is clear from the error message.
You need to force the compiler to evaluate the expression in an Int64 context. A cast will force that:
if DiskFile.Size< Int64(3)*GB then
....
Another option is to make the constant have type Int64:
const
GB = Int64(1073741824);
Although I think I'd write it like this:
const
KB = Int64(1024);
MB = 1024*KB;
GB = 1024*MB;
So long as GB is a 64 bit type then you can revert to:
if DiskFile.Size < 3*GB then
....
I'd like to elaborate on my second paragraph above. How can we tell that the compiler performs the arithmetic in 32 bit signed integer context? The following program suggests that this is so:
{$APPTYPE CONSOLE}
const
C1 = 715827882; // MaxInt div 3
C2 = C1+1;
begin
Writeln(3*C1);
Writeln(3*C2);
Readln;
end.
The first expression, 3*C1 compiles, the second fails with E2099. The first expression does not overflow a signed 32 bit integer, the second does.
When looking at the documentation, it is unclear whether the true constant 1073741824 should be of type Integer or Cardinal. The compiler could choose either. It seems that the compiler, when presented with a choice between signed and unsigned types, chooses signed types.
But then one might imagine that the following program would behave in the same way, but with Smallint and Word taking the place of Integer and Cardinal:
{$APPTYPE CONSOLE}
const
C1 = 10922; // high(Smallint) div 3
C2 = C1+1;
begin
Writeln(3*C1);
Writeln(3*C2);
Readln;
end.
But no, this program compiles. So, at this point I am giving up on the documentation which appears to bear little relationship to the actual behaviour of the compiler.
My best guess is that a integral true constant is handled as follows:
If it is within the range of Integer, it is of type Integer.
Otherwise, if it is within the range of Cardinal, it is of type Cardinal.
Otherwise, if it is within the range of Int64, it is of type Int64.
Otherwise, if it is within the range of UInt64, it is of type UInt64.
Otherwise it is a compiler error.
Of course, all of this assumes that the compilers rules for evaluating constant expressions follow the same rules as the rest of the language. I'm not certain that is the case.

Related

set of WideChar: Sets may have at most 256 elements

I have this line:
const
MY_SET: set of WideChar = [WideChar('A')..WideChar('Z')];
The above does not compile, with error:
[Error] Sets may have at most 256 elements
But this line does compile ok:
var WS: WideString;
if WS[1] in [WideChar('A')..WideChar('Z')] then...
And this also compiles ok:
const
MY_SET = [WideChar('A')..WideChar('Z'), WideChar('a')..WideChar('z')];
...
if WS[1] in MY_SET then...
Why is that?
EDIT: My question is why if WS[1] in [WideChar('A')..WideChar('Z')] compiles? and why MY_SET = [WideChar('A')..WideChar('Z'), WideChar('a')..WideChar('z')]; compiles? aren't they also need to apply to the set rules?
A valid set has to obey two rules:
Each element in a set must have an ordinal value less than 256.
The set must not have more than 256 elements.
MY_SET: set of WideChar = [WideChar('A')..WideChar('Z')];
Here you declare a set type (Set of WideChar) which has more than 256 elements -> Compiler error.
if WS[1] in [WideChar('A')..WideChar('Z')]
Here, the compiler sees WideChar('A') as an ordinal value. This value and all other values in the set are below 256. This is ok with rule 1.
The number of unique elements are also within limits (Ord('Z')-Ord('A')+1), so the 2nd rules passes.
MY_SET = [WideChar('A')..WideChar('Z'), WideChar('a')..WideChar('z')];
Here you declare a set that also fulfills the requirements as above. Note that the compiler sees this as a set of ordinal values, not as a set of WideChar.
A set can have no more than 256 elements.
Even with so few elements the set already uses 32 bytes.
From the documentation:
A set is a bit array where each bit indicates whether an element is in the set or not. The maximum number of elements in a set is 256, so a set never occupies more than 32 bytes. The number of bytes occupied by a particular set is equal to
(Max div 8) - (Min div 8) + 1
For this reason only sets of byte, (ansi)char, boolean and enumerations with fewer than 257 elements are possible.
Because widechar uses 2 bytes it can have 65536 possible values.
A set of widechar would take up 8Kb, too large to be practical.
type
Capitals = 'A'..'Z';
const
MY_SET: set of Capitals = [WideChar('A')..WideChar('Z')];
Will compile and work the same.
It does seem a bit silly to use widechar if your code ignores unicode.
As written only the English capitals are recognized, you do not take into account different locales.
In this case it would be better to use code like
if (AWideChar >= 'A') and (AWideChar <= 'Z') ....
That will work no matter how many chars fall in between.
Obviously you can encapsulate this in a function to save on typing.
If you insist on having large sets, see this answer: https://stackoverflow.com/a/2281327/650492

MAX / MIN function in Objective C that avoid casting issues

I had code in my app that looks like the following. I got some feedback around a bug, when to my horror, I put a debugger on it and found that the MAX between -5 and 0 is -5!
NSString *test = #"short";
int calFailed = MAX(test.length - 10, 0); // returns -5
After looking at the MAX macro, I see that it requires both parameters to be of the same type. In my case, "test.length" is an unsigned int and 0 is a signed int. So a simple cast (for either parameter) fixes the problem.
NSString *test = #"short";
int calExpected = MAX((int)test.length - 10, 0); // returns 0
This seems like a nasty and unexpected side effect of this macro. Is there another built-in method to iOS for performing MIN/MAX where the compiler would have warned about mismatching types? Seems like this SHOULD have been a compile time issue and not something that required a debugger to figure out. I can always write my own, but wanted to see if anybody else had similar issues.
Enabling -Wsign-compare, as suggested by FDinoff's answer is a good idea, but I thought it might be worth explaining the reason behind this in some more detail, as it's a quite common pitfall.
The problem isn't really with the MAX macro in particular, but with a) subtracting from an unsigned integer in a way that leads to an overflow, and b) (as the warning suggests) with how the compiler handles the comparison of signed and unsigned values in general.
The first issue is pretty easy to explain: When you subtract from an unsigned integer and the result would be negative, the result "overflows" to a very large positive value, because an unsigned integer cannot represent negative values. So [#"short" length] - 10 will evaluate to 4294967291.
What might be more surprising is that even without the subtraction, something like MAX([#"short" length], -10) will not yield the correct result (it would evaluate to -10, even though [#"short" length] would be 5, which is obviously larger). This has nothing to do with the macro, something like if ([#"short" length] > -10) { ... } would lead to the same problem (the code in the if-block would not execute).
So the general question is: What happens exactly when you compare an unsigned integer with a signed one (and why is there a warning for that in the first place)? The compiler will convert both values to a common type, according to certain rules that can lead to surprising results.
Quoting from Understand integer conversion rules [cert.org]:
If the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, the operand with unsigned integer type is converted to the type of the operand with signed integer type.
Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.
(emphasis mine)
Consider this example:
int s = -1;
unsigned int u = 1;
NSLog(#"%i", s < u);
// -> 0
The result will be 0 (false), even though s (-1) is clearly less then u (1). This happens because both values are converted to unsigned int, as int cannot represent all values that can be contained in an unsigned int.
It gets even more confusing if you change the type of s to long. Then, you'd get the same (incorrect) result on a 32 bit platform (iOS), but in a 64 bit Mac app it would work just fine! (explanation: long is a 64 bit type there, so it can represent all 32 bit unsigned int values.)
So, long story short: Don't compare unsigned and signed integers, especially if the signed value is potentially negative.
You probably don't have enough compiler warnings turned on. If you turn on -Wsign-compare (which can be turned on with -Wextra) you will generate a warning that looks like the following
warning: signed and unsigned type in conditional expression [-Wsign-compare]
This allows you to place the casts in the right places if necessary and you shouldn't need to rewrite the MAX or MIN macros

Calculating factorial on FORTRAN with integer variables. Memory overflow

I'm doing a program with FORTRAN that is a bit special. I can only use integer variables, and as you know with these you've got a memory overflow when you try to calculate a factorial superior to 12 or 13. So I made this program to avoid this problem:
http://lendricheolfiles.webs.com/codigo.txt
But something very strange is happening. The program calculates the factorial well 4 or 5 times and then gives a memory overflow message. I'm using Windows 8 and I fear it might be the cause of the failure, or if it's just that I've done something wrong.
Thanks.
Try compiling with run-time subscript checking. In Fortran segmentation faults are generally caused either by subscript errors or by mismatches between actual and dummy arguments (i.e., between arguments in the call to a procedure and the arguments as declared in the procedure). I'll make a wild guess from glancing at your code that you have have a subscript error -- let the compiler find it for you by turning on run-time subscript checking. Most Fortran compilers have this as an compilation option.
P.S. You can also do calculations like this by using already written packages, e.g., the arbitrary precision arithmetic software of David Bailey, et al., available in Fortran 90 at http://crd-legacy.lbl.gov/~dhbailey/mpdist/
M.S.B.'s answer has the gist of your problem: your array indices go out of bounds at a couple of places.
In three loops, cifra - 1 == 0 is out of bounds:
do cifra=ncifras,1,-1
factor(1,cifra-1) = factor(1,cifra)/10 ! factor is (1:2, 1:ncifras)
factor(1,cifra) = mod(factor(1,cifra),10)
enddo
! :
! Same here:
do cifra=ncifras,1,-1
factor(2,cifra-1) = factor(2,cifra)/10
factor(2,cifra) = mod(factor(2,cifra),10)
enddo
!:
do cifra=ncifras,1,-1
sumaprovisional(cifra-1) = sumaprovisional(cifra-1)+(sumaprovisional(cifra)/10)
sumaprovisional(cifra) = mod(sumaprovisional(cifra),10)
enddo
In the next case, the value of cifra - (fila - 1) goes out of bounds:
do fila=1,nfilas
do cifra=1,ncifras
! Out of bounds for all cifra < fila:
sumando(fila,cifra-(fila-1)) = factor(1,cifra)*factor(2,ncifras-(fila-1))
enddo
sumaprovisional = sumaprovisional+sumando(fila,:)
enddo
You should be fine if you rewrite the first three loops as do cifra = ncifras, 2, -1 and the inner loop of the other case as do cifra = fila, ncifras. Also, in the example program you posted, you first have to allocate resultado properly before passing it to the subroutine.

compilers - Instruction Selection for type declarations in AST

I'm learning compilers and creating a code generator for a simple language that deals with two types: characters and integers.
After the user input has been scanned by the scanner and then parsed by the parser, I get an AST representation of the input. I have made a code generation for an even simpler language which only processes expressions with integers, operators and variables.
However with this new language I sometimes get a subtree for a type declaration, like this:
(IS TYPE (x) (INT))
which says x is of type INT.
Should there be a case in my code generator which deals with these type declarations? Or is this simply for the semantic analyzer to type check, so I should just assume the types have been checked and ignore this part of the tree and simply assign the value for x?
Both situations are possible, you need to describe more about your language, to see if you really need to add that feature to your code generator, or skip it as unnecessary, and avoid extra work with this difficult and interesting topic of designing a programming language.
Is you "code generator" a program that recieves as an input code in a programming language (maybe small one) and outputs code in another programming language (maybe small one) ?
This tool is usually called a "translator".
Is you "code generator" a program that receive as an input a programming language and outputs assembler / bytecode like programming language ?
This tool is usually called a "compiler".
Note: "pile" is a synonym for "stack".
Usually an A.S.T., stores the type of an operation, or function call. Example, in c:
...
int a = 3;
int b = 5;
float c = (float)(a * b);
...
The last line, generates an A.S.T. similar to this, (skip A.S.T. for other lines):
..................................................................
..................................................................
......................+--------------+............................
......................| [root] |............................
......................| (no type) = |............................
......................+------+-------+............................
.............................|....................................
.................+-----------+------------+.......................
.................|........................|.......................
...........+-----+-----+....+-------------+-------------+.........
...........| (int) c |....| (float) (cast operation) |.........
...........+-----------+....+-------------+-------------+.........
..........................................|.......................
....................................+-----+-----+.................
....................................| (int) () |.................
....................................+-----+-----+.................
..........................................|.......................
....................................+-----+-----+.................
....................................| (int) * |.................
....................................+-----+-----+.................
..........................................|.......................
..............................+-----------+-----------+...........
..............................|.......................|...........
........................+-----+-----+...........+-----+-----+.....
........................| (int) a |...........| (float) b |.....
........................+-----------+...........+-----------+.....
..................................................................
..................................................................
Note that the "(float)" cast its like an operator or a function,
similar to your question.
Good Luck.
If this is a declaration
(IS TYPE (x) (INT))
then x should be laid out in memory. In the case of C and automatic variables, local auto variables are allocated on stack. To allocate needed size of stack you should know sizes of all local vars and sizes are from types.
If this variable is stored in a register, you should select a register of needed size (think about x86 with: AL, AX, EAX, RAX - the same register with different sizes), if your target has such.
Also, type is needed when there is an ambiguous operation in AST, which can operate on different data sizes (e.g. char, short, int - or 8-bit, 16-bit, 32-bit, etc). And for some assemblers, size of data is encoded into instruction itself; so codegen should remember sizes of variables.
Or, if the type of operation was not recorded in AST, the ADD:
(ADD (x) (y))
may mean both float and int additions (ADD or FADD instructions), so types of x and y are needed in codegen to select right variant.

How to define -1 as a uint64 in a match clause?

let myuint64 = 10uL
match myuint64 with
| -1 -> ()
| _ -> ()
How do I define the given -1 as a uint64 value?
> match 0UL-1UL with
- |System.UInt64.MaxValue -> "-1"
- |_ -> "???"
- ;;
val it : string = "-1"
Let me leave alone the fact that you can't really represent a negative value with a data type that can only store positive values (and zero of course).
If, on the other hand, you were storing it in a signed value, -1 would be stored as all bits set.
So basically, I will assume you want to find a way to represent -1 as a bit-wise value that will be compatible with -1 as a signed value.
The value would then be, in C# and C/C++ syntax, 0xffffffffffffffff. Exactly how to specify that in F# I don't know.
I don't know F# at all, but if it's anything like any other languages, a UInt64 can't be -1. Ever. UInt means unsigned integer, which means it can only represent positive values.
To expand on other answers:
When a type starts with a u it means unsigned. What signed/unsigned means is this:
Numbers are stored using a certain number of bits. In the case of int64 and uint64, 64 bits are used. If the number is signed, the 1st bit is not used as part of the number itself, only the other 63 are. That bit is used to say whether the number is negative. If the number is unsigned, then all bits including the 1st bit are used as part of the number and the number is always non-negative (ie: is positive or 0).
Well you could assign it -1 and on most architectures store the 2's complement in there. The signed and unsigned stuff are really only for the type checking. There is no negative sign in hardware.
I have no idea if f# type checker is smart enough to know that a lexical constant -1 is a negative number and should not be put in a uint64.
C definitely does not care.
#include <stdio.h>
#include <inttypes.h>
main()
{
uint64_t x = -1;
printf("0x%x\n", x); // 0xffffffff
}
if F# will convert it for you then -1UL would work. If not then you can specify it as 0xFFFFFFFFFFFFFFFFUL and add a comment to remember that it's -1.
Don't have the F# tools installed at the moment so I cannot verify this.
If you want to go with a signed int:
-1: int64
but you can't match a negative number to a uint, as others have stated.

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