Integer overflow: Difference between revisions

<br>

If you mask, you can test it in your program:

L 3,=F'1' 1

AR 2,3 add register3 to register2

BO OVERFLOW branch on overflow

....

On the other hand,

you will have the S0C8 system abend code : '''fixed point overflow exception'''

with the same program, if you unmask bit 20 by:

O 1,BITFPO unmask Fixed Overflow

SPM 1 Set Program Mask

...

DS 0F alignment

=={{header|6502 Assembly}}==

===8-Bit Overflow===

These flags will automatically be set or cleared depending on the results of a calculation that can affect them.

CLC

ADC #$01

CLC

ADC #$01

Keep in mind that not all instructions affect the flags in the same way. The only arithmetic instructions that affect the overflow flag are <code>ADC</code> and <code>SBC</code>. Therefore, signed overflow can be "missed" by the CPU very easily if it occurs in other ways:

INX ;although X went from $7F to $80, INX does not affect the overflow flag!

ORA #%10000000 ;accumulator crossed from below $7F to above $80, but ORA doesn't affect the overflow flag.

The same is true for unsigned overflow, but less so since the zero flag can be used as a substitute in these cases.

INX ;the carry flag is not affected by this unsigned overflow, but the zero flag will be set

; so we can detect overflow that way instead!

By default, the CPU will continue with the wrong result, unless you specifically program a branch based on overflow after the calculation. This is because on a hardware level the CPU has no knowledge of whether you intend your data to be signed or unsigned (this is still true even on modern computers).

Unlike in [[Z80 Assembly]], the 6502 has no 16-bit registers or built-in 16-bit arithmetic instructions. It ''can'' perform 16-bit or higher addition and subtraction, by separating the number into 8-bit pieces and operating on them separately. Unfortunately, this means that the 6502's flags cannot look at the number as a whole; only the individual bytes. As a result, the CPU will detect "overflow" when any of the bytes cross the $7F-$80 boundary, regardless of whether the byte is the most significant byte or not. This is another reason why the ability to selectively ignore overflow is handy, as it only counts as signed overflow when the most significant byte crosses the $7F-$80 boundary.

;The result will be stored at $14 and $15.

ADC $13 ;high byte of second operand

STA $15 ;high byte of result

=={{header|68000 Assembly}}==

Overflow happens when certain arithmetic operations result in the most significant byte of the register crossing over from 0x7F to 0x80. (Which byte of the 32-bit register is treated as "most significant" depends on the data size of the last instruction. See the example below)

ADD.W #1,D0 ;DOESN'T SET THE OVERFLOW FLAG, SINCE AT WORD LENGTH WE DIDN'T CROSS FROM 7FFF TO 8000

Like the 6502, the 68000 will continue with the wrong result unless you tell it to stop. As with the majority of computer architectures, whether a value is "signed" or "unsigned" is not actually a property of the value itself, but of the comparators used to evaluate it. Otherwise even unsigned arithmetic would produce overflow errors! There are a few options for handling overflow errors:

In Ada, both predefined and user-defined integer types are in a given range, between Type'First and Type'Last, inclusive. The range of predefined types is implementation specific. When the result of a computation is out of the type's range, the program <b>does not continue with a wrong result, but</b> instead <b>raises an exception</b>.

procedure Overflow is

A := A + 1; -- line 49 -- this will later raise a CONSTRAINT_ERROR

end loop;

{{out}}

Other implementations are at liberty to take any action they wish, including to continue silently with a "wrong" result or to throw a catchable exception (though the latter would require at least one addition to the standard prelude so as to provide the handler routine(s).

print (max int);

print (1+max int)

{{out}}

<pre>+2147483647

Note that, unlike many other languages, there is no presupposition that Algol 68 is running on a binary computer. The second example code below shows that for variables of mode '''long int''' arithmetic is fundamentally decimal in Algol 68 Genie.

print (long max int);

print (1+ long max int)

END

{{out}}

=={{header|Applesoft BASIC}}==

The integer variable type is a signed 16-bit integer with a range from -32767 to 32767. When an integer variable is assigned a value less than -32767 or greater than 32767, an "?ILLEGAL QUANTITY ERROR" message is displayed and no change is made to the current value of the variable. All of the expressions for assigning the values use floating point.

{{out}}<pre>?ILLEGAL QUANTITY ERROR</pre>

{{out}}<pre>?ILLEGAL QUANTITY ERROR</pre>

{{out}}<pre>?ILLEGAL QUANTITY ERROR</pre>

{{out}}<pre>?ILLEGAL QUANTITY ERROR</pre>

It is possible using a POKE statement to assign the value -32768 which would normally be out of range.

{{out}}<pre>-32768</pre>

Arturo has unlimited-precision integers, without the possibility of an overflow, all with the same <code>:integer</code> type.

big64bit: 9223372036854775808

print big32bit * 2

{{out}}

Since AutoHotkey treats all integers as signed 64-bit, there is no point in demonstrating overflow with other integer types.

A AutoHotkey program does <b>not</b> recognize a signed integer overflow and the program <b>continues with wrong results</b>.

{{out}}

<pre>Testing signed 64-bit integer overflow with AutoHotkey:

Overflow does <b>not</b> trigger an exception (because Axe does not support exceptions). After an overflow the program <b>continues with wrong results</b> (specifically, the value modulo 65536).

Disp 40000+40000▶Dec,i

Disp 32767-65535▶Dec,i

{{out}}

For those with a finite integer range, though, the most common stack cell size is a 32 bit signed integer, which will usually just wrap when overflowing (as shown in the sample output below). That said, it's not uncommon for the last expression to produce some kind of runtime error or OS exception, frequently even crashing the interpreter itself.

v,,,9"="*84 .: ,,"+"*84 .: **:*" }}" ,+55 .-\0-1<

>:+. 55+, ::0\- :. 48*"-",, \:. 48*"="9,,, -. 55v

v.*: ,,,,,999"="*84 .: ,,"*"*84 .: *+8*7"s9" ,+<

{{out}}

An overflow for signed integer arithmetic is undefined behavior.

A C program does <b>not</b> recognize a signed integer overflow and the program <b>continues with wrong results</b>.

int main (int argc, char *argv[])

printf("%lu\n", 4294967296LU * 4294967296LU);

return 0;

{{out}}

The default behavior can be changed with a compiler flag.

public class IntegerOverflow

}

{{out}}

<pre>

{{works with | g++ | 4.7}}

#include <cstdint>

#include <limits>

<< 4294967296LU * 4294967296LU << '\n';

return 0;

{{out}}

By default, Clojure throws Exceptions on overflow conditions:

(+ 5000000000000000000 5000000000000000000)

(- -9223372036854775807 9223372036854775807)

{{out}} for all of the above statements:

<pre>ArithmeticException integer overflow clojure.lang.Numbers.throwIntOverflow</pre>

If you want to silently overflow, you can set the special *unchecked-math* variable to true or use the special operations, unchecked-add, unchecked-multiply, etc..

(* -1 (dec -9223372036854775807)) ;=> -9223372036854775808

(+ 5000000000000000000 5000000000000000000) ;=> -8446744073709551616

; Note: The following division will currently silently overflow regardless of *unchecked-math*

; See: http://dev.clojure.org/jira/browse/CLJ-1253

Clojure supports an arbitrary precision integer, BigInt and alternative math operators suffixed with an apostrophe: +', -', *', inc', and dec'. These operators auto-promote to BigInt upon overflow.

=={{header|COBOL}}==

COBOL uses decimal arithmetic, so the examples given in the specification are not directly relevant. This program declares a variable that can store three decimal digits, and attempts to assign a four-digit number to it. The result is that the number is truncated to fit, with only the three least significant digits actually being stored; and the program then proceeds. This behaviour may sometimes be what we want.

PROGRAM-ID. PROCRUSTES-PROGRAM.

DATA DIVISION.

MOVE 1002 TO X.

DISPLAY X UPON CONSOLE.

{{out}}

<pre>002</pre>

A small example:

program-id. overflowing.

goback.

{{out}}

=={{header|Computer/zero Assembly}}==

Arithmetic is performed modulo 256; overflow is not detected. This fragment:

ADD one

ff: 255

causes the accumulator to adopt the value 0. With a little care, the programmer can exploit this behaviour by treating each eight-bit word as either an unsigned byte or a signed byte using two's complement (although the instruction set does not provide explicit support for negative numbers). On the two's complement interpretation, the code given above would express the computation "–1 + 1 = 0".

Additionally, standard functions are available to perform arithmetic on int, long, uint, ulong values that modify a boolean value to signal when an overflow has occurred.

import std.stdio;

immutable r = muls(46_341, 46_341, overflow);

writeln("\n", r, " ", overflow);

{{out}}

<pre>Signed 32-bit:

=={{header|FreeBASIC}}==

For the 64-bit integer type a FreeBASIC program does <b>not</b> recognize a signed integer overflow and the program <b>continues with wrong results</b>.

' The suffixes L, LL, UL and ULL are added to the numbers to make it

Print

Print "Press any key to quit"

{{out}}

[http://play.golang.org/p/jsPWC8KGzD Run this in the Go playground].

A Go program does <b>not</b> recognize an integer overflow and the program <b>continues with wrong results</b>.

import "fmt"

u64 = 4294967296

fmt.Printf(" %d * %d: %d\n", u64, u64, u64*u64)

{{out}}

<pre>32 bit signed integers

Groovy does not recognize integer overflow in any bounded integral type and the program '''continues with wrong results'''. All bounded integral types use ''2's-complement'' arithmetic.

assert -(-2147483647-1) != 2147483648g

println(-(-2147483647-1))

println(3037000500g * 3037000500g)

assert (-9223372036854775807g-1g).intdiv(-1) == 9223372036854775808g

Output:

=={{header|Haskell}}==

Haskell supports both fixed sized signed integers (Int) and unbounded integers (Integer). Various sizes of signed and unsigned integers are available in Data.Int and Data.Word, respectively. The Haskell 2010 Language Report explains the following: "The results of exceptional conditions (such as overflow or underflow) on the fixed-precision numeric types are undefined; an implementation may choose error (⊥, semantically), a truncated value, or a special value such as infinity, indefinite, etc" (http://www.haskell.org/definition/haskell2010.pdf Section 6.4 Paragraph 4).

import Data.Word

import Control.Exception

f ((10000000000000000000 + 10000000000000000000) :: Word64)

f ((9223372036854775807 - 18446744073709551615) :: Word64)

{{out}}

<pre>-2147483648

Also, negative numbers do not use - for the negative sign in J (a preceding - means to negate the argument on the right - in some cases this is the same kind of result, but in other cases it's different). Instead, use _ to denote negative numbers. Also, J does not use / for division, instead J uses % for division. With those changes, here's what the results look like in a 32 bit version of J:

2.14748e9

2000000000 + 2000000000

_9.22337e18

4294967296 * 4294967296

And, here's what it looks like in a 64 bit version of J:

2147483648

2000000000 + 2000000000

_9.22337e18

4294967296 * 4294967296

That said, note that the above was with default 6 digits of "printing precision". Here's how things look with that limit relaxed:

32 bit J:

2147483648

2000000000 + 2000000000

_9223372036854775800

4294967296 * 4294967296

64 bit J:

2147483648

2000000000 + 2000000000

_9223372036854775800

4294967296 * 4294967296

Finally, note that both versions of J support arbitrary precision integers. These are not the default, for performance reasons, but are available for cases where their performance penalty is acceptable.

The type int is a signed 32-bit integer and the type long is a 64-bit integer.

A Java program does <b>not</b> recognize an integer overflow and the program <b>continues with wrong results</b>.

public static void main(String[] args) {

}

{{out}}

''The task''

def compare:

if type == "string" then "\n\(.)\n"

[4294967296 * 4294967296, "18446744073709551616"]

===jq 1.6===

=={{header|Julia}}==

'''Plain Integer Types and Their Limits'''

S = subtypes(Signed)

U = subtypes(Unsigned)

@printf("%8s: [%s, %s]\n", s, typemin(s), typemax(s))

@printf("%8s: [%s, %s]\n", u, typemin(u), typemax(u))

{{out}}

<pre>Integer limits:

Julia does not throw an explicit error on integer overflow.

for t in S

over = typemax(t) + one(t)

@printf("%8s → %-25s (%s)\n", t, over, typeof(over))

{{out}}

A Kotlin program does <b>not</b> recognize a signed integer overflow and the program <b>continues with wrong results</b>.

/* Kotlin (like Java) does not have unsigned integer types but we can simulate

println((2147483647L - 4294967295L.toUInt()).toUInt())

println((65537L * 65537L).toUInt())

{{out}}

=={{header|Ksh}}==

#!/bin/ksh

(( LONG_INT = 2**63 -1 )) ; print " LONG_INT (2^63 -1) = $LONG_INT"

(( LONG_INT = 2**63 )) ; print " LONG_INT (2^63) : $LONG_INT"

{{out}}<pre>

SHORT_INT (2^15 -1) = 32767

Lingo uses 32-bit signed integers.

A Lingo program does <b>not</b> recognize a signed integer overflow and the program <b>continues with wrong results</b>.

-- -2147483648

put (-2147483647-1) / -1

=={{header|Lua}}==

Lua 5.3+ supports integer and floating sub-types of its generic number type. The ''standard'' implementation is 64-bit signed, under/overflow is not recognized.

minint, maxint = math.mininteger, math.maxinteger

print("min, max int64 = " .. minint .. ", " .. maxint)

print("min-1 underflow = " .. (minint-1) .. " equals max? " .. tostring(minint-1==maxint))

{{out}}

<pre>min, max int64 = -9223372036854775808, 9223372036854775807

=={{header|M2000 Interpreter}}==

Long A

Try ok {

Print Hex$(Eval(DataMem, 0!b))="BBBBAAAA"

Print Eval(DataMem, 0!b)=Eval(DataMem, 0!a2)*0x10000+Eval(DataMem, 0!a1)

=={{header|Mathematica}}/{{header|Wolfram Language}}==

=={{header|Nim}}==

Catching an overflow (when --panics is off) is done this way:

var x: int32 = -2147483647

x = -(x - 1) # Raise overflow.

echo x

except OverflowDefect:

It is possible to tell the compiler to not generate code to detect overflows by using pragmas “push” and “pop”:

try:

var x: int32 = -2147483647

except OverflowDefect:

echo "Overflow detected" # Not executed.

It is also possible to suppress all overflow checks by using compile option <code>--overflowChecks:off</code>. Also, compiling with option <code>-d:danger</code> suppress these checks and several others.

This program presents the behavior when overflow checks are suppressed. Remember that for signed integers, this is not the normal behavior and that the result is always wrong when an overflow occurs.

{.push overflowChecks: off.}

var a: int32

echo " 9223372036854775807 - 18446744073709551615 gives ", d # 9223372036854775808.

d = 4294967296u64 * 4294967296u64

{{out}}

=={{header|PARI/GP}}==

Machine-sized integers can be used inside a <code>Vecsmall</code>:

{{out}}

<pre>%1 = Vecsmall([1])

Using Perl 5.18 on 64-bit Linux with use integer:

The Perl 5 program below does <b>not</b> recognize a signed integer overflow and the program <b>continues with wrong results</b>.

use strict;

use warnings;

say(3037000500 * 3037000500);

say((-9223372036854775807-1) / -1);

{{out}}

<pre>

Integer overflow is handled by automatic promotion to atom (an IEEE float, 64/80 bit for the 32/64 bit implementations respectively),

which triggers a run-time type check if stored in a variable declared as integer, eg:

<span style="color: #004080;">integer</span> <span style="color: #000000;">i</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">1000000000</span> <span style="color: #0000FF;">+</span> <span style="color: #000000;">1000000000</span>

{{out}}

<pre>

{{works with|8080 PL/M Compiler}} ... under CP/M (or an emulator)<br>

8080 PL/M does not check for overflow, incrementing the largest integer values wraps around to 0 (numbers are insigned in 8080 PL/M) and the program <b>continues with wrong results</b>.

/* CP/M SYSTEM CALL */

CALL PRNL;

{{out}}

<pre>

https://docs.microsoft.com/en-us/dotnet/api/system.decimal?view=netframework-4.8#remarks

try {

# All of these raise an exception, which is caught below.

$Error.Exception

}

=={{header|PureBasic}}==

CPU=x64, OS=Windows7

#MAX_ASCII=255 ;=MAX_CHAR Ascii-Mode

say("Quad",q1,q2,SizeOf(q1))

{{out}}

<pre>

Python 2.X has a 32 bit signed integer type called 'int' that automatically converts to type 'long' on overflow. Type long is of arbitrary precision adjusting its precision up to computer limits, as needed.

Type "copyright", "credits" or "license()" for more information.

>>> for calc in ''' -(-2147483647-1)

Expression: '46341 * 46341' evaluates to 2147488281 of type <type 'long'>

Expression: '(-2147483647-1) / -1' evaluates to 2147483648 of type <type 'long'>

===Python 3.x===

Python 3.X has the one 'int' type that is of arbitrary precision. Implementations ''may'' use 32 bit integers for speed and silently shift to arbitrary precision to avoid overflow.

Type "copyright", "credits" or "license()" for more information.

>>> for calc in ''' -(-2147483647-1)

Expression: '46341 * 46341' evaluates to 2147488281 of type <class 'int'>

Expression: '(-2147483647-1) / -1' evaluates to 2147483648.0 of type <class 'float'>

Note: In Python 3.X the division operator used between two ints returns a floating point result, (as this was seen as most often required and expected in the Python community). Use <code>//</code> to get integer division.

The unsafe operations expects fixnums in the arguments, and that the result is also a fixnum. They don't autopromote the result. They are faster but they should be used only in special cases, where the values known to be bounded. We can use them to see the behavior after an overflow. In case of a overflow they have undefined behaviour, so they may give different results or change without warning in future versions. (I don't expect that they will change soon, but there is no official guaranty.)

(require racket/unsafe/ops)

(/ -1073741824 -1) ;==> 1073741824

The 64-bit version is similar. The fixnum are effectively 63-bits signed integers.

The Raku program below does <b>not</b> recognize a signed integer overflow and the program <b>continues with wrong results</b>.

{{out}}

<pre>-9223372036854775808

For newer versions of REXX, the &nbsp; '''signal on lostDigits''' &nbsp; statement can be used to accomplish the same results &nbsp;

<br>(for detecting a loss of significance [digits]).

numeric digits 9 /*the REXX default is 9 decimal digits.*/

call showResult( 999999997 + 1 )

else _=' [underflow]' /*did it underflow? */

say right(x, 20) _ /*show the result. */

'''output''' &nbsp; using the default input(s): <br><br>

Output note: &nbsp; (as it happens, all of the results below are numerically correct)

Bignum objects are created automatically when integer calculations would otherwise overflow a Fixnum.

When a calculation involving Bignum objects returns a result that will fit in a Fixnum, the result is automatically converted.

=> 4611686018427387903

2.1.1 :002 > a.class

2.1.1 :004 > (b-1).class

=> Fixnum

Since Ruby 2.4 these different classes have disappeared: all numbers in above code are of class Integer.

The following code will always panic when run in any mode

// The following will panic!

let i32_1 : i32 = -(-2_147_483_647 - 1);

let i64_4 : i64 = 3_037_000_500 * 3_037_000_500;

let i64_5 : i64 = (-9_223_372_036_854_775_807 - 1) / -1;

In order to declare overflow/underflow behaviour as intended (and, thus, valid in both debug and release modes), Rust provides two mechanisms:

<br>

// The following will never panic!

println!("{:?}", 65_537u32.checked_mul(65_537)); // None

println!("{:?}", 65_537i32.saturating_mul(65_537)); // 2147483647

println!("{:?}", 65_537i32.wrapping_mul(-65_537)); // -131073

Second, a generic <code>Wrapping<T></code> one-element tuple type is provided which implements the same basic operations as the <code>wrapping_...</code> methods, but allows you to use normal operators and then use the <code>.0</code> field accessor to retrieve the value once you are finished.<ref>{{Cite web |url=https://doc.rust-lang.org/std/num/struct.Wrapping.html |title=Struct std::num::Wrapping |website=The Rust Standard Library |access-date=2019-11-18}}</ref>

{{works with|Java|8}}

Math.addExact works for both 32-bit unsigned and 64-bit unsigned integers, but Java does not support signed integers.

def requireOverflow(f: => Unit) =

println("Test - Expect Undetected overflow:")

requireOverflow(++(1,1)) // Undetected overflow

=={{header|Seed7}}==

The type [http://seed7.sourceforge.net/manual/types.htm#integer integer] is a 64-bit signed integer type.

All computations with the type integer are checked for overflow.

const proc: writeResult (ref func integer: expression) is func

writeResult(3037000500 * 3037000500);

writeResult((-9223372036854775807-1) div -1);

{{out}}

{{trans|Raku}}

Sidef has unlimited precision integers.

{{out}}

<pre>

However, to emulate wrong behavior (eg. when interfacing to external programs or document formats), it can be emulated.

{{works with|Smalltalk/X}}

2147483647 add_32: 1 -> -2147483648

4294967295 + 1. -> 4294967296

16rFFFFFFFF add_32u: 1. -> 0

=={{header|Swift}}==

// However, you can opt-in to overflow behavior with the overflow operators and continue with wrong results

println(uInt64)

uInt64 = 4294967296 &* 4294967296

{{out}}

<pre>

=={{header|Standard ML}}==

PolyML

poly: : error: Overflow exception raised while converting ~9223372036854775807 to int

Int.maxInt ;

2147483648 * 2147483648 ;

Exception- Overflow raised

=={{header|Tcl}}==

Tcl (since 8.5) uses logical signed integers throughout that are “large enough to hold the number you are using” (being internally anything from a single machine word up to a bignum). The only way to get 32-bit and 64-bit values in arithmetic is to apply a clamping function at appropriate points:

expr {$x<0 ? -((-$x) & 0x7fffffff) : $x & 0x7fffffff}

}

Tcl 8.4 used a mix of 32-bit and 64-bit numbers on 32-bit platforms and 64-bit numbers only on 64-bit platforms. Users are recommended to upgrade to avoid this complexity.

=={{header|True BASIC}}==

PRINT -(-2147483647-1) !-2147483648

PRINT 2000000000 + 2000000000 !4000000000

!returns the largest number that can be represented in your computer

END WHEN

=={{header|VBScript}}==

<br>- Yes, because typename(2147483647)="Long" and typename(2147483648)="Double", so we have switched from fixed binary integer to double floating point. But thanks to mantissa precision there is no harm. The integer overflow is when you reach 10^15, because you are now out of the integer set : (1E+15)+1=1E+15 !?.

<br>A good way to test integer overflow is to use the vartype() or typename() builtin functions.

i=(-2147483647-1)/-1

wscript.echo i

i1=1000000000000000-1 '1E+15-1

i2=i1+1 '1E+15

{{out}}

<pre>

{{works with|Visual Basic|VB6 Standard}}

Overflow is well handled, except for a strange bug in the computation of f the constant -(-2147483648).

Dim i As Long '32-bit signed integer

i = -(-2147483647 - 1) '=-2147483648 ?! bug

i = 46341 * 46341 'Run-time error '6' : Overflow

i = (-2147483647 - 1) / -1 'Run-time error '6' : Overflow

'''Error handling - method 1'''

On Error Resume Next

i = 2147483647 + 1

Debug.Print i 'i=0

'''Error handling - method 2'''

On Error GoTo overflow

i = 2147483647 + 1

overflow:

Debug.Print "Error: " & Err.Description '-> Error: Overflow

'''Error handling - method 3'''

i = 2147483647 + 1 'Run-time error '6' : Overflow

Debug.Print i

=={{header|Visual Basic .NET}}==

'''32-bit signed integer'''

Pre-compilation error:

'Error: Constant expression not representable in type 'Integer'

for:

i = 0 - (-2147483647 - 1)

i = -(-2147483647L - 1)

i = -2147483647 - 2147483647

i = 46341 * 46341

Execution error:

'An unhandled exception of type 'System.OverflowException' occurred

'Additional information: Arithmetic operation resulted in an overflow.

for:

'''32-bit unsigned integer'''<br>

In Visual Basic .Net there is no specific UInteger constants as in C.

Pre-compilation error:

'Error: Constant expression not representable in type 'UInteger'

for:

i = 3000000000 + 3000000000

i = 2147483647 - 4294967295

Execution error:

'An unhandled exception of type 'System.OverflowException' occurred

'Additional information: Arithmetic operation resulted in an overflow.

for:

'''64-bit signed integer'''

Pre-compilation error:

'Error: Constant expression not representable in type 'Long'

for:

i = 5000000000000000000 + 5000000000000000000

i = -9223372036854775807 - 9223372036854775807

i = 3037000500 * 3037000500

Execution error:

'An unhandled exception of type 'System.OverflowException' occurred

'Additional information: Arithmetic operation resulted in an overflow.

for:

'''64-bit unsigned integer'''<br>

In Visual Basic .Net there is no specific ULong constants as in C.

And 'Long' constants are not good enough.

Pre-compilation error:

'Error: Overflow

for:

i = 10000000000000000000 + 10000000000000000000

Pre-compilation error:

'Error: Constant expression not representable in type 'Long'

for:

Execution error:

'An unhandled exception of type 'System.OverflowException' occurred

'Additional information: Arithmetic operation resulted in an overflow.

for:

'''how the exception is catched'''

Try

i = -2147483647 : i = -(i - 1)

Catch ex As Exception

Debug.Print("Exception raised : " & ex.Message)

{{out}}

<pre>

However, within this overall framework, Wren also has an unsigned 32-bit integer sub-system when dealing with bitwise operations. All values are converted internally to such integers before the corresponding C bitwise operation is performed (Wren's VM is written in C) and can therefore overflow without warning. Fortunately, we can easily observe these effects by performing the operations required by the task and then (for example) right shifting them by 0 places.

System.print("Unsigned 32-bit:")

{{out}}

continues with wrong results.

[N:= -(-2147483647-1);

IntOut(0, N); CrLf(0);

N:= (-2147483647-1)/-1;

IntOut(0, N); CrLf(0);

{{out}}

* <code>PO</code> parity odd, no overflow

add 1

Like other CPUs, the Z80 has no way of knowing whether a value is intended to be signed or unsigned, and unless you explicitly have a jump, call, or return based on overflow after a calculation, the CPU '''will continue with the wrong result.'''

zkl uses C's 64 bit integer math and the results are OS dependent. Integers are signed. GMP can be used for big ints.

A zkl program does <b>not</b> recognize an integer overflow and the program <b>continues with wrong results</b>.

println(-(-9223372036854775807-1));

println(5000000000000000000+5000000000000000000);

println(-9223372036854775807 - 9223372036854775807);

println(3037000500 * 3037000500);

{{out}}

Linux/BSD/clang