An enumerated type is essentially a named integral value. That enumerated type is associated with an underlying integral type that can represent all the named values. That underlying integral type is required to be able to represent all the unique named values, but its actual type is implementation defined. In...

The integers you are trying to use are larger than 2^53. JavaScript cannot represent those integers precisely. Lets have a look at the console: > var i = 9007199254740993; undefined > i++ 9007199254740992 > i++ 9007199254740992 > i++ 9007199254740992 ... As you can see, due to loss of precision, the...

c,integer,overflow,integer-overflow

Let's take this example: 1 << 32 If we assume 32-bit int, C clearly says it is undefined behavior. Period. But any implementation can define this undefined behavior. gcc for example says (while not very explicit in defining the behavior): GCC does not use the latitude given in C99 only...

java,c#,language-lawyer,integer-overflow

On platforms were signed values that overflow are defined as wrapping, signed and unsigned values will behave in isomorphic fashion when fed to the +, -, *, &, ^, |, <<, and ~ operators, or when performing an unchecked cast to a smaller type. They will behave differently when used...

java,long-integer,integer-overflow

1 << 31 is an int, not a long. And what you have is not really an overflow, but Integer.MIN_VALUE (sign bit set, all the rest zeroes). Unless you suffix them appropriately, any numeric constant in Java is an int by default. If you want a long, you have to...

python,arrays,numpy,integer-overflow

total_img = 255*(min - total_img)/(min - max) You literally just plug in total_img instead of px, and it does the operation elementwise. If you want to store the result into the original total_img instead of replacing it, you can slice-assign: total_img[:] = 255*(min - total_img)/(min - max) but note that...

function multiply31($num) { return (int)(($num << 5) - $num); } Turn the multiply into add seems work....

assembly,integer-overflow,carryflag

If you have a look at the pseudo algorithm of IMUL, you'll see IF OperandSize = 16 THEN TMP_XP ← AX ∗ SRC (* Signed multiplication; TMP_XP is a signed integer at twice the width of the SRC *) DX:AX ← TMP_XP[31:0]; SF ← TMP_XP[15]; IF SignExtend(TMP_XP[15:0]) = TMP_XP THEN...

java,math,expression,integer-overflow

((x + 231) mod 232) - 231 Is this what you're looking for? That should be the result of any mathematical operation on a machine that uses 32-bit signed 2's complement integers. That is, if the mathematical value of an operation returns x, the above formula gives the integer that...

The simple answer is to use strtoul() instead. The longer answer is that even if all you needed was signed 32 bit integers or were happy with 31 bits for unsigned, the atoi() function is a poor fit for what you appear to be doing. As you have already noted,...

javascript,bit-manipulation,unsigned,bit-shift,integer-overflow

The shift operators always effectively has a right operand in the range 0-31. From the Mozilla docs: Shift operators convert their operands to 32-bit integers in big-endian order and return a result of the same type as the left operand. The right operand should be less than 32, but if...

c++,c,cross-platform,standards,integer-overflow

What guarantees (in C or C++ standard) might help to devise the code? C specifies signed integer representation as using 1 of 3 forms: sign and magnitude, two’s complement, or ones’ complement. Given these forms, only division by 0 and two’s complement division of INT_MIN/-1 may overflow. What (cross-platform)...

python,c,linux,integer-overflow,iostat

The following function will work for 32-bit counters: def calc_32bit_diff(old, new): return (new - old + 0x100000000) % 0x100000000 print calc_32bit_diff(1, 42) print calc_32bit_diff(2147483647, -2147483648) print calc_32bit_diff(-2147483648, 2147483647) This obviously won't work is the counter wraps around more than once between two consecutive reads (but then no other method would...

c,percentage,integer-overflow,integer-division

Diagnosis The value you expect is, presumably, 91. The problem appears to be that your compiler is using 16-bit int values. You should identify the platform on which you're working and include information about unusual situations such as 16-bit int types. It is reasonable for us to assume 32-bit or...

If you look at the source for Word8's Num instance, you'll see that everything is done by converting to a Word# unboxed value and performing operations on that, and then narrowing down to a 8-bit value. I suspected that doing comparison on that Word# value would be more efficient, so...

c,type-conversion,integer-overflow

This answer assumes that float is an IEEE-754 single precision float encoded as 32-bits, and that an int is 32-bits. See this Wikipedia article for more information about IEEE-754. Floating point numbers only have 24-bits of precision, compared with 32-bits for an int. Therefore int values from 0 to 16777215...

You have left < right by definition. As a consequence, right - left > 0, and furthermore left + (right - left) = right (follows from basic algebra). And consequently left + (right - left) / 2 <= right. So no overflow can happen since every step of the operation...

sql,sql-server,integer-overflow,gaps-and-islands

Try This: with cte as ( select start = (cast(m.API_WellNo as bigint) + 1) , [stop] = ca.[stop] from tblWellMaster m cross apply ( select top 1 [stop]=(cast(x.API_WellNo as bigint) -1) from tblWellMaster x where x.API_WellNo > m.API_WellNo order by x.API_WellNo ) as ca where not exists ( select 1...

symfony2,doctrine2,integer-overflow

I've solved the problem. First, I've defined a custom id generator: class MyIdGenerator extends AbstractIdGenerator { public function generate(EntityManager $em, $entity) { $stmt = $em->getConnection()->prepare(sprintf('SELECT %s.nextval id FROM DUAL', $entity->getGeneratorName())); $stmt->execute(); $rs = $stmt->fetch(); return $rs['ID']; } } As you can see, in all my entities I've defined a function...

[to long for a comment] Nested functions are not supported in C. Properly indented C sources might look like this: #include <stdio.h> #include <stdlib.h> #include <string.h> /*boolean */ #define true 1 #define false 0 typedef int bool; void ShowUsage() { printf("Integer Overflow Check before performing an arithmetic.\n" "=======================================================\n" "Usage:\n" "Integer...

python,matlab,numpy,integer-overflow

So... I got curious and dug around a little. As I already mentioned in the comments, a "largest finite value" kind of exists in IEEE 754, if you consider the exception status flags. A value of infinity with the overflow flag set corresponds to your proposed LFV, with the difference...

vb.net,application-settings,integer-overflow

2 billion is the maximum value for the Integer data type, you can use Long instead to allow you to go up to about 9 * 10^18. If you want to use numbers that can be of any size at all, you should look at System.Numerics.BigInteger. You will need to...

c++,long-integer,modulus,integer-overflow

Many compilers offer a 128-bit integral type. For example, with g++ you can make a function static inline int64_t mulmod(int64_t x, int64_t y, int64_t m) { return ( (__int128_t)x * y) % m; } Aside: if you can, try to stick to unsigned integer types when you're doing modular arithmetic....

python,algorithm,python-3.x,integer-overflow,pow

Is there a version of math.pow that doesn't overflow? Try using the built-in exponentiation operator, **. AFAIK it doesn't have the same limitations that math.pow does. >>> math.pow(8,342) Traceback (most recent call last): File "<stdin>", line 1, in <module> OverflowError: math range error >>> 8**342...

c++,terminal,user-input,integer-overflow,numeric-limits

You want to input a string and use strtoll on it.

clojure,fibonacci,integer-overflow

The default integer type in Clojure is long. If you want to specify that an integer literal should be considered a clojure.lang.BigInt just add an N right after the number. (defn fib-pair [[a b]] [b (+ a b)]) (nth (map first (iterate fib-pair [1N 1N])) 500) ;= 225591516161936330872512695036072072046011324913758190588638866418474627738686883405015987052796968498626N You can...

The problem is with this line: b, t, nl = 0, 0, 0; It uses comma operator on both sides of assignment, hence only nl is intialized with zero. There are no side-effects for evaluation of b, t on left side* and two trailing zeros on the right side of...

It is possible to simulate 32-bit integer by "abusing" the bitwise operators available in JavaScript (since they can only return integers within that range). To convert to a signed 32-bit integer: x = (a * b) | 0; To convert to an unsigned 32-bit integer: x = (a * b)...

The operetor precedence rules put the c-style cast (long int) at the same priority as * and with assiociativity from right to left. So first the multiplication is done and then the casting. Try else if( ( (( long int ) workingnumber) * 3 ) + 1 > INT_MAX )...

According to the MIPS instruction reference, the only addition operations which can produce overflow exceptions are the signed addition instructions: ADD ADDI MIPS integers are 32-bit, and since you'll be using unsigned integers, the maximum value is 231-1 (aka 2147483647 or hex 7FFFFFFF). Thus any addition which results in a...

vb.net,memory,byte,integer-overflow

This is not the proper way of adding an offset to an IntPtr. It provides the Add method for doing that. I am assuming that you are already marshaling the BaseAddress field of the MEMORY_BASIC_INFORMATION structure as an IntPtr. So you should change your code to Addy = IntPtr.Add(MemInfo.BaseAddress, MemInfo.RegionSize);...