integer-overflow | 易学教程

Reproduce behavior MAX_VALUE and MIN_VALUE

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问题 The following also applied to other MIN_VALUE and MAX_VALUE , but let's only focus on Integer for now. I know that in Java integers are 32-bit, with Integer.MAX_VALUE = 2147483647 (2 31 -1) and Integer.MIN_VALUE = -2147483648 (-2 31 ). When calculating with these values when you go beyond their bounds, the number wraps around / overflows. So when you do something like Integer.MAX_VALUE + 1 , the result is the same as Integer.MIN_VALUE . Here are some basic arithmetic calculations with MIN

How to check if a number overflows an 'int' [duplicate]

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问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Best way to detect integer overflow in C/C++ I was asked this question in an interview : "Convert a string representation of a number into an integer". But if the number exceeds the max value that can be stored in a 32 bit integer, it must throw an error. My question is how can we check if a number overflows a 32-bit unsigned int ? 回答1: (I looked at the possible duplicate to this question and I'm uncertain how

Metafunction to compute x^n and return the integer limit without overflow if not possible?

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问题 Consider the following code: template <std::intmax_t Base, std::intmax_t Exponent> struct integer_power_bounded { static_assert(Exponent >= 0, "Error in 'integer_power_bounded': 'Exponent >= 0' is false"); static constexpr std::intmax_t value = /* something */; }; template <std::intmax_t Base> struct integer_power_bounded<Base, 0> { static constexpr std::intmax_t value = 1; }; Instead of /* something */ , I would like to return std::numeric_limits<std::intmax_t>::min() or std::numeric_limits

How can i define a List of checked integers

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I have a List of Integers defined as List<int> myIntList = new List<int>(); As usual I will add value to the list using myIntList.Add() method. The problem I am facing is that the values in the list are dynamic(result of some calculation) that may exceed the maximum value that an integer can hold. Consider the following scenario: int x = int.MaxValue; myIntList.Add(x + 1); This will add -2147483648 to the list instead of throwing an exception. I need to throw an exception here. I know myIntList.Add(checked(x + 1)); will do the job perfectly or I can even enclose the myIntList.Add() within

How to compile and run an optimized Rust program with overflow checking enabled

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问题 I'm writing a program that's quite compute heavy, and it's annoyingly slow to run in debug mode. My program is also plagued by integer overflows, because I'm reading data from u8 arrays and u8 type spreads to unexpected places via type inference, and Rust prefers to overflow rather than to promote integers to larger types. Building in release mode disables overflow checks: cargo run --release How can I build Rust executable with optimizations and runtime overflow checks enabled as well? 回答1:

How to suppress overflow-checking in PowerShell?

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PowerShell seems to perform bounds-checking after arithmetic operations and conversions. For instance, the following operations fail: [byte]$a = 255 $a++ $a = [byte]256 Is there any way to enforce overflows or the typecast without resorting to a manual calculation via modulo or C# and Add-Type? The behavior you want in PowerShell is achievable, though, it's a bit of a hack; and maybe there's a better way. If you just want cryptographic functionality though, it's worth calling out, that there's a TON of that already built-in to the BCL, and it's fully accessible from PowerShell (MD5, SHA, RSA,

why left+(right-left)/2 will not overflow?

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In this article: http://googleresearch.blogspot.sg/2006/06/extra-extra-read-all-about-it-nearly.html , it mentioned most quick sort algorithm had a bug (left+right)/2, and it pointed out that the solution was using left+(right-left)/2 instead of (left+right)/2 . The solution was also given in question Bug in quicksort example (K&R C book)? My question is why left+(right-left)/2 can avoid overflow? How to prove it? Thanks in advance. You have left < right by definition. As a consequence, right - left > 0 , and furthermore left + (right - left) = right (follows from basic algebra). And

What would happen if I were to make more references to Objects than 32 bits can account for?

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So I just learned when you declare a variable of type Object ( i.e. Object a; ), a 32-bit space is allocated for that variable. Inside this variable/reference, there is a memory address to an actual Object. Now let's pretend I have a large enough amount of memory to do this. What would happen if I created more than 4,294,967,296 (2 32 ) variables of type Object and tried assigning them to a distinct Object? Would some variables/references get the same memory addresses due to integer overflow? Meaning it's impossible to have references to more than 4,294,967,296 Objects in memory? So I just

what's the difference between mid=(beg+end)/2 and mid=beg+(end-beg)/2 in binary search?

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It is a problem from C++ primer fifth edition problem 3.26, I don't know the difference between them ? May be the second one can avoid overflow. May be the second one can avoid overflow. Exactly. There's no guarantee that beg+end is representable; but in the second case the intermediate values, as well as the expected result, are no larger than end , so there is no danger of overflow. The second form can also be used for affine types like pointers and other random-access iterators, which can be subtracted to give a distance, but not added together. In general case the both expressions are

R: simple multiplication causes integer overflow

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问题 In a longer script I have to multiply the length of a vector A (2614) with the numbers of rows of a dataframe B (1456000). If I do that directly with length(A) * nrow(B) I get the message NAs produced by integer overflow although there's no problem when I multiply the same numbers: 2614 * 1456000 [1] 3805984000 The only way to get the multiplication to work is round(length(A)) * nrow(B) or length(A) * round(nrow(B)) . But the numbers produced by length and nrow must be integers anyhow!