integer-overflow

How disastrous is integer overflow in C++?

天大地大妈咪最大 提交于 2019-11-27 21:12:44
I was just wondering how disastrous integer overflow really is. Take the following example program: #include <iostream> int main() { int a = 46341; int b = a * a; std::cout << "hello world\n"; } Since a * a overflows on 32 bit platforms, and integer overflow triggers undefined behavior, do I have any guarantees at all that hello world will actually appear on my screen? I removed the "signed" part from my question based on the following standard quotes: (§5/5 C++03, §5/4 C++11) If during the evaluation of an expression, the result is not mathematically defined or not in the range of

How to check for signed integer overflow in C without undefined behaviour?

大城市里の小女人 提交于 2019-11-27 19:43:26
There's (1): // assume x,y are non-negative if(x > max - y) error; And (2): // assume x,y are non-negative int sum = x + y; if(sum < x || sum < y) error; Whichs is preferred or is there a better way. Integer overflow is the canonical example of "undefined behaviour" in C (noting that operations on unsigned integers never overflow, they are defined to wrap-around instead). This means that once you've executed x + y , if it overflowed, you're already hosed. It's too late to do any checking - your program could have crashed already. Think of it like checking for division by zero - if you wait

Overflow when multiplying Integers and assigning to Long

烈酒焚心 提交于 2019-11-27 16:03:23
If I type the following into the immediate window I get Runtime error '6': Overflow. MsgBox 24 * 60 * 60 Why is this? This also fails: Dim giveTime As Long giveTime = 24 * 60 * 60 Why is this? giveTime is declared as a Long type, so 24 × 60 × 60 = 86400 should comfortably fit. This is a really odd VBA quirk. I'm amazed I've never bumped into this. Dim x As Long x = 24 * 60 * 60 ' Overflow x = 32767 + 1 ' Overflow. x = 32768 + 1 ' Works fine! So it looks like the * and + operators return an Integer in the first two examples. Sure enough, in the help file for the * operator (similar for the +

Often big numbers become negative

萝らか妹 提交于 2019-11-27 15:32:13
Since I started using eclipse for project euler, I noticed that big numbers sometime become a seemingly random negative numbers. I suppose this has something to do with passing the boudry of the type. I'll be glad if you could explain to me how these negative numbers are generated and what is the logic behind it. Also, how can I avoid them (preferable not with BigInteger class). Danke!=) Goatcat This image shows what you're looking for. In your case it's obviously larger numbers, but the principle stays the same. Examples of limits in java are: int: −2,147,483,648 to 2,147,483,647. long: -9

How is integer overflow exploitable?

|▌冷眼眸甩不掉的悲伤 提交于 2019-11-27 13:39:49
问题 Does anyone have a detailed explanation on how integers can be exploited? I have been reading a lot about the concept, and I understand what an it is, and I understand buffer overflows, but I dont understand how one could modify memory reliably, or in a way to modify application flow, by making an integer larger than its defined memory.... 回答1: It is definitely exploitable, but depends on the situation of course. Old versions ssh had an integer overflow which could be exploited remotely. The

Why does Java think that the product of all numbers from 10 to 99 is 0?

陌路散爱 提交于 2019-11-27 10:24:29
The following block of codes gives the output as 0. public class HelloWorld{ public static void main(String []args){ int product = 1; for (int i = 10; i <= 99; i++) { product *= i; } System.out.println(product); } } Please can somebody explain why this happens? Here is what the program does at each step: 1 * 10 = 10 10 * 11 = 110 110 * 12 = 1320 1320 * 13 = 17160 17160 * 14 = 240240 240240 * 15 = 3603600 3603600 * 16 = 57657600 57657600 * 17 = 980179200 980179200 * 18 = 463356416 463356416 * 19 = 213837312 213837312 * 20 = -18221056 -18221056 * 21 = -382642176 -382642176 * 22 = 171806720

If a 32-bit integer overflows, can we use a 40-bit structure instead of a 64-bit long one?

跟風遠走 提交于 2019-11-27 09:29:19
问题 If, say, a 32-bit integer is overflowing, instead of upgrading int to long , can we make use of some 40-bit type if we need a range only within 2 40 , so that we save 24 (64-40) bits for every integer? If so, how? I have to deal with billions and space is a bigger constraint. 回答1: Yes, but... It is certainly possible , but it is usually nonsensical (for any program that doesn't use billions of these numbers): #include <stdint.h> // don't want to rely on something like long long struct bad

Why, In Java arithmetic, overflow or underflow will never throw an Exception?

有些话、适合烂在心里 提交于 2019-11-27 09:07:27
During Java Arithmetic operation , JVM do not throw Underflow or Overflow Exception. So many time we come across unexpected results and wondering what went wrong. While in case of .NET technology we have Overflow and Undeflow exception. So my question is that , why Java was design not to throw this exception during arithmetic operation This was likely a combination of factors: The big languages prior to Java used unchecked arithmetic. Well-known algorithms prone to numerical overflow tended to account for the potential overflow already without relying on checked arithmetic. Checked arithmetic

How to handle arbitrarily large integers

假装没事ソ 提交于 2019-11-27 08:25:14
I'm working on a programming language, and today I got the point where I could compile the factorial function(recursive), however due to the maximum size of an integer the largest I can get is factorial(12). What are some techniques for handling integers of an arbitrary maximum size. The language currently works by translating code to C++. If you need larger than 32-bits you could consider using 64-bit integers (long long), or use or write an arbitrary precision math library, e.g. GNU MP . If you want to roll your own arbitrary precision library, see Knuth's Seminumerical Algorithms, volume 2

Why does this multiplication integer overflow result in zero?

扶醉桌前 提交于 2019-11-27 06:58:44
问题 After answering this question, I was confused about why the overflowing integer in this code resulted in 0 instead of a negative number. It's strange, why such an exact number? Why 0? public class IntegerOverflow { public static void main(String[] args) { int x = 10; int i = 0; for (i = 0; i <= 5; i++) { x = x * x; System.out.println(x); } } } Output: 100 10000 100000000 1874919424 0 0 回答1: This will only happen if the starting value of x is even. According to the JLS §15.17.1: If an integer