Integer overflow occurs when the result of an operation is larger than the maximal value that can be represented by the underlying integer type.
Questions tagged [integer-overflow]
1056 questions
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votes
3 answers
How to represent integers bigger as (10^6)! to use in equations to solve in java
How can we solve equations having N! constants in it , where N can be of range 1<=N<=10^6
BigInteger can only perform upto 128 bits right?
Even when one do logarithm on both sides, it leaves values bigger than BigInteger.
cypronmaya
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votes
1 answer
Why the behavior of double to int typecasting different in x86_64 and aarch64 during the integer overflow case?
I have the following sample code which I am compiling and executing on x86_64 and aarch64 platforms but ending up with different outputs.
#include
int main()
{
double d = 2147483649;
int i = d;
std::cout << i << std::endl;
…
Ajith Gupta
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votes
1 answer
Is it possible to calculate the value after overflow?
I understand that if 1 is added to INT_MAX it will overflow and we will get the negative maximum. It is connected in a cycle. In the code below can I calculate the value after overflow.
#include
using namespace std;
int main()
{
…
Shivam Rana
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votes
1 answer
Overflow after multiplying positive int8s and casting to uint32
I am trying to multiply 8-bit positive integers and cast the result to uint32.
a := int8(12)
b := uint32(a * a) // 4294967184
In the above code multiplying causes an overflow.
But if I cast each int8 to uint32/uint8 before multiplying, I get the…
Hashib Khondokar
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- 1
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votes
1 answer
Overflow in long long int
I am facing an overflow problem for this case and getting output as 7392445620511834112 , I wanted to multiply 2 large values of b and c and want to store that value.
#include
using namespace std;
#define ll long long int
int…
Rahul Rangi
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votes
1 answer
Difference between x + x and 2*x in Java
I was reading the answers to a Leetcode problem: https://leetcode.com/problems/divide-two-integers/. This problem requires not using multiplication. After reading a few answers, I somehow had a misconception that x+x is better than 2*x in terms of…
whatsnext
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votes
1 answer
Integer Overflow Exploit
I have this code, which have some vulnerability, but I can't seem to exploit it.
For now, this is what I've noticed:
1) if argv[1] = 3 and argc = 3, then it overflows and writes argv[2] into memory of array[3] in "place_int_array" function.
2) if…
Elyasaf755
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-2
votes
1 answer
Subtracting two integers causes integer-underflow in device code
In my cuda device code I am doing a check where I subtracting the thread's id and the blockDim to see weather or not the data I might want to use is in range. But when this number goes bellow 0 it seems to wrap back around to be max…
Liiht
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votes
2 answers
Is there an integer i where i<2 && i>10?
if(i<2 && i>10){
//code here will never be reached or what?
}
Just in case of an integer overflow maybe?
thestruggleisreal
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votes
2 answers
Can the C++ sizeof() function be used to prevent integer overflows?
I'm building mathematics functions that I plan to use for cryptography.
Your algorithm is useless if the code is vulnerable to exploitation.
Buffers are fairly easy to protect against overflows; but what about integers?
I won't share my Galois…
Ahab Devoid
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votes
1 answer
Errors involving precedence of modulus operator and brackets with large numbers in C++
The first part of the just computes some mathematical formula, stroed in ans1, ans2 and ans3. Many sites give % and * as having the same priority in the precedence order. Should we then consider them left to right in an expression? Only difference…
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votes
1 answer
python integer overflow solution
I am fairly new to Python and am currently creating a "RSA Encryption" program to send Secret messages to peers. I have the program completed, but run into issues with computations. I keep getting an overflow error because the numbers I am trying to…
Christian Potts
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votes
1 answer
Why does i *= 2 seem to converge to 0 for signed integers?
Consider the following code:
#include
int main(int argc, char* argv[])
{
int i = /* something */;
for (std::size_t n = 0; n < 100; ++n) {
i *= 2;
std::cout << i << std::endl;
}
}
Overflowing on signed integers…
Vincent
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votes
1 answer
A result that I can't figure out
I've been learning C recently.
I have difficulty understanding the result of the code below.
Why is b 255 at last?
unsigned char a=1;
int b=0;
do
{
b++;
a++;
}while(a!=0);
Manhooo
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votes
2 answers
overflow in implicit constant conversion warning while assigning string size(string::npos) to an integer variable
How to get rid of the the warning below?
size_t filesize = getFilesize(strLogFileName.c_str());
// Open file
int fd = open(strLogFileName.c_str(), O_RDONLY, 0);
assert(fd != -1);
// Execute mmap
char* mmappedData =
(char *) mmap(NULL, filesize,…
user2256825
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