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What it is the best general way for testing overflow in Long operations?


What it is the best general way for testing overflow in Long operations?

By : user3042353
Date : November 27 2020, 05:23 PM
I wish this help you
Within the universe of Long the difference between a number and its Double is always 0
code :
println(Long.MAX_VALUE)
println(BigDecimal(Long.MAX_VALUE.toDouble()))
9223372036854775807
9223372036854775808
var l2= Long.MAX_VALUE
var d2 =l2.toDouble()
if (d2-l2==0.0) println("MaxLong")
val product = a*b
if ((b != 0 && product/b != a) || (a == Long.MIN_VALUE && b == -1)) {
    println("Overflow")
} else {
    // can use product here
    println("OK")
}
fun multiplyExact(x: Long, y: Long): Long? = 
    try { java.math.multiplyExact(x, y) } catch (e: ArithmeticException) { null }
val largeNumber = Long.MAX_VALUE - 600
val smallNumber = 100L
// prints true, even though there's no overflow
println((largeNumber.toDouble() + smallNumber.toDouble()) - (largeNumber + smallNumber) != 0.0)


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long ans = ((long) INT_MIN) * 2 - 1; causing a warning because of integer overflow

long ans = ((long) INT_MIN) * 2 - 1; causing a warning because of integer overflow


By : Vitaliy Victorovich
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , This means that the calculation will overflow the range of long. Signed overflow yields undefined behavior.
The only correlation between the range of values representable by long and the value INT_MIN is that INT_MIN is representable as a long. There is no guarantee that one less than twice INT_MIN is representable as a long.
Arithmetic/Logical operations with long long data type

Arithmetic/Logical operations with long long data type


By : Syam S
Date : March 29 2020, 07:55 AM
I hope this helps . Maybe what you are seeking is exemplified by this (which returns true/1 or false/0 depending on whether x is greater than (gt) y):
code :
int ll_compare_gt(long long x, long long y)
{
    return(x > y);
}
C++ overflow detection for large numbers (unsigned long long)

C++ overflow detection for large numbers (unsigned long long)


By : user3518209
Date : March 29 2020, 07:55 AM
Does that help y < 0 will always be false, and anything xor 0 will always be that thing (Did you miss that < is evaluated at higher precedence than ^?).
Therefore unless x + y mod happens to equal 0, you will throw (and probably appears to always throw in practice unless you contrived specific inputs).
I ran into a weird bug in c++ where a statement calculating an addition of 2 small integers overflow into a long long va

I ran into a weird bug in c++ where a statement calculating an addition of 2 small integers overflow into a long long va


By : Aneeb Qureshi
Date : March 29 2020, 07:55 AM
Hope this helps Obviously, on your machine, your size_t is a 32-bit integer, whereas long long is 64 bit. size_t always is an unsigned type, so you get:
code :
cout << a.size()      -   1
//        ^ unsigned      ^ promoted to unsigned
//      output as uint32_t
//                ^ (!)
a.size() - k - 1
// ^ promoted to long long, as of smaller size!
// -> overall expression is  int64_t
//                          ^ (!)
overflow encountered in long scalers while stress testing for maximum pair-wise product

overflow encountered in long scalers while stress testing for maximum pair-wise product


By : user3591221
Date : March 29 2020, 07:55 AM
should help you out The error you're getting has to do with your datatype, as is discussed in this similar question. I think the solution is to specify the datatype as a 64-bit datatype. You can do this when creating your vector:
code :
vector = list(np.float64(np.random.randint(0,1000000, n)))
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