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Avoid Overflow when Calculating π by Evaluating a Series Using 16-bit Arithmetic?


Avoid Overflow when Calculating π by Evaluating a Series Using 16-bit Arithmetic?

By : Mathias
Date : October 25 2020, 07:10 PM
may help you . Take a look at related QA:
Baking-Pi Challenge - Understanding & Improving
code :
//---------------------------------------------------------------------------
AnsiString str_hex2dec(const AnsiString &hex)
    {
    char c;
    AnsiString dec="",s;
    int i,j,l,ll,cy,val;
    int  i0,i1,i2,i3,sig;
    sig=+1; l=hex.Length();
    if (l) { c=hex[l]; if (c=='h') l--; if (c=='H') l--; }
    i0=0; i1=l; i2=0; i3=l;
    for (i=1;i<=l;i++)      // scan for parts of number
        {
        char c=hex[i];
        if (c=='-') sig=-sig;
        if ((c=='.')||(c==',')) i1=i-1;
        if ((c>='0')&&(c<='9')) { if (!i0) i0=i; if ((!i2)&&(i>i1)) i2=i; }
        if ((c>='A')&&(c<='F')) { if (!i0) i0=i; if ((!i2)&&(i>i1)) i2=i; }
        if ((c>='a')&&(c<='f')) { if (!i0) i0=i; if ((!i2)&&(i>i1)) i2=i; }
        }

    l=0; s=""; if (i0) for (i=i0;i<=i1;i++)
        {
        c=hex[i];
             if ((c>='0')&&(c<='9')) c-='0';
        else if ((c>='A')&&(c<='F')) c-='A'-10;
        else if ((c>='a')&&(c<='f')) c-='A'-10;
        for (cy=c,j=1;j<=l;j++)
            {
            val=(s[j]<<4)+cy;
            s[j]=val%10;
            cy  =val/10;
            }
        while (cy>0)
            {
            l++;
            s+=char(cy%10);
            cy/=10;
            }
        }
    if (s!="")
        {
        for (j=1;j<=l;j++) { c=s[j]; if (c<10) c+='0'; else c+='A'-10; s[j]=c; }
        for (i=l,j=1;j<i;j++,i--) { c=s[i]; s[i]=s[j]; s[j]=c; }
        dec+=s;
        }
    if (dec=="") dec="0";
    if (sig<0) dec="-"+dec;

    if (i2)
        {
        dec+='.';
        s=hex.SubString(i2,i3-i2+1);
        l=s.Length();
        for (i=1;i<=l;i++)
            {
            c=s[i];
                 if ((c>='0')&&(c<='9')) c-='0';
            else if ((c>='A')&&(c<='F')) c-='A'-10;
            else if ((c>='a')&&(c<='f')) c-='A'-10;
            s[i]=c;
            }
        ll=((l*1234)>>10);  // num of decimals to compute
        for (cy=0,i=1;i<=ll;i++)
            {
            for (cy=0,j=l;j>=1;j--)
                {
                val=s[j];
                val*=10;
                val+=cy;
                s[j]=val&15;
                cy=val>>4;
                }
            dec+=char(cy+'0');
            for (;;)
                {
                if (!l) break;;
                if (s[l]) break;
                l--;
                }
            if (!l) break;;
            }
        }

    return dec;
    }
//---------------------------------------------------------------------------
AnsiString pi_BBP() // https://en.wikipedia.org/wiki/Bailey–Borwein–Plouffe_formula
    {
    const int N=100;        // 32*N bit uint arithmetics
    int sh;
    AnsiString s;
    uint<N> pi,a,b,k,k2,k3,k4;

    for (pi=0,sh=(N<<5)-8,k=0;sh>=0;k++,sh-=4)
        {
        k2=k*k;
        k3=k2*k;
        k4=k3*k;
        a =k2* 120;
        a+=k * 151;
        a+=     47;
        b =k4* 512;
        b+=k3*1024;
        b+=k2* 712;
        b+=k * 194;
        b+=     15;
        a<<=sh;
        pi+=a/b;
        }
    pi<<=4;
    s=pi.strhex();
    s=s.Insert(".",2);
    return str_hex2dec(s);
    }
//---------------------------------------------------------------------------
ref: 3.1415926535897932384626433832795028841971693993751058209749445923078164062862089986280348253421170679821480865132823066470938446095505822317253594081284811174502841027019385211055596446229489549303819644288109756659334461284756482337867831652712019091456485669234603486104543266482133936072602491412737245870066063155881748815209209628292540917153643678925903600113305305488204665213841469519415116094330572703657595919530921861173819326117931051185480744623799627495673518857527248912279381830119491298336733624406566430860213949463952247371907021798609437027705392171762931767523846748184676694051320005681271452635608277857713427577896091736371787214684409012249534301465495853710507922796892589235420199561121290219608640344181598136297747713099605187072113499999983729780499510597317328160963185950244594553469083026425223082533446850352619311881710100031378387528865875332083814206171776691473035982534904287554687311595628638823537875937519577818577805321712268066130019278766111959092164201989
BPP: 3.14159265358979323846264338327950288419716939937510582097494459230781640628620899862803482534211706798214808651328230664709384460955058223172535940812848111745028410270193852110555964462294895493038196442881097566593344612847564823378678316527120190914564856692346034861045432664821339360726024914127372458700660631558817488152092096282925409171536436789259036001133053054882046652138414695194151160943305727036575959195309218611738193261179310511854807446237996274956735188575272489122793818301194912983367336244065664308602139494639522473719070217986094370277053921717629317675238467481846766940513200056812714526356082778577134275778960917363717872146844090122495343014654958537105079227968925892354201995611212902196086403441815981362977477130996051870721134999999837297804995105973173281609631859502445945534690830264252230825334468503526193118817101000313783875288658753320838142061717766914730359825349042875546873115956286388235378759375195778185778048187


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SQL Server giving arithmetic overflow when calculating avg

SQL Server giving arithmetic overflow when calculating avg


By : Aljerdani
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Internally SQL Server is summing the values (to divide by the count later) and storing them in the columns data type - in this case an int - which isn't large enough to hold the sum - if you cast the value as a bigint first it will sum the values also storing those values in a bigint - which is probably large enough, then the average calculation can proceed.
code :
select avg(cast(number as bigint)) from LargeNumbers
Overflow Error evaluating arithmetic expression with integers in VBA

Overflow Error evaluating arithmetic expression with integers in VBA


By : user3237752
Date : March 29 2020, 07:55 AM
hop of those help? If my logic is correct, Dim or 'Dimension' is a way of telling the application that you expect use a variable of a certain type, and that type pertains to a certain amount of 'bits' (of memory).
This reserves a section of the system's memory, which has been allocated a certain amount of bits dependant on the variable type that you have instructed in your code. These bits then define how many (If you're familiar with C++ or similar then you will probably already know all this...)
code :
unsigned 16 bits =         0 - 65,536    (2^16)
signed 16 bits   =   -32,768 - 32,767
32,767           =    111111111111111    (Binary)
32,768           =    1000000000000000 <--- note the extra bit
Dim t as Integer
t = 1639
Debug.Print (20 * t - 1)
20 * t = 20 * 1639 = 32,780
32,780 = 1000000000001100 (Binary)
Sub SO()

Dim t As Integer, c As Long

t = 1639
c = 20

Debug.Print (20 * (t - 1))  '// No Error
Debug.Print (c * (t - 1))   '// No Error
Debug.Print ((c * t) - 1)   '// No Error

c = (20 * t - 1)            '// Error
Debug.Print (20 * t - 1)    '// Error

End Sub
Avoid arithmetic overflow on aggregate function

Avoid arithmetic overflow on aggregate function


By : jikang sun
Date : March 29 2020, 07:55 AM
I hope this helps you . Casting to bigint before the aggregate then try_casting the result back to int would be preferable to me
But a way without that is
code :
SET ARITHABORT OFF
SET ANSI_WARNINGS OFF

  DECLARE @tbl TABLE ( [i] INT );

    INSERT  INTO @tbl
            ( [i] )
    VALUES  ( POWER(2, 30) ),
            ( POWER(2, 30) );

    SELECT  SUM([t].[i])
    FROM    @tbl AS [t];
c++ overflow while calculating an arithmetic expression

c++ overflow while calculating an arithmetic expression


By : Majiet ze Liar
Date : March 29 2020, 07:55 AM
will be helpful for those in need When you compile your example code, the compiler usually throws a warning identifying exactly the part of your expression that overflows
code :
main.cpp:4:24: warning: integer overflow in expression [-Woverflow]
     std::cout << 30000 * (5 * 30000 - 22207) / 2 + 50488389 << std::endl;
                  ~~~~~~^~~~~~~~~~~~~~~~~~~~~
#include <iostream>
#include <cmath>

int main(int argc, char* argv[]) {
    std::cout << std::trunc(30000.0 * (5.0 * 30000.0 - 22207.0) / 2.0 + 50488389.0) << std::endl;
                              // ^^     ^^        ^^        ^^     ^^           ^^

    return 0;
}
std::cout  << std::fixed << std::setprecision(0) 
        // ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
           << std::trunc(30000.0 * (5.0 * 30000.0 - 22207.0) / 2.0 + 50488389.0) << std::endl;
Avoid Instant.toEpochMilli() Arithmetic Overflow

Avoid Instant.toEpochMilli() Arithmetic Overflow


By : Jeremy Fr
Date : March 29 2020, 07:55 AM
This might help you You can use java.lang.Math.addExact. It will throw an ArithmeticException if overflow occurs. It was added in Java 8.
EDIT
code :
private Instant capped(Instant instant) {
    Instant[] instants = {Instant.ofEpochMilli(Long.MIN_VALUE), instant, Instant.ofEpochMilli(Long.MAX_VALUE)};
    Arrays.sort(instants);
    return instants[1];
}
millis = capped(deadline).toEpochMilli();
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