CODEWITHSUNDEEP
c#type-conversionmultiplication
Ka0s
Ka0s

Reputation: 398

Multiplication in C# error

when multiplying a set of int's and converting the result to a long I get a different answer as opposed to when I multiply a set of doubles and convert the result to a long. eg:

int a = 5;
int b = 5;
int c = 7;
int d = 6;
int ee = 6;
int f = 8;
int g = 9;
int h = 6;
int i = 6;
int j = 4;
int k = 8;
int l = 9;
int m = 5;
long x = a * b * c * d * ee * f * g * h * i * j * k * l * m;

double aa = 5;
double ab = 5;
double ac = 7;
double ad = 6;
double aee = 6;
double af = 8;
double ag = 9;
double ah = 6;
double ai = 6;
double aj = 4;
double ak = 8;
double al = 9;
double am = 5;
long y = (long)(aa * ab * ac * ad * aee * af * ag * ah * ai * aj * ak * al * am);

Can anyone tell me why this happens? Thank you

Upvotes: 5

Views: 1154

Answers (3)

Afzal Ahmad
Afzal Ahmad

Reputation: 626

I give it a brief search and run some tests even on scripts as well and I found the following outcomes, in script format I test it as,

<script>
        var a = 5;
        var b = 5;
        var c = 7;
        var d = 6;
        var ee = 6;
        var f = 8;
        var g = 9;
        var h = 6;
        var i = 6;
        var j = 4;
        var k = 8;
        var l = 9;
        var m = 5;
        var x = a * b * c * d * ee * f * g * h * i * j * k * l * m;
        var y = eval(a * b * c * d * ee * f * g * h * i * j * k * l * m);
        document.writeln("RESULT X : " + x);
        document.write("<br>");
        document.writeln("RESULT Y : " + y);
    </script>

and the outcome was 23514624000 for both x and y, but in c# as you try to multiply all integers and in result an integer is formed but we also know that int.MAX = 2147483647 which is less then thw actuall result, So in result when assigning to long x= a * b * c * d * ee * f * g * h * i * j * k * l * m; it truncates the original values, and in other result it is not, if you want same result in both you can use following code,

int a = 5;
int b = 5;
int c = 7;
int d = 6;
int ee = 6;
int f = 8;
int g = 9;
int h = 6;
int i = 6;
int j = 4;
int k = 8;
int l = 9;
int m = 5;
long x = (long)a * b * c * d * ee * f * g * h * i * j * k * l * m;

double aa = 5;
double ab = 5;
double ac = 7;
double ad = 6;
double aee = 6;
double af = 8;
double ag = 9;
double ah = 6;
double ai = 6;
double aj = 4;
double ak = 8;
double al = 9;
double am = 5;
long y = (long)(aa * ab * ac * ad * aee * af * ag * ah * ai * aj * ak * al * am);
Console.WriteLine(x);
Console.WriteLine(y);
Console.ReadKey();

Now you will get the same result i.e 23514624000

Upvotes: 2

Jakob Olsen
Jakob Olsen

Reputation: 823

That's probably because the result of your multiplication is too big to hold in an Int32 - yet it can be held in a double.

To be a bit more specific look at x and y in hexadecimal:

x = 0x000000007994b000
y = 0x000000057994b000

An Int32 can only hold the lower 8 hex values. This is why x is the wrong number. The lowest 8 hex-values are correct, but the top is cut off.

Upvotes: 5

Dmitrii Bychenko
Dmitrii Bychenko

Reputation: 186668

You've got an integer overflow, since whenever you multiply two int you have int: this is interpreted as Int32 so far

 a * b * c * d * ee * f * g * h * i * j * k * l * m

and only than converted into Int64 (long). To correct the implementation, declare all the variables initially being long e.g.

  long a = 5;
  long b = 5; // actually, not necessary, since a * b will be long, but safer
  long c = 7;
  ...
  long m = 5;

  long x = a * b * c * d * ee * f * g * h * i * j * k * l * m;

Upvotes: 13

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