integer overflow in expression, when multiplying 13 digits

Question

This is just a test C program that I want it to calculate the product of 13 digits

#include <stdio.h>
#include <stdlib.h>
int main ()
{
    long prod; 
    prod = 5*5*7*6*6*8*9*6*6*4*8*9*5;
    printf ("%ld\n", prod);
    system("pause");
    return 0;
}

but it just says

[Warning] integer overflow in expression [-Woverflow]

The program runs but it displays a very different product value than the real one... I tried to change the int and the long int to float and double and not sure what to do next.

What compiler are you using? What does sizeof(long) and sizeof(long long) give you? — brothir, Jan 14 '19 at 02:40
See also [Why write 1,000,000,000 as 1000*1000*1000 in C?](https://stackoverflow.com/q/40633059/2410359) and [There are reasons not to use 1000 * 1000 * 1000.](https://stackoverflow.com/a/40637622/2410359) — chux - Reinstate Monica, Jan 14 '19 at 02:48
**If** you're using a 64 bit machine, `long` is the same size as `long long` [usually, for posix OSes: linux, BSD, macOS--the exception is WinX]. So, if you changed `5*5*...` into `5lu*5*...` it should produce the same answer as Jerry's below. That change forces the rvalue calculations to be done with `long` instead of `int` (i.e.) the lvalue `prod` does _not_ influence the size of the `rvalue`--the conversion is done _after_ the calculations. But, using `long long` is the safer bet for all architectures/systems. — Craig Estey, Jan 14 '19 at 03:33

score 4 · Accepted Answer · answered Jan 14 '19 at 02:42

4

Changing the type of the result won't help in a case like this.

The problem is that since all the input values are ints, only int sized multiplication is ever done (then, if you assign that to a double or whatever, the result is converted to double--but only after it has overflowed.

If you start with a larger type, then the multiplication will be done with that larger type:

#include <stdio.h>
#include <stdlib.h>
int main ()
{
    unsigned long long prod; 
    prod = 5ull*5*7*6*6*8*9*6*6*4*8*9*5;
    printf ("%llu\n", prod);
    system("pause");
    return 0;
}

answered Jan 14 '19 at 02:42

Jerry Coffin

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Thank you, but what if i wanted to do the same to an array – Adham Elshabasy Jan 14 '19 at 02:54
@EddieElshabasy: In that case you might want something like this: `unsigned long long result = 1; for (int i=0; i<10; i++) result *= (unsigned long long)input[i];` – Jerry Coffin Jan 14 '19 at 05:09

integer overflow in expression, when multiplying 13 digits

1 Answers1