C - How to print calculated numbers that have 15 digits?

Question

I tried to print the result of 99,999 * 99,999 * 99,999, the result should have been 999,970,000,299,999 but what got printed instead was 18446744072244127711. I tried to print 20000 * 20000 * 20000 and the result was pretty much the same. Can you guys tell me how to get the real result?

#include <stdio.h>

int main()
{
    int num;

    printf("Insert number : ");
    scanf("%ld", &num);

    fflush(stdin);

    unsigned long long int total = num * num * num;
    printf("Result : %llu", total);

    getchar();
    return 0;
}

[`fflush(stdin);` is undefined behavior, don't do it.](https://stackoverflow.com/a/38325926/2173917) — Sourav Ghosh, Feb 21 '18 at 14:26
`scanf("%ld",&num)` is undefined behavior too, as `num` is `int`. — Eugene Sh., Feb 21 '18 at 14:26
*Can you guys tell me how to get the real result* - use calculator? Or pen-and-paper? — Eugene Sh., Feb 21 '18 at 14:27
because when you multiply int values, the result is calculated in a temporary int variable. then for the assignment a cast to long long occurs. but the result which does not fit in int was already computed so an overflow will happen — berendeanicolae, Feb 21 '18 at 14:33
@berendeanicolae So casting will be sufficient. The `num` itself is likely not overflowing (well, it's an assumption one should check). — Eugene Sh., Feb 21 '18 at 14:34
@SouravGhosh bro can you tell me what is undefined behavior ? why cannot use fflush? i just know that one — Kenny, Feb 21 '18 at 14:36
@Kenny Undefined behavior = doing something dangerous that isn't covered by the C language. Most often, but not always, undefined behavior manifests itself as some sort of severe run-time bug. — Lundin, Feb 21 '18 at 15:03
See also [Why write 1,000,000,000 as 1000*1000*1000](https://stackoverflow.com/a/40637622/2410359) — chux - Reinstate Monica, Feb 21 '18 at 15:27

Eric Postpischil · Accepted Answer · 2018-02-21T15:09:11.827

One way to do it is:

#include <inttypes.h>
#include <stdint.h>
#include <stdio.h>

int main(void)
{
    uint64_t x = 99999;
    printf("%" PRIu64 "\n", x*x*x);
}

You may also be able to do this with unsigned long or unsigned long long types, but these depend on your C implementation (the compiler you are using).

The arithmetic you attempted overflowed the supported range of the int type you were using. When programming, it is important to ensure that the calculations you do will not overflow the range of the types you use.

Additionally, you must be careful about types in C code:

When using scanf, match the conversion specifier to the type. For example, when scanning a decimal numeral for an int object, you should use %d, not %ld. %ld is for long int.
In unsigned long long int total = num*num*num;, the unsigned long long int affects the type of total. It does not affect the type used in the calculation num*num*num. In that expression, the type of num is used. To use unsigned long long int in the calculation, you would have to convert one or all of the operands to unsigned long long int, as by using a cast: (unsigned long long int) num * num * num.
By assigning 99999 to a uint64_t object, which I named x, I ensured that x*x*x was calculated with uint64_t arithmetic. (uint64_t is a 64-bit unsigned integer.) Another way to do this would be with UINT64_C(99999) * UINT64_C(99999) * UINT64_C(99999). UINT64_C is a macro defined in stdint.h that essentially means “treat this constant as having type uint64_t or wider (actually uint_least64_t).” This could also be done with UINT64_C(99999) * 99999 * 99999 because C will automatically convert the narrower-type integer operands to the wider-type integer operand. It could also be done with (uint64_t) 99999 * 99999 * 99999.
PRIu64 is a macro that provides the right printf conversion specifier for uint64_t. It is defined in inttypes.h.

Pedantically, `UINT64_C` gives a literal of type `uint_least64_t` and not `uint64_t`. — Lundin, Feb 21 '18 at 14:57
Nice answer (as always). But I would make it clear that `(unsigned long long int) num * num * num` is only casting the first coefficient. The other two are implicitly converted. — Bathsheba, Feb 21 '18 at 15:20

Lundin · Answer 2 · 2018-02-21T15:00:31.557

Bugs:

%ld on an int invokes undefined behavior.
fflush(stdin) invokes undefined behavior since stdin is an input stream.
num*num*num is carried out on operands that have type int, so you get overflow if you attempt 99999 * 99999 * 99999 since this is a larger number than 2³¹-1 (assuming 32 bit int).

Fixed code:

#include <stdio.h>

int main(void)
{
    int num;
    printf("Insert number : ");
    scanf("%d",&num); getchar();
    long long int total = (long long)num*num*num;
    printf("Result : %llu",total);
    getchar();
    return 0;
}

C - How to print calculated numbers that have 15 digits?

2 Answers2