Buggy transfer of single long long numbers to int array

Question

I’m trying to grab a Long Long Int and split each place number into it’s own spot in an array, in order of course, with array[0] being the largest number.

So for instance, if the number was 314, then array[0] = 3, array[1] =1, and array[2] = 4.

This is part of a calculator project for a microcontroller where I’m writing the graphics library (for fun) and using arrays to display each line.

The issue is, it needs to be able to deal with really large numbers (9,999,999,999+), and I’m having dramas with the large stuff. If the Long Long is < 1,000,000, it will writes all the numbers perfectly, but the more numbers I add, they all start to be written wrong towards the end.

For instance, 1,234,567,890 displays as 1,234,567,966.

Here’s the snippet of code I’m using:

long long int number = 1234567890;
int answerArray[10];
int numberLength = 10;

for(writeNumber = 0; writeNumber < numberLength; writeNumber++)
{
    answerArray[writeNumber] = ((int)(number / pow(10, (numberLength - 1 - writeNumber))) % 10;
}

I’m fairly sure this has to do with either the “%” and multiple data types, because any number within the Int range works perfectly.

Can you see where I’m going wrong? Is there a better way achieve my goal? Any tips for large numbers?

pow is for floating point exponentiation; it will lose precision with long long arguments, and floating point is not something you want to use on a microcontroller. Typically you would rewrite this to successively divide and mod by 10 instead. — Nate Eldredge
– Nate Eldredge, Commented Oct 27, 2020 at 23:51
@NateEldredge amazing i didn’t know that, so you think just do it 10 at a time? — Eman
– Eman, Commented Oct 27, 2020 at 23:53

Philipp Claßen · Accepted Answer · 2020-10-28 00:07:14Z

The signature of pow is

double pow(double x, double y);

When you call the function, the computation will implicitly use floating point. That is why it is no longer exact as pure integer operations.

In addition, you have to be careful how you cast to int.

In your question, you have

((int)(number / pow(10, (numberLength - 1 - writeNumber))) % 10;

The parentheses do not match, so I will assume you meant:

(int)(number / pow(10, (numberLength - 1 - writeNumber))) % 10;

However, here you cast a number that may exceed the range of int before you apply the modulo 10 operation. That can result in an integer overflow. The code is doing the same as if you had written:

((int)(number / pow(10, (numberLength - 1 - writeNumber)))) % 10;

To avoid the overflow, it would be better to perform the modulo operation first. However, you are dealing implicitly with double at this point (because of pow), so it is not ideal either. It is best to stick with pure integer operations to avoid these pitfalls.

Daniel Walker · Accepted Answer · 2020-10-28 00:26:26Z

2

Your issue is that you're casting what is potentially a very large number to an int. Look at the iteration when writeNumber is numberLength-1. In that case, you're dividing a long long by 1 and then forcing the result into an int. Once number becomes larger than 2^31-1, you're going to run into problems.

You should remove the cast altogether as well as the call to pow. Instead, you should iteratively grab the next digit by modding out by 10 and then dividing number (or a copy of it) by 10.

E.g.,

int index = sizeof(answerArray)/sizeof(answerArray[0]);
for (long long x=number; x>0; x /= 10) {
    answerArray[--index] = x%10;
}

edited Oct 28, 2020 at 0:26

answered Oct 27, 2020 at 23:55

Daniel Walker

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4 Comments

Eric Postpischil Over a year ago

sizeof(answerArray) should be the number of elements, sizeof answerArray / sizeof *answerArray.

Eman Over a year ago

Thank you! And modding a long long int will still work fine because it’s a whole number?

Daniel Walker Over a year ago

It will indeed.

Eric Postpischil Over a year ago

As long as it is non-negative. For negative operands, % can produce negative results.

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