How to detect integer overflow? [duplicate]

Question

Write a program that reads and stores a series of integers and then computes the sum of the first N integers. First ask for N, then read the values into a vector, then calculate the sum of the first N values. For example: “Please enter the number of values you want to sum:” 3 “Please enter some integers (press '|' to stop):” 12 23 13 24 15 | “The sum of the first 3 numbers ( 12 23 13 ) is 48.” Handle all inputs. For example, make sure to give an error message if the user asks for a sum of more numbers than there are in the vector.

Modify the program from exercise 8 to write out an error if the result cannot be represented as an int.

So in exercise 9 it says to write out an error message if the result cannot be represented as an int, but if I'm only adding up integers here the only thing I can think of here that would cause a number to not be able to be represented as an int would be if there's some integer overflow and the integer goes above INT_MAX or below INT_MIN but how exactly can I detect this?

This is my code:

#include<vector>
#include<cmath>
#include<string>
#include<iostream>
#include<algorithm>
#include<stdexcept>
using namespace std;

    int main()
    {
        int N = 0;
        vector<int> values;
        cout << "Please enter the number of values you want to sum." << '\n';
        cin >> N;
        cout << "Please enter some numbers" << '\n';
        for (int tempnum; cin >> tempnum; )
        {
    
            values.push_back(tempnum);
    
    
        }
        if (N > values.size())
        {
    
            cout << "How am I gonna add up more numbers than you gave me?" << '\n';
        }
        else
        {
            int total = 0;
            for (int i = 0; i < N; i++)
            {
                total += values[i];
            }
            
    
        }
    
    }

But I can't just check if total is above or below INT_MAX or INT_MIN because INT_MAX +1 or something returns some random negative value so I'm not really sure how to do this

Note that the question says integers and not int. So you are not restricted to using int in the computations. If on your system long is larger than an int you can use this and check the result of each addition against the maximum possible int. — Richard Critten
– Richard Critten, Commented Oct 3, 2021 at 12:50
But then right after exercise 9 in exercise 10 it says: 10. Modify the program from exercise 8 to use double instead of int. Also, make a vector of doubles containing the N–1 differences between adjacent values and write out that vector of differences. Which I'm gonna assume basically means that I do have to use ints in 8 and 9. — jaylse2
– jaylse2, Commented Oct 3, 2021 at 12:53
duplicates: How to catch integer overflow in C++, automatic overflow detection in C++? — phuclv
– phuclv, Commented Oct 3, 2021 at 14:08

Benjamin Maurer · Accepted Answer · 2021-10-03 13:42:58Z

1

You can do this with some arithmetic:

So you want to know if acc + x > INT_MAX, where acc is the accumulator (the sum so far) and x is the next element. Adding them together may overflow though. But acc > INT_MAX - x is equivalent and x is already stored in an int. So this can also not underflow, bc. x can be at most INT_MAX.

I.e., just transform the inequation to avoid overflow. In this case, subtract x from both sides.

That only works for positive x though, thanks to user interjay for pointing this out.

With negative x, this becomes a subtraction:

acc + (-x) > INT_MAX
acc - x    > INT_MAX

This can not overflow, because we are subtracting from acc. But it may underflow, so the question becomes:

acc - x < INT_MIN
acc     < INT_MIN + x

So:

int acc = 0;
for (int i = 0; i < N; ++i) {
    int x = values[i];
    if (x >= 0 && acc > (INT_MAX - x)) {
        cout << "ERROR: Overflow.";
        return 1;
    } else if (x < 0 && acc < (INT_MIN - x)) {
        cout << "ERROR: Underflow.";
        return 1;
    } else {
        acc += x;
    }
}

edited Oct 3, 2021 at 13:42

answered Oct 3, 2021 at 12:56

Benjamin Maurer

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3 Comments

interjay Over a year ago

This can overflow if x < 0. You'd need to handle negative x differently.

Benjamin Maurer Over a year ago

Ah dang, good point

lalala Over a year ago

Intmin plus a negative number always underflows. There is a typo in your second if.

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