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Chapter 2 exercise answer bug #267

@ZenoTan

Description

@ZenoTan

The exercise answer could not handle negative values:

template<typename ... T>
auto average(T ... t) {
    return (t + ... ) / sizeof...(t);
}
int main() {
    std::cout << average(1, 2, 3, 4, 5, -6, -7, -8, -9, -10) << std::endl;
}

It would return 1844674407370955159 on my side since it deduced the return value to be unsigned int.

Activity

Prathamesh01110

Prathamesh01110 commented on Oct 1, 2023

@Prathamesh01110

The issue you're facing is due to integer promotion in C++. When you perform operations on values of different types, C++ promotes them to a common type before performing the operation. In your case, the values you're passing to the average function include both positive and negative integers, which leads to integer promotion.

The problem is that the sizeof...(t) part of your code is evaluated as an unsigned integer (size_t), which causes all the values to be promoted to unsigned integers for the division operation. This is why you're getting an unexpected result.

To fix this issue, you can explicitly cast the result of the division to the desired type (e.g., double) to ensure that the division is performed with the correct type. Here's an updated version of your code:

#include <iostream>

template<typename ... T>
auto average(T ... t) {
    return static_cast<double>((t + ...)) / sizeof...(t);
}

int main() {
    std::cout << average(1, 2, 3, 4, 5, -6, -7, -8, -9, -10) << std::endl;
}

By casting the result of (t + ...), you ensure that the division is performed using a floating-point type, which can handle both positive and negative values correctly. This will give you the expected average, including negative values.

Delta456

Delta456 commented on Oct 1, 2023

@Delta456
Contributor

I have made a PR for fixing this bug. See #268

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        Chapter 2 exercise answer bug · Issue #267 · changkun/modern-cpp-tutorial