Why Does Function Overloading Become Ambiguous with the Most Negative Integer in C ?

Function Overloading Ambiguity with the Most Negative Integer

In C , function overloading allows multiple functions with the same name but different parameter lists. However, encountering the most negative integer value can result in an ambiguity error due to implicit type conversions.

Understanding the Issue

The following code snippet demonstrates the issue:

<code class="cpp">void display(int a) { cout << "int" << endl; }
void display(unsigned a) { cout << "unsigned" << endl; }

int main() {
    int i = -2147483648;
    cout << i << endl; // prints -2147483648
    display(-2147483648); // error: ambiguous function call
}

When the most negative integer (-2147483648) is passed to display(-2147483648), the compiler generates an "ambiguous function overloads" error. This occurs because the compiler cannot determine which function to call: display(int) or display(unsigned).

The Role of Integer Literals

The behavior observed in the given example stems from the absence of negative integer literals in C . Integer literals must begin with a non-zero digit. In the case of -2147483648, the negative sign is interpreted as the unary minus operator, resulting in the expression -1 * (2147483648).

Type Promotion

Since 2147483648 exceeds the range of int, it is promoted to long int. However, this promotion creates an ambiguity between the parameter types of display(int) and display(unsigned), leading to the error.

Portable Solution

To obtain the minimum or maximum values for a type in a portable manner, one can use:

<code class="cpp">std::numeric_limits<type>::min(); // or max()</code>

By utilizing this method, ambiguous function overloads are avoided, ensuring correct type handling regardless of the implementation.

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