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Ambiguous Function Overload for the Most Negative Integer Value
In C , function overloading allows multiple functions with the same name but different parameters. However, ambiguity arises when the compiler cannot determine which overloaded function to invoke based on the given arguments. This issue can occur when working with integer types, particularly when dealing with the most negative value.
Consider the following code snippet:
<code class="c++">void display(int a) { cout << "int" << endl; } void display(unsigned a) { cout << "unsigned" << endl; } int main() { int i = -2147483648; cout << i << endl; //will display -2147483648 display(-2147483648); }
According to our understanding of function overloading, any value within the integer range (4 bytes in this case) should invoke the display(int) function. Values outside this range would lead to ambiguity. However, compiling this code results in the following error:
call of overloaded `display(long int)` is ambiguous
This error arises when passing the most negative integer value (-2147483648) to the display function. Strangely, printing the same value directly (as seen on line 6) produces the correct result: -2147483648.
The Subtlety: Lack of Negative Literals in C
The key to understanding this behavior lies in the absence of negative literals in C . All integer literals in C are regarded as unsigned by default, meaning they do not have a sign prefix (- or ). As a result, -2147483648 is actually treated as -1 * (2147483648).
Implications for Overloaded Functions
Since 2147483648 exceeds the range of an integer (4 bytes), it gets promoted to a long integer. This means that the compiler tries to invoke the display(long int) function, which conflicts with the existing display(int) overload. Hence, the ambiguity arises.
Resolution
To avoid this ambiguity, it is recommended to use the std::numeric_limits class to obtain type-specific minimum and maximum values:
<code class="c++">std::cout << std::numeric_limits<int>::min() << endl; // displays -2147483648</code>
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