Why Does Passing the Most Negative Integer to an Overloaded Function Cause Ambiguity in C ?

Ambiguity in Overload Resolution for Most Negative Integer

In C , function overloading allows multiple functions to have the same name but different parameter types. However, when resolving overloaded function calls, ambiguity can arise if the compiler cannot determine the most appropriate function to invoke.

Consider the following code demonstrating function overloading for displaying integer and unsigned integer values:

<code class="cpp">void display(int a) { cout <p>According to our understanding, any integer value within the specified range for int should invoke display(int), while values outside this range would cause ambiguity. However, when attempting to pass the most negative int value (-2147483648) to display, the compiler raises an error:</p>
<pre class="brush:php;toolbar:false">call of overloaded display(long int) is ambiguous

This surprising behavior stems from the absence of negative integer literals in C . Integer literals defined in the C standard do not include the '-' character. Instead, when encountering a '-' followed by a number, the compiler treats it as the unary negation operator applied to a positive integer literal.

In our case, '-2147483648' is interpreted as '-1 * 2147483648'. Since '2147483648' exceeds the range of an int, it is promoted to a long int. This type promotion leads to ambiguity because now both display(int) and display(long int) are viable candidates for the function call.

To avoid such ambiguities, it is recommended to use the std::numeric_limits class to retrieve the minimum or maximum values for different data types in a portable manner:

<code class="cpp">std::cout ::min() </code>

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