Why Does Pointer Decay Override Template Deduction in C Function Overloading?

Pointer Decay Takes Precedence Over Template Deduction

When writing code that operates on strings, it is common to encounter a dilemma where attempting to overload a function to accommodate both array-based and non-array-based string representations can result in unexpected behavior.

In this specific scenario, an initial function foo is defined to print the length of an array of characters:

template <size_t n>
void foo(const char (&s)[N]) {
    std::cout <p>This function accepts arrays of characters and outputs the array length. However, when extending <strong>foo</strong> to also handle non-array strings, ambiguity arises:</p>
<pre class="brush:php;toolbar:false">void foo(const char* s) {
    std::cout <p>The intention is for the second <strong>foo</strong> overload to handle non-array strings, but surprisingly, this extension leads to the original overload being bypassed for both array and non-array strings:</p><pre class="brush:php;toolbar:false">foo("hello") // prints raw, size=5

This unexpected behavior stems from the concept of pointer decay. Arrays are essentially pointers to their first element, and converting an array to a pointer is a very inexpensive operation. As a result, the compiler prioritizes the overload that accepts a pointer parameter, even though it requires an implicit conversion from the array argument.

To ensure the desired behavior, where the array-based overload is used for arrays, it is necessary to introduce an additional overload:

template <typename t>
auto foo(T s)
  -> std::enable_if_t<:is_convertible char const>{}>
{
    std::cout <p>This overload uses template deduction and restricts its applicability to types that can be converted to character pointers. Through partial ordering, the compiler now correctly selects the appropriate <strong>foo</strong> function based on the argument type.</p></:is_convertible></typename>

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