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Can C Deduce Template Types from Function Return Types?

Mary-Kate Olsen
Mary-Kate OlsenOriginal
2024-11-06 12:23:02299browse

Can C   Deduce Template Types from Function Return Types?

Template Deduction Based on Function Return Type in C

In C , it can be desirable to utilize template deduction to simplify code that instantiates generic functions based on the data types of the function's arguments. Consider the following example:

<code class="cpp">GCPtr<A> ptr1 = GC::Allocate();
GCPtr<B> ptr2 = GC::Allocate();</code>

Instead of specifying the generic type parameters explicitly, the goal is to achieve this deduction using the return type of the GC::Allocate() function. However, C does not allow type deduction based on the return type.

<code class="cpp">class GC
{
public:
    template<typename T>
    static GCPtr<T> Allocate();
};</code>

Despite the return type being generic, the compiler requires the explicit specification of the template type parameters and when instantiating the GC::Allocate() function.

To mitigate this limitation, an auxiliary function can be introduced:

<code class="cpp">template <typename T>
void Allocate(GCPtr<T>& p) {
    p = GC::Allocate<T>();
}</code>

Using this function, the original goal can be achieved as follows:

<code class="cpp">GCPtr<A> p;
Allocate(p);</code>

Whether this syntax offers any significant advantage over the explicit type specification is subjective.

Note: In C 11, it is possible to omit one of the type declarations using the auto keyword:

<code class="cpp">auto p = GC::Allocate<A>(); // p is of type GCPtr<A></code>

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