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Can C Template Deduction Be Simplified Using Function Return Type?

Susan Sarandon
Release: 2024-11-06 15:46:02
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Can C   Template Deduction Be Simplified Using Function Return Type?

Simplifying Template Deduction Based on Function Return Type

In C , template deduction plays a crucial role in inferring the parameter types for generic functions. However,有时候,如何 simplified template deduction by utilizing only the function's return type can be desirable.

Consider the following example:

<code class="cpp">class GC
{
public:
    template <typename T>
    static GCPtr<T> Allocate();
};</code>
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Here, the Allocate function takes a generic type parameter T and returns a pointer to an object of type T. Using template deduction, it would be possible to simplify the syntax for allocating objects as follows:

<code class="cpp">GCPtr<A> ptr1 = GC::Allocate(); // Desired output</code>
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Unfortunately, this is not possible in C because the return type of a function is not involved in template deduction. Instead, template signatures are matched based on the function call arguments.

Alternative Solution:

One workaround is to utilize a helper function that hides the explicit type specification:

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

int main()
{
   GCPtr<A> p = 0;
   Allocate(p);
}</code>
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In this approach, the Allocate function accepts a reference to the pointer as an argument, and the actual allocation is handled internally. This allows for simplified usage without sacrificing flexibility.

C 11 Enhancement:

In C 11, template deduction rules were extended, enabling the omission of one of the type declarations:

<code class="cpp">auto p = GC::Allocate<A>(); // p is of type GCPtr<A></code>
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This further simplifies the syntax, making it more convenient for certain scenarios.

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