Can Template Deduction Work Based on a Function's Return Type in C ?

Template Deduction for Function Based on Its Return Type?

In C , template deduction provides a convenient way to determine template arguments based on the arguments provided to a function call. However, there are certain limitations to template deduction, such as the inability to deduce type arguments based on the return type of a function.

The Issue:

The original question seeks to eliminate the need to explicitly specify type arguments when calling the Allocate() function in the following code:

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

The Answer:

Unfortunately, template deduction cannot be used to deduce the type arguments based on the return type. Instead, it is the other way around: the return type is determined after the template signature has been matched.

Workaround:

To bypass this limitation, the Allocate() function can be wrapped in a helper function that hides the type argument from the caller:

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

int main() {
   GCPtr<a> p = 0;
   Allocate(p);
}</a></t></t></typename></code>

This allows the caller to use the Allocate() function without explicitly specifying the type argument:

<code class="cpp">GCPtr<a> p = 0;
Allocate(p);</a></code>

Additional Note:

C 11 introduces the auto keyword, which allows the compiler to deduce the type from the initializer. This further simplifies the code:

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

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