Home >Backend Development >C++ >Why Does Template Deduction Fail with Implicit Type Conversion for User-Defined Types?

Why Does Template Deduction Fail with Implicit Type Conversion for User-Defined Types?

Linda Hamilton
Linda HamiltonOriginal
2024-10-28 14:28:301048browse

Why Does Template Deduction Fail with Implicit Type Conversion for User-Defined Types?

Implicit Type Conversion and Template Deduction

When attempting to implicitly convert an int to a Scalar object within a template function call, the template argument deduction may fail. This occurs because template argument deduction does not automatically take into account user-defined conversions, such as the one between int and Scalar.

In the provided code:

<code class="cpp">func(a, 2);</code>

the compiler attempts to implicitly convert the int 2 to a Scalar object, but the template argument deduction fails as it does not consider user-defined conversions. To resolve this, the argument must be explicitly converted at the caller site:

<code class="cpp">func(a, Scalar<int>{2}); // C++14</code>

Alternatively, if C 17 is used, a deduction guide can be defined for Scalar, allowing the following syntax:

<code class="cpp">func(a, Scalar{2});</code>

Finally, explicit instantiation of the template function with the specified type argument can also bypass the need for implicit conversion:

<code class="cpp">func<int>(a, 2); // Assuming Scalar<T>::Scalar(T) is not explicit</code>

The above is the detailed content of Why Does Template Deduction Fail with Implicit Type Conversion for User-Defined Types?. For more information, please follow other related articles on the PHP Chinese website!

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn