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Editorial Team
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Asked: 2026-05-23T14:27:24+00:00

Given a template function, that does not use the template parameter for input directly.

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Given a template function, that does not use the template parameter for input directly. How would C++ type inference work? For instance given

template<typename T>
void f(T::value_type){}

when (if at all) will type inference work for this function?

Is there any other place except of template<typename T1,...>void f(T1,T2,...) where type inference might occur?

As always, quote the standard for extra credit.

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  1. Editorial Team

    I think here’s your answer:

    14.8.2.4 – Deducing template arguments from a type [temp.deduct.type]

    […]

    -3- […]

    In most cases, the types, templates, and non-type values that are used to compose P participate in template argument deduction. That is, they may be used to determine the value of a template argument, and the value so determined must be consistent with the values determined elsewhere. In certain contexts, however, the value does not participate in type deduction, but instead uses the values of template arguments that were either deduced elsewhere or explicitly specified. If a template parameter is used only in nondeduced contexts and is not explicitly specified, template argument deduction fails.

    -4- The nondeduced contexts are:

    • The nested-name-specifier of a type that was specified using a qualified-id.

    • A type that is a template-id in which one or more of the template-arguments is an expression that references a template-parameter.

    When a type name is specified in a way that includes a nondeduced context, all of the types that comprise that type name are also nondeduced. However, a compound type can include both deduced and nondeduced types. [Example: If a type is specified as A<T>::B<T2>, both T and T2 are nondeduced. Likewise, if a type is specified as A<I+J>::X<T>, I, J, and T are nondeduced. If a type is specified as void f(A<T>::B, A<T>), the T in A<T>::B is nondeduced but the T in A<T> is deduced. ]

    Your T::value_type is a qualified-id of a type, so types in its nested-name-specifier are nondeduced and must be specified explicitly.

    Edit: this information is from ISO/IEC 14882:1998.

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