I'd like to know if there's a standard way that the compiler interprets
a template argument. That is, let's say you pass a Foo instance as a
template argument to some function. Is the instance itself passed, or
does the compiler assume it's a Foo reference? For example:
=============== =============== ==
template<typena me I> someFunc(I i) { /* ... */ }
Foo f;
someFunc(f);
=============== =============== ===
In someFunc(), does I = Foo, or I = Foo&, or I = const Foo & ?
Thanks,
--Steve 2 1413
mrstephengross wrote: I'd like to know if there's a standard way that the compiler interprets a template argument. That is, let's say you pass a Foo instance as a template argument to some function. Is the instance itself passed, or does the compiler assume it's a Foo reference? For example:
=============== =============== == template<typena me I> someFunc(I i) { /* ... */ }
Foo f;
someFunc(f); =============== =============== ===
In someFunc(), does I = Foo, or I = Foo&, or I = const Foo & ?
The first. You can make it the second or third by adding those
parameters to your template, e.g.,
template<typena me I> someFunc(I& i) { /* ... */ }
Cheers! --M
Can I also explicitly cast the Foo instance in the someFunc()
invocation?
For example: someFunc(const Foo &)f);
--Steve This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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