template<class T, T i> void f(int[10][i]) { };
int main() {
int a[10][30];
f(a);
}
Why does f(a) fail?
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3 Answers
f(a) fails because a template type argument cannot be deduced from the type of a non-type argument. In this case the compiler cannot deduce the type of the template parameter T.
Try calling it as f<int>(a);
2 Comments
sehe
Mmm suboptimal. It took some tinkering (i.e. reading up), but there is a good way to do this. (The second template param actually should never have been of type T)
Prasoon Saurav
Why a downvote? The question asked is why
f(a) fails and my post manages to answer that.Try this:
template<class T, T i> void f(T[10][i]) { }; // note the 'T'
int main() {
int a[10][30];
f(a);
}
.. this enables the compiler to deduce the type of T, which is totally impossible in your sample (because T is not used at all).
answered Apr 24, 2011 at 15:14
Alexander Gessler
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Comments
template< std::size_t N > void f(int (&arr)[10][N])
{
}
int main() {
int a[10][30];
f(a);
}
This one works (http://codepad.org/iXeqanLJ)
Useful backgrounder: Overload resolution and arrays: which function should be called?
