David Abrahams wrote:

"Allen Yao" <yaozhen@ustc.edu> writes:

...

It seems that gcc does not allow this kind of template argument deduction:

template <int N> void f(int a[N]) { for (int i = 0; i < N; i++) cout << a[i] << " "; cout << endl; }

int main() { int a[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};

f(a); }

The error message is:

error: no matching function for call to `f(int[10])'

I would like to know how to fix this. Is there any workaround? Or if I am missing something?

You're missing something. Your declaration of f is equivalent to:

template <int N> void f(int* a);

try:

template <int N> void f(int a(&)[N]) ^^^ instead.

I tried both the original problem and the proposed solution on VC 7.1 For the original problem I got the following error messages: c:\Krishna\Console\Console7.cpp(242) : error C2784: 'void f(int [N])' : could not deduce template argument for 'int [N]' from 'int [10]' c:\Krishna\Console\Console7.cpp(207) : see declaration of 'f' And when I tried the proposed solution, I got the following error messages: c:\Krishna\Console\Console7.cpp(209) : error C2988: unrecognizable template declaration/definition c:\Krishna\Console\Console7.cpp(209) : error C2059: syntax error : '&' c:\Krishna\Console\Console7.cpp(209) : error C2090: function returns array c:\Krishna\Console\Console7.cpp(241) : error C2783: 'void f(int *(__cdecl *)(void))' : could not deduce template argument for 'N' c:\Krishna\Console\Console7.cpp(209) : see declaration of 'f' I did manage to find a work-around, which goes like this: template<class T> struct F; template<class T, int N> struct F<T[N]> { void operator()(T a[N]) { for (int i = 0; i < N; i++) cout << a[i] << " "; cout << endl; } }; template<class T> void f(T& a) { F<T>()(a); } int main() { int a[10] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}; f(a); } Does it help? - Krishna Achuthan