I have a struct:
typedef struct { int nNum; string str; }KeyPair; Let's say I initialize my struct:
KeyPair keys[] = { {0, "tester"}, {2, "yadah"}, {0, "tester"} }; I want to use the initialized values in a function. How do I pass this array struct as a function parameter?
I have:
FetchKeys( KeyPair *pKeys) { //get the contents of keys[] here... } How about?
template<int N> void FetchKeys(KeyPair const (&r)[N]){} EDIT 2:
Or even
template<int N> void FetchKeys(KeyPair const (*p)[N]) with the call as
FetchKeys(&keys); 
You can do it as @MSalters mentioned, or you can create a std::vector<KeyPair> and pass it to the function. Here is a sample code:
using namespace std; struct KeyPair { int nNum; string str; }; void fetchKeys(const vector<KeyPair>& keys) { //Go through elements of the vector vector<KeyPair>::const_iterator iter = keys.begin(); for(; iter != keys.end(); ++iter) { const KeyPair& pair = *iter; } } int main() { KeyPair keys[] = {{0, "tester"}, {2, "yadah"}, {0, "tester"} }; //Create a vector out of the array you are having vector<KeyPair> v(keys, keys + sizeof(keys)/sizeof(keys[0])); //Pass this vector to the function. This is safe as vector knows //how many element it contains fetchKeys(v); return 0; } 
Should be
// Definition void FetchKeys( KeyPair *pKeys, int nKeys) { //get the contents of keys[] here... } // Call FetchKeys(keys, sizeof(keys)/sizeof(keys[0])); 
size_t instead of int to represent object sizes and array indices.
In c/c++ the name of the array (of any type) represents the address of the first element of the array, so keys and &keys [0] are same. You can pass any one of them for KeyPair*.
void foo( int (*array)[5] ) {} int main() { int array[5] = {0}; foo( &array ); int array2[2] = {0}; /*foo(&array2)*/ } The code will compile for the first case as the array has 5 elements, but will fail in the second (commented out) case as the number of elements differ. The test verifies that the array is not a pointer to the first element, but has the size information --which is only available at compile time, and before the array decays into a pointer.You just callFetchKeys(keys);
EDIT
Pay attention to declare FetchKeys' return type.
EDIT 2
If you also need the number of items, you add size as FetchKeys input parameters:
void FetchKeys(KeyPair*, size_t size); and call FetchKeys(keys, sizeof(keys)/sizeof(*keys));
BTW, state all your question by editing your first post if you can.
Depending on what you want to do you can even use boost range and pass it to function as a pair of iterators:
void FetchKeys(KeyPair *begin, KeyPair *end) FetchKeys(boost::begin(keys), boost::end(keys)); See this answer: How can I pass an array by reference to a function in C++?
Wrap it in a structure, nice and easy..
#include <iostream> struct foo { int a; int b; }; template <typename _T, size_t _size> struct array_of { static size_t size() { return _size; } _T data[_size]; }; template <typename _at> void test(_at & array) { cout << "size: " << _at::size() << std::endl; } int main(void) { array_of<foo, 3> a = {{ {1,2}, {2,2}, {3,2} }}; test(a); } EDIT: URGH, I can't see the toolbar to format the code correctly, hopefully the tags works..
_T) are reserved and using them results in undefined behaviour. Do you know whether C++ has this same restriction, even for templates?
i use VS 2008, and this works fine for me.
#include "stdafx.h" typedef struct { int nNum; CString str; }KeyPair; void FetchKeys( KeyPair *pKeys); int _tmain(int argc, _TCHAR* argv[]) { KeyPair keys[] = { {0, _T("tester")}, {2, _T("yadah")}, {0, _T("tester")} }; FetchKeys(keys); //--> just pass the initialized variable. return 0; } void FetchKeys(KeyPair *pKeys) { printf("%d, %s\n",pKeys[0].nNum, pKeys[0].str); } I don't understand the difficulty. correct me if i'm wrong. To keep it simple, i avoided using vectors, templates and etc. edit: to know size of struct, you can pass one more arg.
