Is there any way to pass an anonymous array as an argument in C++?

Question

I'd like to be able to declare an array as a function argument in C++, as shown in the example code below (which doesn't compile). Is there any way to do this (other than declaring the array separately beforehand)?

#include <stdio.h>  static void PrintArray(int arrayLen, const int * array) {    for (int i=0; i<arrayLen; i++) printf("%i -> %i\n", i, array[i]); }  int main(int, char **) {    PrintArray(5, {5,6,7,8,9} );  // doesn't compile    return 0; } 

Answer 1

If you're using older C++ variants (pre-C++0x), then this is not allowed. The "anonymous array" you refer to is actually an initializer list. Now that C++11 is out, this can be done with the built-in initializer_list type. You theoretically can also use it as a C-style initializer list by using extern C, if your compiler parses them as C99 or later.

For example:

int main() {     const int* p;     p = (const int[]){1, 2, 3}; } 

Answer 2

It's allowed with a typecast in C++11 and in extern "C" with C99:

void PrintArray(size_t len, const int *array) {     for(size_t i = 0; i < len; i++)         printf("%d\n", array[i]); }  int main(int argc, char **argv) {     PrintArray(5, (const int[]){1, 2, 3, 4, 5});     return 0; }