如何获得传递给函数的数组的大小?

(b+1) b[1]

void print_array(int* b, int num_elements)
{
  for (int i = 0; i < num_elements; i++)
    {
      printf("%d", b[i]);
    }
}

sizeof(b) / sizeof(b[0])

arrays are not the same as pointers

template<class T, int N> void f(T (&r)[N]){
}

int main(){
    int buf[10];
    f(buf);
}

std::array here here at()

C99 n

void print_array(int b[static n]);

6.7.5.3.7

pass the size of an array implicitly

void print_array(int n, int b[n]);

#include <cstdio>                                                               

void 
print_array(int b[], size_t N) 
{ 
    for (int i = 0; i < N; ++i) 
        printf("%d ", b[i]);
    printf("\n");
}

template <size_t N>
inline void 
print_array(int (&b)[N]) 
{
    // could have loop here, but inline forwarding to
    // single function eliminates code bloat...
    print_array(b, N);
}                                                                                

int main()                                                                      
{                                                                               
    int a[] = { 1, 2 };                                                         
    int b[] = { };
    int c[] = { 1, 2, 3, 4, 5 };                                                

    print_array(a);                                                             
    // print_array(b);                                                          
    print_array(c);                                                             
}

array_size.cc: In function `int main()':
array_size.cc:19: error: no matching function for call to `print_array(int[0u])'