how to pass arrays as parameters in constructor? c++ - c++

Question

I am trying to make a constructor for class call, in which 4 arrays are passed as parameters. I've tried using *,&, and the array itself; however when I assign the values in the parameters to the variables in the class, I get this error :
call.cpp: In constructor ‘call::call(int*, int*, char*, char*)’:
call.cpp:4:15: error: incompatible types in assignment of ‘int*’ to ‘int [8]’
call.cpp:5:16: error: incompatible types in assignment of ‘int*’ to ‘int [8]’
call.cpp:6:16: error: incompatible types in assignment of ‘char*’ to ‘char [14]’
call.cpp:7:16: error: incompatible types in assignment of ‘char*’ to ‘char [14]’
I would appreciate your help in finding my error and helping me correct it.
here is my code:
.h file
#ifndef call_h
#define call_h
class call{
private:
int FROMNU[8];
int DESTNUM[8];
char INITIME[14];
char ENDTIME[14];
public:
call(int *,int *,char *,char *);
};
#endif
.cpp file
call:: call(int FROMNU[8],int DESTNUM[8],char INITIME[14],char ENDTIME[14]){
this->FROMNU=FROMNU;
this->DESTNUM=DESTNUM;
this->INITIME=INITIME;
this->ENDTIME=ENDTIME;
}

Answer 1

Raw arrays are non-assignable and generally difficult to handle. But you can put an array inside a struct, and assign or initialize that. Essentially that's what std::array is.
E.g. you can do
typedef std::array<int, 8> num_t;
typedef std::array<char, 14> time_t;
class call_t
{
private:
num_t from_;
num_t dest_;
time_t init_;
time_t end_;
public:
call_t(
num_t const& from,
num_t const& dest,
time_t const& init,
time_t const& end
)
: from_t( from ), dest_( dest ), init_( init ), end_( end )
{}
};
But this still lacks some essential abstraction, so it's merely a technical solution.
To improve things, consider what e.g. num_t really is. Is it, perhaps, a telephone number? Then model it as such.
Consider also using standard library containers std::vector and, for the arrays of char, std::string.

Answer 2

Passing a raw array as an argument is possible in C++.
Consider the following code:
template<size_t array_size>
void f(char (&a)[array_size])
{
size_t size_of_a = sizeof(a); // size_of_a is 8
}
int main()
{
char a[8];
f(a);
}

Answer 3

In C/C++ you cannot assign arrays by doing this->FROMNU=FROMNU; thus your method wont work, and is one half of your error.
The other half is that you try to assign a pointer to the array. Even if you pass arrays to a function, they decay to pointers to the first element, despite what you say in the definition.