I'm new to this C++ environment and I am having difficulty with my constructor. Here is my code:
class Student {
char name[20];
long number;
public:
Student (char nm[20], long val) :
name(nm), number(val) {}
When I compile it, it gives me an error saying incompatible types in assignment of char* to char[20].
How can i fix this??
Your constructor argument nm is, actually, not an array! Yes, I know it looks like one, because you wrote char nm[20]. But, actually, it's char* nm. That translation is performed automatically when you write an array type in a function parameter list. Yes, it's stupid. Blame C.
So, the error message is telling you that you cannot assign a pointer to an array. Fair enough. Doesn't really matter anyway, since the language also doesn't let you assign arrays to arrays. Lol.
This is why, since 1998, we've had std::string to fix all these terrible problems:
class Student {
std::string name;
long number;
public:
Student (std::string nm, long val) :
name(nm), number(val) {}
};
If you must use an array, you can do this:
class Student {
std::array<char, 20> name;
long number;
public:
Student (std::array<char, 20> nm, long val) :
name(nm), number(val) {}
};
because std::array, introduced in 2011, is a handy wrapper around raw arrays that can be assigned (and don't have that weird decay to pointers!).
"Ah, but my teacher told me to use raw arrays," I hear you say. A shame, but we can fix that too. Either accept the array by reference, or take in a pointer to it (as you're doing now! but this drops the dimension 20 from the type and makes things unsafe -.-) and do a manual copy of each element from the source to the destination. Certainly not ideal, but it may be what your teacher is expecting if this is homework:
class Student {
char name[20];
long number;
public:
Student (char (&nm)[20], long val) :
number(val)
{
assert(sizeof(nm) == sizeof(name));
std::copy(std::begin(nm), std::end(nm), std::begin(name));
}
};
class Student {
char name[20];
long number;
public:
Student (char* nm, long val) :
number(val)
{
// just have to hope that the input is a pointer to **20** elements!
std::copy(nm, nm+20, name);
}
};
You can't assign to an array.
You can iterate over the array, and copy each item individually - eg. :
Student (char nm[20], long val) :
number(val) {
for (size_t i = 0; i < 20; ++i) {
name[i] = nm[i];
}
}
Alternatively (as suggested in a comment by @LogicStuff), you can use std::string :
class Student {
std::string name;
long number;
public:
Student (const std::string& nm, long val) :
name(nm), number(val) {
}
};
Arrays passed by value as arguments to a function are implicitly converted to pointers to their first elements.
Thus for example the following constructor declarations are equivalent
Student (char nm[20], long val);
Student (char nm[100], long val);
Student (char nm[], long val);
Student (char *nm, long val);
Arrays have neither the copy constructor nor the copy assignment operator. It means that you may not write like this
Student (char nm[20], long val) :
name(nm), number(val) {}
^^^^^^^^
You should change the constructor.
First of all it is better then the first parameter is declared with qualifier const. In this case you may pass to the constructor a string literal
The constructor can look the following way
#include <cstring>
//...
Student ( const char nm[20], long val) : number(val)
{
std::strncpy( name, nm, 20 );
name[19] = '\0';
}
In this declaration I used nm[20] only for exposition.
Otherwise you should use standard class std::string instead of the character array.
In C++ you can think of arrays as simply pointers to their first array-element. Maybe this makes it a bit more clear why you can't assign an array to another since you, thus, try to assign an array to the pointer to an array-element.
Instead of doing so, you can either iterate through the array and copy it value by value or you may use std::memcpy or std::copy to do the job. I'd recommend you to use std::copy.
