I have a class field
char chars[4]
I want to init this field in the constructor.
My constructor need to get char chars[] as a parameter and in the constructor call I want to init my class fields.
Can I do it or I need to call copy?
5 Answers
Use std::copy like this:
class MyClass
{
char m_x[4];
public:
MyClass(const char (&x)[4])
{
std::copy(x, x + 4, m_x);
}
};
You should really be explicit with your types here to enforce that you pass exactly 4 elements.
8 Comments
R. Martinho Fernandes
Being explicit with the types here doesn't enforce anything at all.
MyClass(const char x[4]) is the same as MyClass(const char* x).
Avihai Marchiano
It's not enforce but document the method by the signature, which is good
tenfour
Edited so it does enforce. My bad.
Puppy
Logically, you could actually accept a character array of any size greater than four.
tenfour
Right; I went with the most literal interpretation of the original question. There's no way to know where the flexibility lies: in the passed-in type, or in the stored type. We don't even know what this variable is used for; most likely it's a string which makes most of the answers here worthless.
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You need to call copy.
MyClass::MyClass(const char* param)
{
std::copy(param, param + 4, chars);
}
This is slightly risky code since there is no guarantee that param has four characters to copy.
6 Comments
Avihai Marchiano
Is there any simple alternative. Is there a difference between define the param in the constructor as char* to define as char array? Thank you
john
For a parameter
char* is exactly the same as char[], and const char * is exactly the same as const char[]. But only for parameters.Avihai Marchiano
Can you please explain why only for char
john
No, not only for char, only for parameters. In a parameter
int* is the same as int[] for instance. The reason is that you cannot pass an array as a parameter in C/C++. So the designers of C chose to use the array syntax to mean the same as a pointer for parameters. If you want to say char[] then do it like that. It won't make any difference.
Heisenbug
what about passing a std::vector<char> by reference instead?
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An alternative could be to use a vector<char> which can be initialised in the initialiser list:
class A
{
public:
A(char* a): a_(a, a + strlen(a)) {}
private:
std::vector<char> a_;
};
See demo at http://ideone.com/x9p6I .
This could be made into a template to support different types but would require the number of elements in the array to be provided to the constructor (as strlen() would not be applicable to an array of int). For example:
template <class T>
class A
{
public:
A(const T* a, const size_t count): a_(a, a + count) {}
private:
std::vector<T> a_;
};
See demo at http://ideone.com/hhoaC .
As I understood, you need to create Conversion constructors, it should looks like this:
A::A(const char* string)
{strncpy(chars, string, 4);
}
Comments
Yes. You need to copy the contents from the input parameter.
6 Comments
Puppy
It disgusts me that people post such things as answers to a C++ question.
Tony The Lion
strcpy_s takes 3 arguments, not two! This does not compile! and it's not C++
Etienne de Martel
@john It shouldn't have been. The OP didn't even try the answer before accepting it.
john
@EtiennedeMartel: I guess the OP really doesn't want to call
std::copy for some reason.Avihai Marchiano
I accept the answer because this is the best for char array. In these simple answers it's not matter if it compile because the answer is clear , so if there is compilation error you fix it by yourself.
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std::string,std::vectororstd::arrayinstead.