Re: [Dev-C++] array + pointers???

[Nicolas R. <ni...@bo...>]

Hello Noorez,

> I have asked questions about pointers and arrays but all the answers=20=

> are not that great. Maybe I was unclear so I will explain my questions=20=

> better...

Thank you for the compliments. For great answers, I'd suggest (for the=20=

second time) that you google for some tutorials and reference. For even=20=

better answers, you can try reading the K&R ("The C Programming=20
Language) and/or Stroustroup's book ("The C++ Programming Language").


>   int nArray[5]=3D{1,2,3,4,5};
>   cout << nArray << endl;/*this gives the address of the array*/

Correct, because the name of an array of T is always a T* (a pointer to=20=

T). All arrays are pointers (implicitly).


>   char sArray[6]=3D{'h','e','l','l','o',\0};
>   cout << sArray << endl;/*this give the string hello, but i used the=20=

> name of the array without  subscripts? Don't i get the address of it=20=

> like in the int array?*/

No, you get the string instead of the address because, as I told you=20
before, arrays of char get treated in a special way (due to the fact=20
that, in plain old C, they were the only way to represent strings of=20
characters). If you really want the address, you could cast the pointer=20=

to some other type first. If you don't know what 'to cast' means,=20
please look it up in TFM.

Again, as Per already said,

> cout always assumes that a pointer to a char is not just a pointer to=20=

> a single character but actually a pointer to the first character of a=20=

> zero-terminated string


Next question:

>   char* pszString =3D "hello";/*strangest of all*/
>   /*this looks as if you were trying to put "hello" where the address=20=

> of a char would go.*/

The easy way to think about this is the one I already said: C/C++=20
provide special syntax to assign constant strings into arrays of char.=20=

(It would be rather unpractical if they didn't.)

What's going on behind the scenes there is that the string "hello" gets=20=

created "on the fly" by the compiler as a constant array of chars that=20=

contains {'h', 'e', 'l', 'l', 'o', \0}. Then you create a=20
pointer-to-char called pszString. What does the assignment do? It just=20=

copies the address. So pszString now points to the same place where the=20=

compiler stored the array containing the constant string.


Good luck with your C/C++ learning. You will encounter many, many such=20=

quirks along the way. Please try to use the excellent and free FAQs and=20=

tutorials around. They are your friends.

Regards,
Nicol=E1s