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Control Flow

2 (2.2%)

Arrays, Strings, Pointers, and References

8 (9%)

Functions

4 (4.5%)

Basic object-oriented programming

30 (33.7%)

A bit more advanced OOP (Composition, Operator overloading, Inheritance, Virtual Functions)

18 (20.2%)

Templates

8 (9%)

Other (Explain)

4 (4.5%)

Working with files?  (Streams)

15 (16.9%)

Total Members Voted: 89

[Normandy]

Huh? Why are the Java I/O capabilities failing you? Use the built-in readers/writers and not much can go wrong.

Also, the problem is not in initialization. There's a problem somewhere else in your code. Check your loop bounds. And check to make sure your code will actually compile (you have a syntax error in the code you pasted) before posting it on the forum.

[eerr]

"new RGB(0, 0, 0)[hght*widt];"
This is not valid syntax!

This will not work.

You cannot do that.

The result of RGB(0,0,0) is an RGB object.
The RGB object is not an array, and therefore fails any function of []

x= new RGB(0,0,0);
x[hght*widt];

will produce an error, and is the closest interpretation to what you are trying to do.

perhaps you should try this page?

http://download.oracle.com/javase/1.4.2/docs/api/java/awt/Graphics.html
http://download.oracle.com/javase/1.4.2/docs/api/java/awt/Image.html

[Virex]

"new RGB(0, 0, 0)[hght*widt];"
This is not valid syntax!

True, but the problem is not where you're looking for it. You can't call the constructor like that when initializing an array of objects. Since you're initializing them all to 0,0,0 anyway, you could define a default constructor that does just that. Alternatively:

Code: [Select]

RGB[] foo;
for(int i = 0; i<(hght*widt); i++)
{
    foo[i] = new RGB(0,0,0);
}
should also work[/quote]

[eerr]

"new RGB(0, 0, 0)[hght*widt];"
This is not valid syntax!

True, but the problem is not where you're looking for it. You can't call the constructor like that when initializing an array of objects. Since you're initializing them all to 0,0,0 anyway, you could define a default constructor that does just that. Alternatively:

Code: [Select]

RGB[] foo;
for(int i = 0; i<(hght*widt); i++)
{
    foo[i] = new RGB(0,0,0);
}
should also work

[/quote]

foo is null, but otherwise that will work.

[Normandy]

When you create an array (i.e. new Object[]) all values of the array are initialized to null, not to the default constructor. Keep that in mind.

[Supermikhail]

I'm studying C++ by a book, and I'm kind of lost. Help! Can somebody explain to me why there's *const in a custom String class constructor?

Code: [Select]

String( const char *const );
Is it because you don't want the data that the pointer points to changed, as well as the pointer itself?

Also, do I understand it right? const goes after the parentheses when you don't want a function to change anything?

Code: [Select]

char operator[] ( int offset ) const;
What about before the return value?

Code: [Select]

const char * GetString () const { return itsString; }

I think the placement of const has been the most confusing thing so far, after pointers and references in function declarations.

[Normandy]

Part of this confusion stems from the fact that const char and char const are exactly the same thing. Normally, modifiers work right-to-left, i.e. char const**&* will be a pointer to a reference to a pointer to a pointer to a const char. However, because of readability concerns, const also works like const char**&*. However, when you're talking about doing something like char const* const*&* then you're talking a pointer to a reference to a pointer to a const pointer to a const char.

When declaring function prototypes, you don't need to enter the variable names. In String( const char *const ); the const char *const is simply an argument. You might find it better to think of it like String( const char *const str );. As noted, a const char *const is a const pointer to a const char, also known as a string literal. When you type something like: const char* str = "Hello World";, the string literal is of type const char*. The reason the argument is const char *const is because you can convert from a non-const to a const, but not the other way around. Since the function doesn't need to change either the pointer or the value it's pointing to, it is constant.

As for your second question, correct, that's when the function will not change anything. When it's before the return value, it simply makes the return value const.

[Siquo]

Or, the short answer, literals are always const, so you can use

Code: [Select]

String a = new String("hello");
"hello" is a literal char*, and as such is always const. So indeed, both the data and the pointer should not be changed. Const is something of a security flag, so for instance

Code: [Select]

function a(int& number) {}
a(4)
Is illegal, as you reference the parameter, but as it's a const, you may not change it. Copying every parameter is an option, but might slow down your program (especially when you use large objects as parameters).


Or something like that.

[Supermikhail]

Err.

Normandy, nobody uses such constructions, right? Or do they?

Siquo: What I don't exactly understand - "hello" is a literal char* . But then it's a value with which String is created... Wait. So in that segment first a char* is created, then a new String object is created, and then the char* is copied into the String? I mean what comes first - the char* or the String? Isn't "hello" the value of some variable and so easily changeable?

I must be wrong.

[Siquo]

If it's a literal, it's not the value of some variable (well, under water it is).

The value passed to String is a char*, or: an integer with an address value pointing to the first letter of that string, where the string resides in a temporarily reserved block. Copying char* only copies the address pointing to that first character, not the string itself. Since the string itself is a literal, the address pointer and pointee are both const.

String takes the char* address, and starts reading chars from it until it reaches a null-char, meaning it's the end-of-string. It then (or simultaneously or whenever) copies those chars into its own char array, which it can modify, because the original address block can't be used as the memory will be released as soon as the literal is out-of-scope.

If I remember correctly.

[qwertyuiopas]

Any time you use quoted text in C/C++, it is converted to a pointer (that is why adding two strings together, in the absence of operator overloading,  gives rather unexpected results), and the actual string is probably stored as read-only, if possible. If it is defined as char[], however, it will be writable.
See http://stackoverflow.com/questions/2589949/c-string-literals-where-do-they-go.

[Supermikhail]

What? I'm pretty sure I read in one book that char * and char [] are interchangeable, that char[] is still a pointer to the first member. Did they trick me again?

[Rose]

char * and char[] mean the same thing, more or less

but they aren't the same as "hello" or const char *

[Supermikhail]

Okay. Then, I believe, qwertyuiopas is trying to confuse me further, villainously.

Any time you use quoted text in C/C++, it is converted to a pointer (that is why adding two strings together, in the absence of operator overloading,  gives rather unexpected results), and the actual string is probably stored as read-only, if possible. If it is defined as char[], however, it will be writable.
See http://stackoverflow.com/questions/2589949/c-string-literals-where-do-they-go.

How is adding strings together related to literal character arrays? Strings are supposed to be objects of the String class, aren't they? And how can you define quoted text as char[]? That I don't understand at all.

[Rose]

there's cstrings, which is just a char array, and there's the C++ string class, which is an actual class, with overloaded operators in whatnot.

if something converts to a cstring, the generally refer to char arrays.