C++ Object Default Variable

Question

When using a string object, you can set the object equal to a string literal. How is this possible? The object consists of multiple functions and variables, so how can you just set it equal to one string literal?

Thanks!

Edit:

Sorry for the confusing question.

It makes sense to me that you can set a variable or array equal to a value. For example:

char word[3] = {'c', 'a', 'r'};

But a string is not simply a variable, so what in its class allows you to write a statement setting the object to a value?

string word = "car";

I would love to do this in my objects, but thus far I have been using something like this, where objectTemp is the class & tempF is a member holding the temperature value in Fahrenheit:

objectTemp cat;
cat.tempF = 102;

*How is this possible?* Are you asking how `std::string str = "Hello"` is possible? — Dean Seo, Oct 15 '18 at 23:36
Welcome to stackoverflow! It's not very clear exactly what you're asking. Could you maybe post some specific code that you can't understand, with specific questions? — JMAA, Oct 15 '18 at 23:37
JMAA, I edited the question to try to clarify it, and added some code. — sharkByte, Oct 16 '18 at 00:07
It's worth noting that functions are stored _per-class_ in your program, not per-object. Every `string` has the exact same members functions as any other `string`, and the compiler is well aware of this, and so doesn't store that information with each `string` object, referring to the correct functions instead simply by type inference (ignoring `virtual` functions for now). — alter_igel, Oct 16 '18 at 01:10
@alterigel Oh got it! That makes more sense. Thanks for clearing that up for me :) — sharkByte, Oct 16 '18 at 01:49

score 4 · Answer 1 · edited Oct 15 '18 at 23:59

4

C++ supports operator overload which allows to call the function:

string& operator= (const char* s);

of class std::string. This simply takes a const char* (literal) and create internally the new std::string object.

edited Oct 15 '18 at 23:59

πάντα ῥεῖ

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answered Oct 15 '18 at 23:53

Nikos Xenakis

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So ... `std::string foo = "bar";` calls a function named `operator=()`? – Swordfish Oct 15 '18 at 23:59
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@Swordfish Nope, it calls the _magic_ constructor overload. – πάντα ῥεῖ Oct 16 '18 at 00:00
@πάνταῥεῖ Ha! You said it! ;) – Swordfish Oct 16 '18 at 00:00
Hm. "This simply takes a const char* (literal)" can i only assign string literals then? – Swordfish Oct 16 '18 at 00:19
No, it is just an implementation of many. Find more here: http://www.cplusplus.com/reference/string/string/operator=/ – Nikos Xenakis Oct 16 '18 at 00:22
Hm. Which one is called for `char foo[] = "bar"; std::string qux = foo;`? – Swordfish Oct 16 '18 at 00:24
char[] is nothing else but char* so your foo variable in that case is const char* . You have your answer. – Nikos Xenakis Oct 16 '18 at 00:26
@NikosXenakis "char[] is nothing else but char*" Certainly not. – Swordfish Oct 16 '18 at 00:27
The variable name of a table in c++ is just a pointer nothing else. For example, foo[1] == *(foo + 1*sizeof(char)) for valid memory access. Read about the introduction to c++ tables, or just test it in your code, it will be clear to you then. Have fun :) – Nikos Xenakis Oct 16 '18 at 00:36
@NikosXenakis That `foo[n]` is the same as `*(foo + n)` doesn't change the fact that the type of `foo` is `char[4]` and not `char *` or `char const *`. – Swordfish Oct 16 '18 at 00:39
@Swordfish char[4] just reserves memory in the stack and not in the heap as "new" would do with char*. If you keep having doubts please submit a new question. Maybe someone else is also confused and want things to be cleared. Thanks – Nikos Xenakis Oct 16 '18 at 00:44
@NikosXenakis don't think that this is worth a question on its own if you can just make this clear to me: in `char foo[] = "bar";` the `foo` is a `char*` and `"qux"` is of type `char const*`? – Swordfish Oct 16 '18 at 00:46
if by "qux" you mean a literal string then yes. – Nikos Xenakis Oct 16 '18 at 00:49
@NikosXenakis And `foo`? – Swordfish Oct 16 '18 at 00:50
as you said foo is a pointer to a char – Nikos Xenakis Oct 16 '18 at 00:52
@NikosXenakis You are wrong on both. http://eel.is/c++draft/lex.string#8 http://eel.is/c++draft/dcl.init.string#1 – Swordfish Oct 16 '18 at 01:00
@NikosXenakis `"qux"` is of type `char const[4]` and `foo` in `char foo[] = "bar";` is of type `char[4]`. – Swordfish Oct 16 '18 at 01:01
There is neither type char const[4] nor char[4] in c++ language. const char* and char* are the types you are looking for. Please start from the basics before you get more confused. Happy coding! – Nikos Xenakis Oct 16 '18 at 01:04
@NikosXenakis You didn't even bother to look at the links I gave you. – Swordfish Oct 16 '18 at 01:08
@NikosXenakis "There is neither type char const[4] nor char[4] in c++ language." Mhm. so ... `char const a[4];` ... of what type is `a`? ... `char b[4];` ... of what type is `b`? – Swordfish Oct 16 '18 at 01:18
@NikosXenakis Thanks for the answer! It helped me out :) – sharkByte Oct 16 '18 at 02:08
@NikosXenakis Here you are: https://stackoverflow.com/a/52827290/3975177 – Swordfish Oct 16 '18 at 17:11

score -1 · Accepted Answer · edited Aug 03 '19 at 20:50

-1

That is possible because C++ knows of "magic" functions that are called upon object instantiation called constructors.

The other "magic" function you should look up is the assignment operator.

But a string is not simply a variable, so what in its class allows you to write a statement setting the object to a value?
string word = "car";

As said before, one of the constructors of std::string makes this magic happen:

std::string::string(char const*);

It takes a pointer to constant chars which are interpreted as a "C-string" (a zero-terminated string).

I would love to do this in my objects, but thus far I have been using something like this, where objectTemp is the class & tempF is a member holding the temperature value in Fahrenheit:
objectTemp cat;
cat.tempF = 102;

Write a constructor for your class (=don't be afraid of the magic) which takes a numeric value as its parameter. Eg.

class objectTemp {
    int tempF;
public:
    objectTemp(int tempF)  // constructors are named the same as the struct/class
    : tempF{ tempF }  // initializer-list
    {}  // an empty body
}

usage:

objectTemp cat = 102;

or:

objectTemp cat(102);

or:

objectTemp cat{ 102 };

edited Aug 03 '19 at 20:50

Rabbid76

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answered Oct 15 '18 at 23:36

Swordfish

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There's nothing like _magic_ in c++. Stop spreading such myths please. – πάντα ῥεῖ Oct 15 '18 at 23:44
@πάνταῥεῖ Everyone knows that there is magic in C++ which makes programs written in C++ so slow compared to programs written in C! – Swordfish Oct 15 '18 at 23:47
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_"... which makes programs written in C++ so slow compared to programs written in C!"_ That's just another myth. As mentioned stop telling such please. – πάντα ῥεῖ Oct 15 '18 at 23:50
@πάνταῥεῖ That are facts. You'd probably also call the cost of exception handling a myth. – Swordfish Oct 15 '18 at 23:54
How can this be? For he **IS** the Kwisatz Haderach! – user4581301 Oct 16 '18 at 00:07
Thanks for the example! I couldn't get the exact example code working, but it helped me get this working: `class objectTemp` `{` `int tempF;` `public:` `objectTemp(int temp)` ` {` `tempF = temp;` `}` `};` – sharkByte Oct 16 '18 at 01:06
@sharkByte Wait. What in my example using the initializer list was "not working"? – Swordfish Oct 16 '18 at 01:17
@Swordfish Woops! Nevermind! Sorry I messed up. It works great :). It was an issue with my text editor :P. – sharkByte Oct 16 '18 at 01:43
@sharkByte I'm relieved ;) – Swordfish Oct 16 '18 at 01:50

C++ Object Default Variable

2 Answers2