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This might be off-topic since there is no code, no output, no anything, but here goes.

The only way I see C++ can know "this" is a std::string is because the quotation mark is a language construct, much like the char between single quotes.
But I would be really surprised if it was the actual answer: does that imply std::string is the only way C++ has to interpret text between quotation marks? Then the standard lib would not be just an extension, but a part of C++.
What if I want to develop my own MyString class, and I want to create MyStrings on the fly using quotation marks: would I be able to do it or is the "" syntax inevitably linked to the std::string?
I also thought about "" being an operator (after all C++ has operator()), but I couldn't find anything about that.

gsamaras
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deqyra
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    `"this"` is a [*string literal*](http://en.cppreference.com/w/cpp/language/string_literal), it has nothing to do with `std::string`. You may define [your own string literals though](http://en.cppreference.com/w/cpp/language/user_literal). – user7860670 Oct 01 '17 at 14:06
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    Perhaps you should look more into compiler theory and practice? Making a simple parser is actually not that hard, and would teach you a lot (including how to recognize constant string literals). – Some programmer dude Oct 01 '17 at 14:09
  • @VTT Bookmarked, thanks alot! – deqyra Oct 01 '17 at 14:11

3 Answers3

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Let's clear up some misconceptions. The type of a string literal is a const char[], not a std::string. For example, the type of "this" would be const char[5] (there is a null-terminator).

does that imply std::string is the only way C++ has to interpret text between quotation marks?

No (it's not std::string) and there are multiple prefixes for utf-8 strings and wide character strings, like L"wide".

then the standard lib would not be just an extension, but a part of C++.

That's true, but not for string literals. You have the type of nullptr which is std::nullptr_t. There is also std::byte that gets special treatment by the standard. A std::initializer_list constructor is chosen when using list initialization. The result of sizeof is a std::size_t and of typeid it is a std::type_info. There may be others I can't think of right now.

would I be able to do it or is the "" syntax inevitably linked to the std::string?

It's not a std::string, but yeah, that's not possible. What you can do however is define a user-defined literal. std::string has one:

using namespace std::string_literals;
auto string = "this"s;

static_assert(std::is_same_v<decltype(string), std::string>); // ok
Rakete1111
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3

There are some good answers here, but none have addressed why the question comes up in the first place. Why would one think "foo" might have type std::string?

The answer is std::string has a constructor from const char * which will automatically turn the string literal into a std::string in many situations. C++ has ample mechanisms to allow making library and user defined data structures feel like part of the language in this way. (Operator overloading is another example.)

Consider that e.g. strlen(std::string("foo")) does not compile, so "foo" cannot be of type std::string directly.

Zalman Stern
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  • Thanks for clarifying all of this! *"C++ has ample mechanisms to allow making library and user defined data structures feel like part of the language in this way."* -> **I totally agree**, that's why I'd have found it weird if quotation marks and `std::string` really were an inextensible language construct, hence my question. – deqyra Oct 02 '17 at 00:31
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"this" is a string literal, not an std::string.

When you compile your code, Parsing takes action and goes through your code. When it meets enclosing double quotes, it assumes - by convention - that a string literal is found.

If you want to define your own string literals, then check User-defined literals and What new capabilities do user-defined literals add to C++?

PS: I suggest you take a Compiler's course, since if these issues excite you, you will love it.

gsamaras
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