String Literal in the temporary object?

Question

I am beginner in C++ and its low-level layer (in comparison with Python and so on). I have read on StackOverflow that string literals go into the Read-Only data section (String literals: Where do they go?), but currently I am reading a C++ book (by Bjarne Stroustrup) and it is written in there that temporary objects end their life at the end of an expression, so:

string var1 = "Hello ";  
string var2 = "World";
string var3 = var1 + var2; // string temp = "Hello World" and assign it to var3

After execution, will we have two "Hello World" or only one copy will be left?

Hardware:

OS: Windows 11;    
Compiler: MSVC;  
Standard: C++11

I understand that possibly it can be implementation-defined, but if it is so, please specify for the x86 platform.

It boils down to what is a literal. `"Hello World"` is a literal (aka a string directly embedded in the source code), but `var1 + var2` is not. — HolyBlackCat, Feb 01 '22 at 20:55
There is no "but". Literal strings are not temporary. The "read-only data section" is a platform-specific implementation detail. That temporary objects end their lives in the end of expression is a platform-independent question of language semantics. — molbdnilo, Feb 01 '22 at 21:13

score 2 · Accepted Answer · edited Feb 01 '22 at 23:19

String literals are the objects that are referred to specifically with the syntax

"Hello "

or

"World"

That is what literal means, an atomic syntax for a value. It does not refer to string values in general.

String literals are not temporary objects. They are objects stored in an unspecified way (in practice, often in read-only memory, as you indicated) and have static storage duration, meaning they exist for the duration of the complete program execution.

When you write

string var1 = "Hello ";

you are creating a std::string object, and this object is initialized from the string literal. It is neither itself a string literal, nor does it hold or refer to one. It will simply copy the characters from the literal into some storage it manages itself. The var1 object (if declared at block scope, ie inside a function) has automatic storage duration, meaning that it lives until the end of the scope in which it was declared.

The line

string var3 = var1 + var2;

used to (prior to C++17) create a temporary std::string object, which was returned by the + operator overload of std::string. Then var3 was move-constructed from this temporary. The temporary resulting from var1 + var2 would then live until the end of the full expression, meaning until after the initialization of var3 was finished.

However, since C++17, there is no temporary involved here anymore. Instead, the + operator overload will directly construct the var3 object. Before C++17, this was also allowed to happen as so-called return value optimization.

So, in your code, there doesn't exist any "Hello World" string literal, but there exists exactly one string object that holds the characters Hello World, and if you are using a pre-C++17 version, there may be additionally one temporary std::string object that held the same value, but was destroyed after the initialization of var3.

String Literal in the temporary object?

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