C++ lifetime of a dangling string

Question

In C, you can write

const char *result = "AB";

Is this style supported in C++ by standard? Is the life-time of this constant string guaranteed along with the same scope of the pointer?

Possible duplicate of [String literals: Where do they go?](https://stackoverflow.com/questions/2589949/string-literals-where-do-they-go) — Lanting, Jun 03 '19 at 06:52
String constant will be part of code segment, it is just that when the scope of the pointer ends you are just loosing reference to constant string. — kiran Biradar, Jun 03 '19 at 06:53
@kiranBiradar It's not necissarily the code segment. Some platforms do have three segments: Code (readable and excecutable), read-only data (just readable) and standard data (readable and writeable). But your main point that the literal itself does never "go out of scope" is valid nevertheless. — Michael Karcher, Jun 03 '19 at 07:03

score 1 · Answer 1 · answered Jun 03 '19 at 06:52

Literals string constants have a lifetime of the whole program, and the arrays the strings are stored in never go out of scope.

Note that there's a semantic difference between literal strings in C and C++: In C++ literal strings are stored in arrays of constant characters (therefore the const in const char* is needed). In C they aren't constant arrays (so char * is alright in C). However, it's not allowed to modify a literal string in C, which makes them read only (but not constant).

songyuanyao · Answer 2 · 2019-06-03T07:06:52.590

Is the life-time of this constant string guaranteed along with the same scope of the pointer?

No, the lifetime of string literals has nothing to do with the lifetime of pointers pointing to them; String literals exist in the whole life of the program.

String literals have static storage duration, and thus exist in memory for the life of the program.

[lex.string]

6 After translation phase 6, a string-literal that does not begin with an encoding-prefix is an ordinary string literal. An ordinary string literal has type “array of n const char” where n is the size of the string as defined below, has static storage duration ([basic.stc]), and is initialized with the given characters.

15 Evaluating a string-literal results in a string literal object with static storage duration, initialized from the given characters as specified above. ...

score 1 · Answer 3 · answered Jun 03 '19 at 06:59

Quoting C++17, chapter § 5.13.5 (emphasis mine)

Ordinary string literals and UTF-8 string literals are also referred to as narrow string literals. A narrow string literal has type “array of n const char”, where n is the size of the string as defined below, and has static storage duration

and, for static storage duration, chapter § 6.7.1

All variables which do not have dynamic storage duration, do not have thread storage duration, and are not local have static storage duration. The storage for these entities shall last for the duration of the program.

So, the lifetime of the string literals is the entire execution of the program, it never goes out of scope.

score 0 · Answer 4 · answered Jun 03 '19 at 06:52

0

The string constant (literal) has the same lifetime as the whole program. From way before the pointer is created to long after it's destroyed

answered Jun 03 '19 at 06:52

C++ lifetime of a dangling string

4 Answers4