string literal

Syntax

Explanation

Notes

The null character ('\0', L'\0', char16_t(), etc) is always appended to the string literal: thus, a string literal "Hello" is a const char[6] holding the characters 'H', 'e', 'l', 'l', 'o', and '\0'.

String literals placed side-by-side are concatenated at translation phase 6 (after the preprocessor). That is, "Hello," " world!" yields the (single) string "Hello, world!". If the two strings have the same encoding prefix (or neither has one), the resulting string will have the same encoding prefix (or no prefix).

L"Δx = %" PRId16 // at phase 4, PRId16 expands to "d"
                 // at phase 6, L"Δx = %" and "d" form L"Δx = %d"

Any other combination of encoding prefixes may or may not be supported by the implementation. The result of such a concatenation is implementation-defined.

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

String literals can be used to initialize character arrays. If an array is initialized like char str[] = "foo";, str will contain a copy of the string "foo".

The compiler is allowed, but not required, to combine storage for equal or overlapping string literals. That means that identical string literals may or may not compare equal when compared by pointer.

bool b = "bar" == 3+"foobar" // could be true or false, implementation-defined

Attempting to modify a string literal results in undefined behavior: they may be stored in read-only storage (such as .rodata) or combined with other string literals:

const char* pc = "Hello";
char* p = const_cast<char*>(pc);
p[0] = 'M'; // undefined behavior

In C, string literals are of type char[], and can be assigned directly to a (non-const) char*. C++03 allowed it as well (but deprecated it, as literals are const in C++). C++11 no longer allows such assignments without a cast.

A string literal is not necessarily a C string: if a string literal has embedded null characters, it represents an array which contains more than one string.

const char* p = "abc\0def"; // std::strlen(p) == 3, but the array has size 8

If a valid hex digit follows a hex escape in a string literal, it would fail to compile as an invalid escape sequence, but string concatenation can be used as a workaround:

//const char* p = "\xfff"; // error: hex escape sequence out of range
const char* p = "\xff""f"; // OK: the literal is const char[3] holding {'\xff','f','\0'}

Example

#include <iostream>
 
char array1[] = "Foo" "bar";
// same as
char array2[] = { 'F', 'o', 'o', 'b', 'a', 'r', '\0' };
 
const char* s1 = R"foo(
Hello
World
)foo";
//same as
const char* s2 = "\nHello\nWorld\n";
 
int main()
{
    std::cout << array1 << '\n' << array2 << '\n';
 
    std::cout << s1 << s2;
}

Foobar
Foobar
 
Hello
World
 
Hello
World

See also