C++ Storage class specifiers extern
Example
The extern storage class specifier can modify a declaration in one of the three following ways, depending on context:
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It can be used to declare a variable without defining it. Typically, this is used in a header file for a variable that will be defined in a separate implementation file.
// global scope int x; // definition; x will be default-initialized extern int y; // declaration; y is defined elsewhere, most likely another TU extern int z = 42; // definition; "extern" has no effect here (compiler may warn) -
It gives external linkage to a variable at namespace scope even if
constorconstexprwould have otherwise caused it to have internal linkage.// global scope const int w = 42; // internal linkage in C++; external linkage in C static const int x = 42; // internal linkage in both C++ and C extern const int y = 42; // external linkage in both C++ and C namespace { extern const int z = 42; // however, this has internal linkage since // it's in an unnamed namespace } -
It redeclares a variable at block scope if it was previously declared with linkage. Otherwise, it declares a new variable with linkage, which is a member of the nearest enclosing namespace.
// global scope namespace { int x = 1; struct C { int x = 2; void f() { extern int x; // redeclares namespace-scope x std::cout << x << '\n'; // therefore, this prints 1, not 2 } }; } void g() { extern int y; // y has external linkage; refers to global y defined elsewhere }
A function can also be declared extern, but this has no effect. It is usually used as a hint to the reader that a function declared here is defined in another translation unit. For example:
void f(); // typically a forward declaration; f defined later in this TU
extern void g(); // typically not a forward declaration; g defined in another TU
In the above code, if f were changed to extern and g to non-extern, it would not affect the correctness or semantics of the program at all, but would likely confuse the reader of the code.
