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The 'static' keyword in C ++

I understand that it retains value after exiting the scope (but becomes inaccessible), but I have a few questions.

  • When people say that this is not available outside the scope, does it simply mean that you cannot change the value (it will be erroneous) outside the scope of identification?

  • I was thinking about this code:

    #include "iostream"

void staticExample();

int main()
{
    staticExample();

    return 0;
}

void staticExample()
{
    for (int i = 1; i <= 10; ++i)
    {
        static int number = 1;
        std::cout << number << "\n";

        ++number;
    }
}

and I thought to myself that at each iteration of the loop I set the 'number' variable to 1. As I expected, this was printed 1, 2, 3 .. 10. The compiler recognizes that the line setting it to 1 was a declaration and ignores her "change"?

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5 answers

, . . (IINA? い い, な!)

+2

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#include <iostream> 

void staticExample()
{
    static int number = 1;

    for (int i = 1; i <= 10; ++i)
    {
        std::cout << number << "\n";
        ++number;
    }
}

int main()
{
    staticExample();  // begins counting at 1
    staticExample();  // begins counting at 10

    return 0;
}

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+3

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+1

  • , :

    {
    static int x;
    x = 5;
}
x = 6;  // Compiler error here!  We're outside the scope that x was declared in.

    , . :

    int *p = NULL;
{
    static int x;
    x = 5;
    p = &x;
}
*p = 6;  // This is fine

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0

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Your code example = 1has an initializer, not a job. It is used when a variable is initialized, which, since the object is declared static, occurs only the first time that execution passes through the declaration statement.

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