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Can a local variable s memory be accessed outside its scope? (21 answers)

What is a dangling pointer? (8 answers)

Closed 13 mins ago.

这一方案是否没有界定行为?

struct Storage {
    void store(void (*&&fp)())  { fpp = &fp; }
    void call() const           { (*fpp)(); }

    void (**fpp)();
};

void f();

int main() {
    Storage s;
    s.store(&f);    // I think, the temporary variable &F is freed at the  ; 
    s.call();
}

Answer 1

是的,这是联邦。

我们可以取消一切职能,使之更加简单:

#include <iostream>

struct Storage {
    void store(int* && fp)  { fpp = &fp; }
    void call() const           { std::cout << *fpp; }

    int** fpp;
};

int main() {
    int i = 13;
    Storage s;
    s.store(&i); 
    s.call();
}

你可以进一步消除双重点,但这也应明确。 <代码>fpp是当地变式<代码>fp的参照点,在构造结束后立即超出范围。参照这个要点是分行。

关于补充证据,您的原始代码触发器 Address Sanitizer的错误:

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