Question

我无法弄清楚为什么这个程序的输出就是这样。也许有人可以帮助我。

为什么双指针的引用：0062FB78？

为什么取消引用双指针的引用= 0062FAA0？

这些不应该被翻转吗？

0062FB78是x

的地址

我猜0062FAA0是双指针的地址？

#include <iostream>


void print(int x) {
    std::cout << "value: " << (x) << "\n";
    std::cout << "reference: " << (&x) << "\n";
    //std::cout << (*x) << "\n";
}

void printPointer(int *x) {
    std::cout << "value: " << x << "\n";
    std::cout << "reference: " << &x << "\n";
    std::cout << "dereference:" << *x << "\n";
}

void printDoublePointer(int **x) {
    std::cout << "value: " << x << "\n";
    std::cout << "reference: " << &x << "\n";
    std::cout << "dereference:" << *x << "\n";
    printPointer(*x);
}

void printTripplePointer(int ***x) {
    std::cout << "value:" << x << "\n";
    std::cout << "reference:" << &x << "\n";
    std::cout << "dereference:" << *x << "\n";
    printDoublePointer(*x);
}

void print(char* string) {
    std::cout << "\n" << string << "\n";
}


int main()
{
    int x = 19;
    int *y; // y is a address space
    y = &x; // &y now points to the address of x, *y now has the value of x
    int **doublePointer = &y;
    print(x);
    printPointer(y);
    printDoublePointer(doublePointer);
    print("doublePointer");
    std::cin >> x;
}

x
value: 19
reference: 0062FBB78

y
value: 0062FC7C
reference: 0062FBB78
defererence: 19

doublePointer
value: 0062FC58
reference of double Pointer: 0062FB78
dereference of doble Pointer: 0062FC7C
value of dereferenced double pointer: 0062FC7C
reference of dereferenced double pointer: 0062FAA0
dereference: 19

Answer 1

在回答您的问题之前，我们首先同意在致电y= &x后，y不是x的引用，而是x的地址。

现在，让我们检查对print

的调用

如果密切关注，我们会传递变量by-value，因此此方法实际上会打印值19，但该地址将属于x的临时副本。

如果我们将原型更改为以下原型，则此处打印的x地址将等于方法y

中打印的printPointer地址

void print(int & x) {
    std::cout << __PRETTY_FUNCTION__ << "\n";
    std::cout << "value: " << (x) << "\n";
    std::cout << "reference: " << (&x) << "\n";
}

关于你的其他问题，这些问题也会发生，因为你按指令传递指针而不是按引用传递。

这个简单的程序显示一切正常：

int main()
{
    int x = 19;
    int *y = &x;
    int **z = &y;

    std::cout << x << "\t" << &x << std::endl;
    std::cout << y << "\t" << &y << "\t" << *y << std::endl;
    std::cout << z << "\t" << &z << "\t" << *z << std::endl;
}

引用双指针，有什么价值？

1 个答案: