使用**更改地址的值

Question

我已经读过它意味着指向指针的指针.但在下面的代码中,我能够更改地址的值.

int main() {

    int x = 23;           // initializing variable X = 23
    int *myVar = &x;      // Creating a pointer to the address of X
    *myVar = 566;         // My attempt at changing the value of X address
    cout << x << endl;    // Printing out X with new value

}

它解决了.这怎么可能？**是指地址的价值吗？

Answer 1

这里有*两个不同的东西:

• int *myVar = &x;作为指针的*手段myVar
• *myVar = 566;这*意味着您可以访问指针指向的值

基本上,

int x = 23;           // 'x' is 23
int *myVar = &x;      // 'myVar' points to 'x'
*myVar = 566;         // Assign 566 to the value that 'myVar' is pointing at (which is 'x')
cout << x << endl;    // Print 'x'

你是正确的,**这意味着指向指针(或访问指针指针的值):

int a = 0; //'a' is 0
int* pa = &a; //'pa' is pointing to 'a'
int** ppa = &pa; //'ppa' is pointing to 'pa'

*pa = 1; //'a' is 1
**ppa = 2' //'a' is 2