修改指针在C ++函数中指向的位置 [英] Modify the location a pointer points to in a C++ function

问题描述

我已经坚持修改指针的指针。问题是，我不明白为什么我的代码工作。我想做的是修改指针指向指针在一个函数中的位置。然后在我的主函数中访问该值。我尝试了相当多的尝试，这是我使它工作的唯一方式。

I have gotten stuck on modifying pointers of pointers. The problem is that I don't understand why my code works. What I am trying to do is modify where a pointer-to-pointer points to in a function. Then accessing that value in my main function. I tried quite a few attempts and this is the only way I got it to work.

#include <iostream>
using namespace std;
void changePP(int **ppint) {
    int *n = new int;
    *n = 9; //just a value for demonstration purposes
    *ppint = n; //THE LINE IN QUESTION
    delete n;
}
int main() {
    int **ppint = NULL;
    int *p = new int;
    *p = 4; //another value for demonstrating
    ppint = &p;
    cout << **ppint << endl;
    changePP(ppint);
    cout << **ppint << endl;
}

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4，然后在单独的行上9。但是，我不知道代码中的行 * ppint = n 为什么我必须使用 * 来更改ppint在changePP函数中指向的位置，而不是在main中？为什么我不必在函数中使用& ？我不能找到一个我可以在互联网上理解的解释，我想知道有人能为我解释这个。

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So, the output is 4 and then a 9 on seperate lines. However, I am not sure about the line *ppint = n in the code. Why do I have to use the * to change where ppint points to in the changePP function but not in the main? Also why do I not have to use the & in the function? I can't seem to find an explanation that I can understand on the internet and I was wondering if someone could explain this for me?

推荐答案

注意：这些内存地址仅供参考。

Note: these memory addresses are only for illustration.

0x0A | 4
0x0B | 'p' -> 0x0A // Pointer to the value 4 in location 0x0A
0x0C | 'ppint' -> NULL // Initially a null pointer

执行 ppint =& p; / code>产生以下结果，因为 ppint 是位于 0x0C 的指针。

0x0A | 4
0x0B | 'p' -> 0x0A
0x0C | 'ppint' -> 0x0B

对于 changePP 函数，我将其称为 ppintCopy ，以减少混乱。它是一个副本（即，不同于 ppint 的指针），并修改它只修改副本。

As for the parameter in the changePP function, I will refer to it as ppintCopy for less confusion. It's a copy (i.e., a pointer different than ppint) and modifying it only modifies the copy.

0x0D | 'ppintCopy' -> 0x0C // A pointer that points to `ppint`

执行 ppintCopy =& ; n 将修改 0x0D 处的指针，而不是 main 函数。 但，引用 ppintCopy （即 * ppintCopy ）会产生 0X0C ，它是 main 函数的指针，因此 * ppintCopy = n; 其中指针 0x0C 指向。

Performing ppintCopy = &n would modify the pointer at 0x0D which is not the location of the pointer from the main function. However, deferencing ppintCopy (i.e., *ppintCopy) yields 0X0C which is the pointer from the main function therefore *ppintCopy = n; is changing where the pointer 0x0C points.

为什么我必须使用*改变哪里ppint指向changePP函数，但不在主？

Why do I have to use the * to change where ppint points to in the changePP function but not in the main?

在上面的插图中我希望它更清楚为什么在 main 中工作，以及为什么必须在 changePP 函数中使用不同的语法。

With the above illustration I hope it is more clear why it works in main and why you have to use a different syntax in the changePP function.

为什么我不必使用&在函数中

Also why do I not have to use the & in the function?

在 main 函数中，变量 ppint 的类型为 int ** （即指向指针的指针）和 p 的类型为 int * 。您不能执行 ppint = p; ，因为它们是不同的类型。如果删除一个级别的间接，这可能更容易看到。例如：

In the main function the variable ppint is of type int** (i.e., a pointer to a pointer) and p is of type int*. You cannot perform ppint = p; because the they are different types. This may be easier to see if you remove one level of indirection. For example:

int* pMyInt = NULL;
int myInt = 3;

我认为这是很自我解释，这不能工作（即，他们是不同的类型） p>

I think it's pretty self explanatory that this cannot work (i.e., they are different types).

pMyInt = myInt;

However you can take the address of myInt using the & operator which results in a pointer to int (i.e., int*) and since this type is the same as the type of pMyInt the assignment is now possible.

pMyInt = &myInt;

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