存储指针值 [英] Store pointer value

问题描述

据我所知，当将指针传递给函数时，它仅成为真实指针的副本.现在，我希望更改实指针，而不必从函数返回指针.例如:

As I know, when a pointer is passed into a function, it becomes merely a copy of the real pointer. Now, I want the real pointer to be changed without having to return a pointer from a function. For example:

int *ptr;

void allocateMemory(int *pointer)
{
     pointer = malloc(sizeof(int));
}

allocateMemory(ptr);

另一件事，就是如何将内存分配给2个或多个维数组?不是通过下标，而是通过指针算术.是这样吗?

Another thing, which is, how can I allocate memory to 2 or more dimensional arrays? Not by subscript, but by pointer arithmetic. Is this:

int array[2][3];
array[2][1] = 10;

与以下相同:

int **array;
*(*(array+2)+1) = 10

此外，为什么我必须将指针的内存地址传递给函数，而不是实际的指针本身.例如:

Also, why do I have to pass in the memory address of a pointer to a function, not the actual pointer itself. For example:

int * a;

为什么不:

allocateMemory(*a) 

但是

allocateMemory(a)

我知道我总是必须这样做，但是我真的不明白为什么.请向我解释.

I know I always have to do this, but I really don't understand why. Please explain to me.

最后一件事是，在这样的指针中:

The last thing is, in a pointer like this:

int *a;

是包含实际值的内存地址，还是指针的内存地址?我一直认为a是它指向的实际值的内存地址，但是对此我不确定.顺便说一下，当打印这样的指针时:

Is a the address of the memory containing the actual value, or the memory address of the pointer? I always think a is the memory address of the actual value it is pointing, but I am not sure about this. By the way, when printing such pointer like this:

printf("Is this address of integer it is pointing to?%p\n",a);
printf("Is this address of the pointer itself?%p\n",&a);

推荐答案

我将尝试一次解决这些问题:

I'll try to tackle these one at a time:

现在，我希望更改真正的指针而不必从函数返回指针.

Now, I want the real pointer to be changed without having to return a pointer from a function.

您需要再使用一层间接寻址:

You need to use one more layer of indirection:

int *ptr;

void allocateMemory(int **pointer)
{
    *pointer = malloc(sizeof(int));
}

allocateMemory(&ptr);

这是 c-faq.com/"rel =" noreferrer> comp.lang.c常见问题解答.

Here is a good explanation from the comp.lang.c FAQ.

另一件事，就是如何将内存分配给2个或多个维数组?

Another thing, which is, how can I allocate memory to 2 or more dimensional arrays?

为第一个维度分配一个分配，然后为另一个维度分配一个循环:

One allocation for the first dimension, and then a loop of allocations for the other dimension:

int **x = malloc(sizeof(int *) * 2);
for (i = 0; i < 2; i++)
    x[i] = malloc(sizeof(int) * 3);

同样，此处是

这是

int array[2][3];
array[2][1] = 10;

与以下相同:

int **array;
*(*(array+2)+1) = 10

绝对不是.指针和数组是不同的.但是，您可以有时互换使用它们.从这些问题. com/"rel =" noreferrer> comp.lang.c常见问题解答.

ABSOLUTELY NOT. Pointers and arrays are different. You can sometimes use them interchangeably, however. Check out these questions from the comp.lang.c FAQ.

此外，为什么我必须将指针的内存地址传递给函数，而不是实际的指针本身?

Also, why do I have to pass in the memory address of a pointer to a function, not the actual pointer itself?

为什么不呢?

allocateMemory(*a) 

有两件事-C没有传递引用，除非您自己通过传递指针来实现它，并且在这种情况下还因为a尚未初始化-如果要取消引用它，您会导致不确定的行为.此问题与

It's two things - C doesn't have pass-by-reference, except where you implement it yourself by passing pointers, and in this case also because a isn't initialized yet - if you were to dereference it, you would cause undefined behaviour. This problem is a similar case to this one, found in the comp.lang.c FAQ.

int *a;

是包含实际值的内存地址，还是指针的内存地址?

Is a the address of the memory containing the actual value, or the memory address of the pointer?

这个问题对我来说真的没有意义，但我会尽力解释. a(正确初始化时-在这里不是您的示例)是一个地址(指针本身). *a是指向的对象-在这种情况下，它将是int.

That question doesn't really make sense to me, but I'll try to explain. a (when correctly initialized - your example here is not) is an address (the pointer itself). *a is the object being pointed to - in this case that would be an int.

顺便说一句，当打印这样的指针时:

By the way, when printing such pointer like this:

printf("Is this address of integer it is pointing to?%p\n",a);
printf("Is this address of the pointer itself?%p\n",&a);

在两种情况下均正确.

这篇关于存储指针值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！