问题描述
我有一个编程任务，我偶然发现了这个小问题：当我给函数一个指针作为参数时，我无法改变它指向的地址。我通过返回我希望指针指向的新地址来解决这个问题。但我仍然想知道为什么不能操作指针参数，因为所有内存分配函数都使用返回值而不是参数列表。
我可能做错了什么？或者是否真的不可能改变指针？有没有人有解释？
例子：
void foo（ int * ptr）
{
ptr = malloc（sizeof（int））;
} / *在调用这个函数后，我会
期望指针变成了
新分配的内存，但它保持为NULL * /
int main（）
{
int * ptr = NULL;
foo（ptr）;
返回0;
内存分配函数返回指针值，所以它通常被分配给一个指针
。
传递一个函数指针作为参数是一个不同的故事，
因为改变它不会更改原始值（按值传递）。这是
为什么我们在这些情况下使用指针，给变量
的地址并在函数中更改其值。
如果需要要传递一个指针，并改变它指向的内容，使用
a指针指向一个指针！
void
my_function（int **指针）
{
* pointer =& something; //将改变指针地址
}
int * my_pointer =& something_else;
my_function（&my_pointer）;
I had a programming assignment a while back where I stumbled upon this little problem: when I gave a function a pointer as a parameter, I could not change the address it pointed at. I solved that by returning the new adress I wanted the pointer to point to. But I am still wondering why it's not possible to manipulate a pointer parameter because all memory allocating functions work with a return value as well instead of a parameter list.
Was I possibly doing something wrong? Or is it really not possible to change the pointee? Does anyone have an explanation?
Example:
void foo(int *ptr)
{
ptr=malloc(sizeof(int));
} /* after calling this function I would
expect the pointee to have changed to the
newly allocate memory but it stays NULL*/
int main()
{
int *ptr=NULL;
foo(ptr);
return 0;
}
Memory allocating functions return the pointer value, so it is generallyassigned to a pointer.
Passing a pointer to a function as an argument is a different story,because changing it won't change the original (pass by value). That'swhy we use pointers in these cases, to give the address of a variableand change its value in the function.
If you need to pass a pointer instead, and change what it points to, usea pointer to a pointer!
void
my_function(int **pointer)
{
*pointer = &something; // will change the pointer address
}
int *my_pointer = &something_else;
my_function(&my_pointer);
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