如何更改函数中的指针? [英] How do i change pointer in function?
问题描述
我在 C 中有一个递归函数,我希望返回的结构指针成为函数中的新结构.我对返回的结构有问题,因为它没有改变.这是我的代码结构:
I have a recursive function in C and i want the returned struct pointer to become the new struct in the function. Im having a problem with the returned struct because it doesnt change.this is the structure of my code:
struct location_t{
int x,y;
location_t * next;
int dir;
}
location_t * recursive_foo(location_t * loc, maze_t * m){
int x = loc->x;
int y = loc->y;
int dir = loc->dir;
loc->next = malloc(sizeof(location_t));
location_t * temp = loc->next;
if(m->map[--x][y] != '#' && dir != 0){
temp->x = x;
temp->y = y;
temp->dir = 2;
loc = recursive_foo(temp);
}
if(m->map[--x][y] != '#' && dir != 1){
temp->x = x;
temp->y = y;
temp->dir = 3;
loc = recursive_foo(temp);
}
if(m->map[--x][y] != '#' && dir != 2){
temp->x = x;
temp->y = y;
temp->dir = 0;
loc = recursive_foo(temp);
}
if(m->map[--x][y] != '#' && dir != 3){
temp->x = x;
temp->y = y;
temp->dir = 1;
loc = recursive_foo(temp);
}
return loc;
}
我对返回的结构有问题,因为它没有改变.
Im having a problem with the returned struct because it doesnt change.
这意味着通过相互引用来堆叠这些结构.
It is meant to stack these structs by referring to each other.
推荐答案
mystruct
是一个堆栈变量.换句话说,您是按值传递指针,而不是按引用传递.
mystruct
is a stack variable. In other words, you are passing the pointer by value, instead of passing it by reference.
你目前所做的与以下内容基本相同:
What have you done at the moment is essentially the same as:
int f(int i) {
...
i = <any value>;
...
}
在这种情况下,您只修改了值的副本.
In this case you are modifying only a copy of the value.
在您的程序中,您也在修改指针的副本.在函数之外,指针保持不变.
In your program, you are also modifying a copy of the pointer. Outside of the function the pointer stays not modified.
如果要修改它,需要传递一个指针给它:
If you want to modify it, you need to pass a pointer to it:
location_t * recursive_foo(location_t** loc, maze_t * m){
int x = (*loc)->x;
int y = (*loc)->y;
int dir = (*loc)->dir;
...
*loc = recursive_foo(&temp);
...
return *loc;
}
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