i具有一个C 结构,该结构具有动态分配的数组作为指针。我有扭转数组的功能,但它似乎不起作用(我认为这是因为临时变量指向原始值)。
struct s {
int *array;
int length;
...
s(int n) {
this->array = new int[n];
this->length = n;
}
...
void reverse() {
for (int i = 0; i < this->length; i++) {
int n = this->array[i];
this->array[i] = this->array[this->length - i - 1];
this->array[this->length - i - 1] = n;
}
}
...
}
我认为这是this->array[this->length - i - 1] = this->array[i]
因此,数组保持不变,不会逆转。我不知道如何尊重数组指针或如何将this->array[i]
的值在n中获取。
您的反面不起作用的原因是您要经历整个数组的长度。您只需要经过一半。如果您经历下半场,那么您将取消逆转。
作为一个例子,如果您尝试逆转[1, 2, 3, 4]
,则获得
after i = 0: [4, 2, 3, 1]
after i = 1: [4, 3, 2, 1]
--- reversed ---
after i = 2: [4, 2, 3, 1]
after i = 3: [1, 2, 3, 4]
--- back to original ---
而是使您的循环
for (int i = 0; i < this->length / 2; i++) {
...
}
在旁注上,使用2个索引器将大大简化您的代码:
void reverse()
{
int limit = length / 2;
for ( int front = 0 , back = length - 1; front < limit; front++ , back-- )
{
int n = array[front];
array[front] = array[back];
array[back] = n;
}
}