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[英]I am trying to insert a new starting/head node in a linked list...The commented out function doesn't work while the one below it does the job
[英]Why does sending **head to a function work while reversing an SLL and *head doesn't in C?
我正在使用 Apple Clang 11.00 編譯我的代碼以反轉鏈接列表。 我起初使用此代碼但它不起作用:
void reverseSLL(node *head)
{
node *cur, *next, *prev;
cur = head;
prev = next = NULL;
if (head == NULL) {
printf("SLL does not Exist.\n");
return;
}
while (cur != NULL) {
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
head = prev;
}
然后我切換到這個並且它起作用了:
void reverseSLL(node **head)
{
node *cur, *next, *prev;
cur = *head;
prev = next = NULL;
if (*head == NULL) {
printf("SLL does not Exist.\n");
return;
}
while (cur != NULL) {
next = cur->next;
cur->next = prev;
prev = cur;
cur = next;
}
*head = prev;
}
我不明白為什么會這樣。 有人可以幫忙嗎?
C 中函數的參數是按值傳遞的,這意味着一旦函數返回,對函數體內參數值所做的更改就不起作用。 一個簡單的例子:
void foo(int a)
{
a += 10;
}
int main()
{
int b = 0;
printf("before: %d\n", b);
foo(b);
printf("after: %d\n", b);
return 0;
}
這將打印:
before: 0
after: 0
如果您希望調用函數(在本例中為main
)查看更新后的值,您必須傳遞一個指向int
的指針(您也可以返回更新后的值,但我們將在此處忽略):
void foo(int *a)
{
*a += 10;
}
int main()
{
int b = 0;
printf("before: %d\n", b);
foo(&b); /* Note that we are taking the address of 'b' here */
printf("after: %d\n", b);
return 0;
}
在這種情況下,我們的輸出將是:
before: 0
after: 10
傳遞指針也不例外。 對函數中的指針進行更改不會在調用函數中更改它,因此您必須將指針傳遞給該指針。 這就是您的第二個代碼示例有效的原因。
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