Question

所以目前我在if函数中有一个GetNth语句，我试图测试。但是当我插入一个printf函数时，它会让我注意到它会通过if语句，即使条件不满足，但是，当我删除printf语句时程序正常工作完美无缺。任何解释将不胜感激。

注意！这不是我的代码，我试图研究链表并且正在改变试图学习的代码！

守则：

#include <stdio.h>
#include <stdlib.h>
#include <assert.h>

/* Link list node */
struct node
{
    int data;
    struct node* next;
};

/* Given a reference (pointer to pointer) to the head
 of a list and an int, push a new node on the front
 of the list. */
void push(struct node** head_ref, int new_data)
{
    /* allocate node */
    struct node* new_node =
    (struct node*) malloc(sizeof(struct node));

    /* put in the data  */
    new_node->data  = new_data;

    /* link the old list off the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref)    = new_node;
}

/* Takes head pointer of the linked list and index
 as arguments and return data at index*/
int GetNth(struct node* head, int index)
{
    struct node* current = head;
    int count = 0; /* the index of the node we're currently
                    looking at */
    int a;
    while (current != NULL)
    {
        if (count == index)
            return(current->data);
            a = current->data;
            printf("\n Testing If in linked list, should bring same desired value which is 4 %d \n ",a);
        count++;
        current = current->next;
    }

    /* if we get to this line, the caller was asking
     for a non-existent element so we assert fail */
    assert(0);
}

/* Drier program to test above function*/
int main()
{
    /* Start with the empty list */
    struct node* head = NULL;

    /* Use push() to construct below list
     1->12->1->4->1  */
    push(&head, 1);
    push(&head, 4);
    push(&head, 1);
    push(&head, 12);
    push(&head, 1);
    if (head != NULL)
    {

    }
    /* Check the count function */
    printf("Element at index 3 is %d", GetNth(head, 3));
    getchar();
}

Answer 1

缺少牙箍。

这就是为什么我是“永远添加大括号”的捍卫者。

编辑“解决方案”。

目前的代码是：

while (current != NULL)
{
    if (count == index)
        return(current->data);
        a = current->data;
        printf("\n Testing If in linked list, should bring same desired value which is 4 %d \n ",a);
    count++;
    current = current->next;
}

没有大括号，if语句仅适用于下一条指令，即return(current->data);

如果要在if块中包含多个指令，则必须创建一个带括号的块。

if (count == index)
{
        return(current->data);
        a = current->data;
        printf("\n Testing If in linked list, should bring same desired value which is 4 %d \n ",a);
}

但是，您从返回指令开始，因此永远不会执行以下2行。

在返回之前重新排列指令以进行打印。

if (count == index)
{
     a = current->data;
     printf("\n Testing If in linked list, should bring same desired value which is 4 %d \n ",a);
     return(current->data);
}

Answer 2

首先，即使条件为假，它也不会进入select count(*) total_by_value, value, s1.full_avg, s1.full_total from ratings r, (select avg(value) full_avg, count(*) full_total from ratings) s1 group by s1.full_avg, s1.full_total, value;。

如果没有花括号，

if statement语句只考虑if语句后的紧接下一个语句。

If

如果statement为true，它只考虑return语句。

如果statement为false，则转到if (count == index) return(current->data);之后的下一个语句即

这就是你觉得if语句不起作用的原因。

如果你需要在if循环中使用a = current->data; printf("\n Testing If in linked list, should bring same desired value which is 4 %d \n ",a);，你需要在if循环中使用多个语句的语法。即通过花括号

printf

如果链表中的Statement无法按预期工作

2 个答案: