在实现数据结构“堆栈”期间执行循环中的逻辑错误

Question

我的老师给了我一道作业题，我们必须将push 和pop 元素放在一个堆栈中。

输入应为：-

1. 输入的第一行必须包含 no。堆栈中的元素。

2. 输入的第二行必须包含用户（即）选择是 PUSH（即输入'1'）还是 POP 元素（即输入 '2')。

3. 如果选择了 PUSH 操作作为用户的选择；那么输入的第三行必须包含要压入堆栈的元素。

4. 如果用户希望继续这些操作，下一行输入必须包含'y' 或'n' 作为回复。

测试用例 1

输入

3//(capacity of stack)

1 //(selecting PUSH OR POP)

6//(entering the element which is to entered)

是//(to continue or not continue)

1 //(selecting PUSH OR POP)

4//(entering the element which is to entered)

是//(to continue or not continue)

1 //(selecting PUSH OR POP)

7//(entering the element which is to entered)

是//(to continue or not continue)

2//(selecting PUSH OR POP)

n//(to continue or not continue)

输出

deleted element is

7 4 6

针对上述问题我写了如下代码：-

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

struct Stack
{
    int capacity;
    int top;
    int *array;
};

void push(struct Stack *stack, int a) //function to PUSH a character in the stack.
{
    stack->array[++stack->top] = a;
}

int pop(struct Stack *stack) //function to POP a character in the stack.
{
    return stack->array[stack->top--];
}

int main(void)
{
    struct Stack obj;
    obj.top = -1;

    printf("Enter the capacity of stack\n");
    scanf("%d", &obj.capacity); //Inputting the capacity of the stack.

    obj.array = calloc(obj.capacity, sizeof(int));

    int operation;
    int element;
    char continuation;

    do
    {
        printf("\nEnter 1 if you want to PUSH or 2 for POP\n");
        scanf("%d", &operation);

        printf("\nEnter the element which is to be pushed\n");
        scanf("%d", &element);

        scanf("%*c"); //To ignore any newline in stdin buffer.
        printf("\nEnter 'y' if you want to continue else enter 'n'\n");
        scanf("[a-z]%c", &continuation);

        if(operation == 1)
        {
            if(obj.top < obj.capacity)
            {
                push(&obj, element);
            }
            else
            {
                printf("Error\n");
                return EXIT_FAILURE;
            }
        }

        else if(operation == 2)
        {
            printf("deleted element is\n");
            while(obj.top != -1) //will POP all elements on the stack and print it.
            {
                printf("%d", pop(&obj));
                if(obj.top != 0)
                {
                    printf(" ");
                }
            }
        }

        else
        {
            printf("Wrong operation specified\n");
            return EXIT_FAILURE;
        }
    } while(continuation == 'y');

    return 0;
}

我在上面代码中遇到的问题是输入以下内容后：-

3

1

6

是的

程序关闭（即退出do-while 循环）。但是，当我在代码中更改以下行时：-

char continuation;

到

char continuation = 'y';

在我输入以下内容之前它工作正常：-

3

1

6

是的

1

4

是的

1

7

是的

然后给我一个突然的输出：-

Error

然后退出程序。

我的问题是：-

当我只写char continuation;时，为什么我的代码在第一种情况下不起作用？

当我将char continuation; 更改为char continuation = 'y'; 时，第二种情况下的“逻辑错误”是什么？

Answer 1

我可以通过删除"[a-z]%c" 并将其替换为" %c" 来解决我的问题。

这是我的新（正确）代码。

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

struct Stack
{
    int capacity;
    int top;
    int *array;
};

void push(struct Stack *stack, int a) //function to PUSH a character in the stack.
{
    stack->array[++stack->top] = a;
}

int pop(struct Stack *stack) //function to POP a character in the stack.
{
    return stack->array[stack->top--];
}

int main(void)
{
    struct Stack obj;
    obj.top = -1;

    printf("Enter the capacity of stack\n");
    scanf("%d", &obj.capacity); //Inputting the capacity of the stack.

    obj.array = calloc(obj.capacity, sizeof(int));

    int operation;
    int element;
    char continuation;

    do
    {
        printf("\nEnter 1 if you want to PUSH or 2 for POP\n");
        scanf("%d", &operation);

        if(operation == 1)
        {
            printf("\nEnter the element which is to be pushed\n");
            scanf("%d", &element);

            if(obj.top < obj.capacity)
            {
                push(&obj, element);
            }
            else
            {
                printf("Error\n");
                return EXIT_FAILURE;
            }
        }

        else if(operation == 2)
        {
            printf("deleted element is\n");
            while(obj.top != -1) //will POP all elements on the stack and print it.
            {
                printf("%d", pop(&obj));
                if(obj.top != 0)
                {
                    printf(" ");
                }
            }
        }

        else
        {
            printf("Wrong operation specified\n");
            return EXIT_FAILURE;
        }

        scanf("%*c"); //To ignore any newline in stdin buffer.
        printf("\nEnter 'y' if you want to continue else enter 'n'\n");
        scanf(" %c", &continuation);

    } while(continuation == 'y');

    return 0;
}

注意：- 我也避免按照 cmets 中的建议交错 printf。

但是我仍然无法理解为什么我以前的代码不起作用。

附言

我终于弄错了，应该是"%[a-z]c"，而不是"[a-z]%c"。