C-字符推送和弹出操作 [英] C- Character Push and Pop operation

问题描述

Stack create(int c)
{
    Stack S=(Stack)malloc(sizeof(struct stack));

    S->size=c;
    S->top=-1;
    S->array=(char *)malloc(sizeof(char)*c);

    return S;
}

Stack makeEmpty(void)
{
    Stack *S1=create(100);
    S1[0]->top=-1;
    return S1;
}

char pop(Stack S)
{
    return S->array[S->top--];
};

int main(void)
{
    Stack *S1;
    S1=makeEmpty();
    int j;
    int k=0;
    char result[30];
    for(j=0; j<2; j++)
    {
        char result1=pop(S1);
        strcat(result, result1);
        k++;
    }
}

我跳过了某些部分，例如 typedef struct stack Stack ;
我想做的是在for循环工作时从堆栈中弹出元素。然后，将这些元素存储在一个新数组中，该数组为 result 。为了检查它是否有效，我打印了出来，但是遇到运行时错误。如何存储元素以及如何将其打印出来？

I skipped some parts, like typedef struct stack Stack; What I wanted to do was pop out elements from the stack while for-loop works. Then, store those elements in a new array which is result. To check whether it works or not, I printed out but I had a runtime error. How to store the element and how to print it out?

推荐答案

我已经复制并粘贴了代码，并将其粘贴没有被编译。我认为
您要么不发布您的实际代码，要么不费心编译和阅读编译器警告。帮助您相当困难。在这里，我注意到了一些事情：

I've made copy&paste of your code, and it doesn't get compiled. I think that you are either not posting your actually code nor you don't bother to compile and read the compiler warnings. It's rather difficult to help you. Here some things I noticed:

1。
create 必须返回指向 Stack 而不是对象的指针。

1. create must return a pointer to Stack, not the object.

Stack *create(int c)
{
    Stack *S = malloc(sizeof *S);

    S->size=c;
    S->top=-1;
    S->array = malloc(c);

    return S;
}

2。
makeEmpty

Stack *makeEmpty(void)
{
    Stack *S1=create(100);
    S1->top=-1;
    return S1;
}

3。
pop 应该获得指向 Stack 而不是对象的指针

3. pop should get a pointer to Stack, not the object

char pop(Stack *S)
{
    return S->array[S->top--];
};

在这里，您应该检查堆栈中是否有元素。 int pop（Stack * S，char * val）返回1并在$ b $上的 * val 上写入b成功，然后返回0，否则会更好。

Here you should check whether there are elements on your stack. int pop(Stack *S, char *val) where it returns 1 and writes on *val on success, and returns 0 otherwise would be better.

4。
从您的流行音乐来看，您仅在推送 char 。我不明白您
正在尝试使用 strcat 做什么。无论哪种方式，您都在执行 strcat 错误。您
声明了一个包含100个空格的堆栈，但您只声明了30个空格
表示结果。如果堆栈中有31个以上的元素怎么办？我知道
您仅检查2个元素，但是很容易忽略，并且
对其进行扩展以遍历所有堆栈，而不会更改 result <的内存要求
/ code>。

4. Judging from your pop you are pushing char only. I don't get what you are trying to do with strcat. Either way, you are doing strcat wrong. You are declaring a stack with 100 spaces, but you are only declaring 30 spaces for result. What if you have more than 31 elements on your stack? I know that you are only inspecting 2 elements but it's easy to overlook that and expand it to go through all the stack without changing the memory requirements for result.

也 strcat 是可与C字符串一起使用的函数，这意味着它期望
C字符串。 C字符串必须必须 \0 终止，而不是。您有
类似C字符串的东西，但实际上不是。如果您坚持使用
strcat ，则应这样做：

Also strcat is a function that works with C-Strings, that means it expects C-Strings. A C-String must be \0 terminated, yours are not. You have something that looks like a C-String but it's not. If you insist on using strcat, the you should do it like this:

for(j=0; j<2; j++)
{
    char result1[] = { pop(S1), 0 };
    strcat(result, result1);
}

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