如何用c语言实现链接中的代码

您提到的需求是在C语言中实现类似于链接中提供的C++代码的功能。以下是根据链接内容，将其核心逻辑转成C语言版本的方法概述：

### 1. **定义栈和队列**

在C语言中，可以通过结构体来定义栈和队列，并实现其基本操作。

#### 栈的定义和操作

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

#define MAX_SIZE 100

typedef struct {
    char data[MAX_SIZE];
    int top;
} Stack;

void initStack(Stack *stack) {
    stack->top = -1;
}

int isStackEmpty(Stack *stack) {
    return stack->top == -1;
}

void push(Stack *stack, char item) {
    if (stack->top < MAX_SIZE - 1) {
        stack->data[++stack->top] = item;
    }
}

char pop(Stack *stack) {
    if (!isStackEmpty(stack)) {
        return stack->data[stack->top--];
    }
    return '\0';
}

#### 队列的定义和操作

typedef struct {
    char data[MAX_SIZE];
    int front;
    int rear;
} Queue;

void initQueue(Queue *queue) {
    queue->front = 0;
    queue->rear = -1;
}

int isQueueEmpty(Queue *queue) {
    return queue->front > queue->rear;
}

void enqueue(Queue *queue, char item) {
    if (queue->rear < MAX_SIZE - 1) {
        queue->data[++queue->rear] = item;
    }
}

char dequeue(Queue *queue) {
    if (!isQueueEmpty(queue)) {
        return queue->data[queue->front++];
    }
    return '\0';
}

### 2. **实现字符串逆置**

void reverseStr(char str[], int length) {
    int start = 0;
    int end = length - 1;
    while (start < end) {
        char temp = str[start];
        str[start] = str[end];
        str[end] = temp;
        start++;
        end--;
    }
}

### 3. **将字符串逆序压入栈中**

void invert(Stack *stack, const char sentence[]) {
    int len = strlen(sentence);
    for (int i = len - 1; i >= 0; --i) {
        push(stack, sentence[i]);
    }
}

### 4. **输入检查和递归处理**

void inputAndCheck(Stack *stack, char ch, int index, char test[26][MAX_SIZE]) {
    char upperVoca[MAX_SIZE];
    int vocaVal;
    char final[MAX_SIZE];

    printf("请输入%c字符的解释: ", ch);
    scanf("%s", upperVoca);

    for (int i = 0; i < strlen(upperVoca); i++) {
        vocaVal = upperVoca[i];
        if (isupper(vocaVal) && test[vocaVal - 'A'][0] == '\0') {
            if (vocaVal - 'A' != index) {
                Stack newStack;
                initStack(&newStack);
                inputAndCheck(&newStack, upperVoca[i], vocaVal - 'A', test);
            }
        }
        if (isupper(vocaVal) && test[vocaVal - 'A'][0] != '\0') {
            invert(stack, test[vocaVal - 'A']);
        } else {
            push(stack, upperVoca[i]);
        }
    }

    int i = 0;
    while (!isStackEmpty(stack)) {
        final[i++] = pop(stack);
    }
    reverseStr(final, i);

    strcpy(test[index], final);
    test[index][i] = '\0';
}

### 5. **主函数**

int main() {
    char test[26][MAX_SIZE];
    char language[MAX_SIZE];
    char inBrackets[MAX_SIZE];
    char translatedInBra[MAX_SIZE];
    char translation[MAX_SIZE];
    int flag = 0;
    Stack translatedLanguage;
    Stack finalInterp;

    initStack(&translatedLanguage);
    initStack(&finalInterp);

    for (int i = 0; i < 26; ++i) {
        test[i][0] = '\0';
    }

    printf("请输入魔王语言: ");
    scanf("%s", language);

    int langLength = strlen(language);
    invert(&translatedLanguage, language);

    for (int i = 0; i < langLength; i++) {
        int Char = language[i];
        if (isupper(Char)) {
            if (test[Char - 'A'][0] == '\0') {
                inputAndCheck(&finalInterp, language[i], Char - 'A', test);
            }
        }
    }

    int i = 0;
    int j = 0;
    while (!isStackEmpty(&translatedLanguage)) {
        char tmp = pop(&translatedLanguage);
        if (isupper(tmp)) {
            invert(&translatedLanguage, test[tmp - 'A']);
        } else if (tmp == ')') {
            flag = 1;
        } else if (tmp == '(') {
            flag = 0;
            int inBracketsLength = strlen(inBrackets);
            int counter1 = 0;
            int counter2 = 0;
            while (counter1 != inBracketsLength) {
                translatedInBra[counter2] = inBrackets[inBracketsLength - 1];
                if (counter1 != (inBracketsLength - 1)) {
                    translatedInBra[counter2 + 1] = inBrackets[counter1];
                }
                counter1++;
                counter2 += 2;
            }
            invert(&translatedLanguage, translatedInBra);
        } else {
            if (flag) {
                inBrackets[i++] = tmp;
            }
            if (!flag) {
                translation[j++] = tmp;
            }
        }
    }

    reverseStr(translation, j);
    for (int k = 0; k < j; k++) {
        printf("%c", translation[k]);
    }

    return 0;
}

### 总结
以上代码实现了从魔王语言到人类语言的翻译过程，通过栈和队列的配合，处理了括号和大写字母的解释。希望这些代码能够满足您的需求。如果有任何问题或需要进一步的帮助，请随时提问！