K&R练习1-24

以下是K&R练习1-24的解决方法，其中包含了代码示例：

#include <stdio.h>

#define MAX_LENGTH 1000

int checkBalance(char code[MAX_LENGTH]);
void checkQuotes(char code[MAX_LENGTH], int* inSingleQuotes, int* inDoubleQuotes);

int main() {
    char code[MAX_LENGTH];
    int i, c, inComment = 0, inSingleQuotes = 0, inDoubleQuotes = 0;

    i = 0;
    while ((c = getchar()) != EOF) {
        code[i] = c;
        i++;
    }
    code[i] = '\0';

    if (checkBalance(code) == 0) {
        printf("Parentheses, brackets, and braces are balanced.\n");
    } else {
        printf("Parentheses, brackets, or braces are not balanced.\n");
    }

    return 0;
}

int checkBalance(char code[MAX_LENGTH]) {
    int i, c, prevC, inComment = 0, inSingleQuotes = 0, inDoubleQuotes = 0;
    int parenthesesBalance = 0, bracketsBalance = 0, bracesBalance = 0;

    for (i = 0; code[i] != '\0'; i++) {
        c = code[i];

        if (c == '/' && prevC == '/') {
            inComment = 1;
        } else if (c == '\n') {
            inComment = 0;
        } else if (!inComment) {
            if (c == '\'' && !inDoubleQuotes) {
                checkQuotes(code, &inSingleQuotes, &inDoubleQuotes);
            } else if (c == '"' && !inSingleQuotes) {
                checkQuotes(code, &inSingleQuotes, &inDoubleQuotes);
            } else {
                if (c == '(') {
                    parenthesesBalance++;
                } else if (c == ')') {
                    parenthesesBalance--;
                } else if (c == '[') {
                    bracketsBalance++;
                } else if (c == ']') {
                    bracketsBalance--;
                } else if (c == '{') {
                    bracesBalance++;
                } else if (c == '}') {
                    bracesBalance--;
                }
            }
        }

        prevC = c;
    }

    if (parenthesesBalance == 0 && bracketsBalance == 0 && bracesBalance == 0) {
        return 0;
    } else {
        return -1;
    }
}

void checkQuotes(char code[MAX_LENGTH], int* inSingleQuotes, int* inDoubleQuotes) {
    int i, c, prevC, escaped = 0;

    for (i = 0; code[i] != '\0'; i++) {
        c = code[i];

        if (c == '\\' && prevC != '\\') {
            escaped = 1;
        } else if (!escaped) {
            if (c == '\'' && *inSingleQuotes) {
                *inSingleQuotes = 0;
                break;
            } else if (c == '\'' && !*inSingleQuotes) {
                *inSingleQuotes = 1;
                break;
            } else if (c == '"' && *inDoubleQuotes) {
                *inDoubleQuotes = 0;
                break;
            } else if (c == '"' && !*inDoubleQuotes) {
                *inDoubleQuotes = 1;
                break;
            }
        } else {
            escaped = 0;
        }

        prevC = c;
    }
}

以上代码实现了一个简单的代码检查器，用于检查代码中的圆括号、方括号和大括号是否平衡。代码首先通过getchar()函数从标准输入中读取代码，并存储在字符数组code中。然后，通过checkBalance()函数检查代码中的圆括号、方括号和大括号是否平衡，如果平衡则输出相应的提示信息。

checkBalance()函数遍历code数组，通过检查各种符号的出现次数来判断是否平衡。同时，它还使用checkQuotes()函数来处理单引号和双引号中的情况。

checkQuotes()函数遍历code数组，通过检查引号的出现次数和引号中的转义字符来判断引号是否成对出现。

请注意，此示例代码仅检查圆括号、方括号和大括号的平衡情况，不检查其他类型