字符串坐标提取场景下如何使用try-catch处理异常？

问题1：处理非预期分隔符的异常 #

当字符串中出现&这类替代预期的,或*的符号时，可在分割、格式校验逻辑中主动识别异常符号，抛出带提示的异常，再通过try-catch捕获输出对应信息。

实现逻辑 #

1. 先清理字符串首尾的(、)、;及多余空格
2. 遍历字符串检查是否存在非预期符号，同时验证每个坐标段是否符合数字 * SCALE的格式
3. 发现异常符号时抛出包含错误提示的异常，在catch块中输出具体问题

代码示例 #

#include <iostream>
#include <string>
#include <stdexcept>
#include <algorithm>

// 预处理字符串，去除首尾无关字符
std::string preprocess(const std::string& str) {
    size_t start = str.find_first_not_of("( ");
    size_t end = str.find_last_not_of("); ");
    if (start == std::string::npos || end == std::string::npos) {
        throw std::invalid_argument("无效的坐标格式");
    }
    return str.substr(start, end - start + 1);
}

// 检查单个坐标段的格式是否合法
void check_segment_format(const std::string& segment) {
    size_t star_pos = segment.find('*');
    if (star_pos == std::string::npos) {
        size_t wrong_pos = segment.find_first_not_of("0123456789 SCALE");
        if (wrong_pos != std::string::npos) {
            char wrong_char = segment[wrong_pos];
            throw std::invalid_argument("该行应使用'*'而非'" + std::string(1, wrong_char) + "'");
        }
        throw std::invalid_argument("坐标段缺少'*'分隔符");
    }
}

// 分割坐标段并检查逗号分隔是否合法
std::vector<std::string> split_segments(const std::string& processed_str) {
    std::vector<std::string> segments;
    size_t pos = 0;
    size_t next_pos;
    while ((next_pos = processed_str.find(',', pos)) != std::string::npos) {
        std::string seg = processed_str.substr(pos, next_pos - pos);
        seg.erase(std::remove_if(seg.begin(), seg.end(), isspace), seg.end());
        segments.push_back(seg);
        pos = next_pos + 1;
    }
    std::string last_seg = processed_str.substr(pos);
    last_seg.erase(std::remove_if(last_seg.begin(), last_seg.end(), isspace), last_seg.end());
    segments.push_back(last_seg);

    // 检查是否存在非逗号的分隔符
    size_t wrong_sep_pos = processed_str.find_first_not_of("0123456789 * SCALE,");
    if (wrong_sep_pos != std::string::npos) {
        char wrong_char = processed_str[wrong_sep_pos];
        throw std::invalid_argument("该行应使用','而非'" + std::string(1, wrong_char) + "'");
    }
    return segments;
}

int main() {
    // 错误输入示例1
    std::string str1 = "(94 * SCALE & 10 * SCALE, 62 * SCALE, 10 * SCALE);";
    try {
        std::string processed = preprocess(str1);
        std::vector<std::string> segments = split_segments(processed);
        for (const auto& seg : segments) {
            check_segment_format(seg);
        }
    } catch (const std::invalid_argument& e) {
        std::cout << e.what() << std::endl; // 输出：该行应使用','而非'&'
    }

    // 错误输入示例2
    std::string str2 = "(94 & SCALE , 10 * SCALE, 62 * SCALE, 10 * SCALE);";
    try {
        std::string processed = preprocess(str2);
        std::vector<std::string> segments = split_segments(processed);
        for (const auto& seg : segments) {
            check_segment_format(seg);
        }
    } catch (const std::invalid_argument& e) {
        std::cout << e.what() << std::endl; // 输出：该行应使用'*'而非'&'
    }

    return 0;
}

问题2：处理字符串无法转换为整数的异常 #

stoi转换失败时会抛出std::invalid_argument（非数字格式）或std::out_of_range（数字超出整数范围）异常，直接捕获这些异常并输出对应提示即可。

实现逻辑 #

1. 从坐标段中提取数字部分字符串
2. 用try包裹stoi转换操作
3. 捕获转换异常，抛出带具体信息的提示

代码示例 #

#include <iostream>
#include <string>
#include <stdexcept>
#include <algorithm>

std::string preprocess(const std::string& str) {
    size_t start = str.find_first_not_of("( ");
    size_t end = str.find_last_not_of("); ");
    if (start == std::string::npos || end == std::string::npos) {
        throw std::invalid_argument("无效的坐标格式");
    }
    return str.substr(start, end - start + 1);
}

std::vector<std::string> split_segments(const std::string& processed_str) {
    std::vector<std::string> segments;
    size_t pos = 0;
    size_t next_pos;
    while ((next_pos = processed_str.find(',', pos)) != std::string::npos) {
        std::string seg = processed_str.substr(pos, next_pos - pos);
        seg.erase(std::remove_if(seg.begin(), seg.end(), isspace), seg.end());
        segments.push_back(seg);
        pos = next_pos + 1;
    }
    std::string last_seg = processed_str.substr(pos);
    last_seg.erase(std::remove_if(last_seg.begin(), last_seg.end(), isspace), last_seg.end());
    segments.push_back(last_seg);
    return segments;
}

int extract_coordinate(const std::string& segment) {
    size_t star_pos = segment.find('*');
    if (star_pos == std::string::npos) {
        throw std::invalid_argument("坐标段缺少'*'分隔符");
    }
    std::string num_str = segment.substr(0, star_pos);
    num_str.erase(std::remove_if(num_str.begin(), num_str.end(), isspace), num_str.end());
    try {
        return std::stoi(num_str);
    } catch (const std::invalid_argument& e) {
        throw std::invalid_argument("无法将'" + num_str + "'转换为整数");
    } catch (const std::out_of_range& e) {
        throw std::invalid_argument("数字'" + num_str + "'超出整数范围");
    }
}

int main() {
    // 错误输入示例：数字部分为非整数
    std::string str = "(abc * SCALE, 10 * SCALE, 62 * SCALE, 10 * SCALE);";
    try {
        std::string processed = preprocess(str);
        std::vector<std::string> segments = split_segments(processed);
        for (const auto& seg : segments) {
            int coord = extract_coordinate(seg);
            std::cout << "提取的坐标值：" << coord << std::endl;
        }
    } catch (const std::invalid_argument& e) {
        std::cout << e.what() << std::endl; // 输出：无法将'abc'转换为整数
    }

    return 0;
}

内容的提问来源于stack exchange，提问作者tushar