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下面是一段代码，我试图从提供的目录路径中查找具有匹配模式的文件。

期望列出所有具有模式匹配的文件，例如在“/usr/local”路径下，存在以下文件 abc.txt axy.txt bcd.txt azz.txt bby.txt

使用模式匹配代码，我期待以下输出

       abc.txt
       axy.txt
       azz.txt

#include <glob.h> 
#include <string.h> 
#include <vector>
#include <stdexcept>
#include <string>
#include <sstream>
#include <iostream>
    
    using namespace std;
    vector<string> glob(const string& pattern) {
    
        // glob struct resides on the stack
        glob_t glob_result;
        memset(&glob_result, 0, sizeof(glob_result));
    
        // do the glob operation
        //int return_value = glob(pattern.c_str(), 0, globerr, &glob_result);
       int return_value = glob(pattern.c_str(), GLOB_TILDE, NULL, &glob_result);
        if(return_value != 0) {
            globfree(&glob_result);
            stringstream ss;
            ss << "glob() failed with return_value " << return_value << endl;
            throw std::runtime_error(ss.str());
        }
    
       // collect all the filenames into a std::list<std::string>
        vector<string> filenames;
        for(size_t i = 0; i < glob_result.gl_pathc; ++i) {
            filenames.push_back(string(glob_result.gl_pathv[i]));
        }
    
        // cleanup
        globfree(&glob_result);
    
        // done
        return filenames;
    }

    int main(int argc, char **argv) {
        vector<string> res;
        res= glob("a");
        return 0;
      }
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1 回答 1

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正如评论中提到的，你只会匹配一个名为 exact 的文件a，globfree当你不应该并抛出一个你没有捕获的异常时。

#include <glob.h>

#include <cerrno>
#include <cstring>
#include <iostream>
#include <stdexcept>
#include <string>
#include <vector>

std::vector<std::string> glob(const std::string& pattern) {
    glob_t glob_result = {0}; // zero initialize

    // do the glob operation
    int return_value = ::glob(pattern.c_str(), GLOB_TILDE, NULL, &glob_result);

    if(return_value != 0) throw std::runtime_error(std::strerror(errno));

    // collect all the filenames into a std::vector<std::string>
    // using the vector constructor that takes two iterators
    std::vector<std::string> filenames(
        glob_result.gl_pathv, glob_result.gl_pathv + glob_result.gl_pathc);

    // cleanup
    globfree(&glob_result);

    // done
    return filenames;
}

int main() {
    try { // catch exceptions
        std::vector<std::string> res = glob("*a*"); // files with an "a" in the filename

        for(size_t i = 0; i< res.size(); ++i)
            std::cout << res[i] << '\n';

    } catch(const std::exception& ex) {
        std::cerr << ex.what() << std::endl;
    }
}
于 2021-01-18T15:38:32.017 回答