如何正确调用getopt函数 [英] How to call correctly getopt function

问题描述

从 http://code.google.com/p/darungrim/source/browse/trunk/ExtLib/XGetopt.cpp?r=17

`check.cpp: In function ‘int main()’:`

check.cpp：14：55：错误：从'const char **'到'char * const *'的无效转换[-fpermissive]

/usr/include/getopt.h:152:12：error：初始化int getopt（int，char * const *，const char *）的参数2 '[-fpermissive]

#include <iostream>
#include <cstring>
#include <string>
#ifdef USE_UNISTD
#include <unistd.h>
#else
#include "XGetopt.h"
#endif
using namespace std;

int main() {

string text="-f  input.gmn -output.jpg";
int argc=text.length();
cout<<"argc: "<<argc<<endl;
char const * argv = text.c_str();
cout<<"argv: "<<argv<<endl;
int c = getopt (argc, &argv, "f:s:o:pw:h:z:t:d:a:b:?");
cout<<"c: "<<c<<endl;
return 0;
}

推荐答案

：

1. 参数列表不是字符串。它是一个字符串列表。不要被shell或其他程序要求参数列表作为单个字符串混淆。在一天结束时，这些程序会将一个字符串拆分成参数数组，并运行一个可执行文件（参见 execv ）。


2. 在参数列表中始终有一个隐含的第一个参数，它是一个程序名称。

1. Argument list is not a string. It is a list of strings. Don't get confused by shell or other programs that ask for a list of arguments as a single string. At the end of day, those programs would split a string into arrays of arguments and run an executable (see execv, for example).
2. There is always an implicit first argument in argument list that is a program name.

这是您的代码，修正：

#include <string>
#include <iostream>
#include <unistd.h>

int main()
{
    const char *argv[] = { "ProgramNameHere",
                           "-f", "input.gmn", "-output.jpg" };
    int argc = sizeof(argv) / sizeof(argv[0]);
    std::cout << "argc: " << argc << std::endl;
    for (int i = 0; i < argc; ++i)
        std::cout << "argv: "<< argv[i] << std::endl;
    int c;

    while ((c = getopt(argc, (char **)argv, "f:s:o:pw:h:z:t:d:a:b:?")) != -1) {
        std::cout << "Option: " << (char)c;
        if (optarg)
            std::cout << ", argument: " << optarg;
        std::cout << '\n';
    }
}

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