检索命令行参数时出错

Question

要运行我的程序，您必须输入：

(program name) -s (a number) -f (the file name) 

我遇到的问题是文件名在string，命令参数在char。

这是我的代码，以便我可以更清楚地说明我的问题：

if(argc==1){
   cout << "Please Enter it in the Form of (prog -s (number of state) -f (file name)\n";
}   
if(argc==2){
   cout << "Please Enter it in the Form of (prog -s (number of state) -f (file name)\n";
}   
if(argc==3){
   cout << "Please Enter it in the Form of (prog -s (number of state) -f (file name)\n";
}   
if(argc==4){
   cout << "Please Enter it in the Form of (prog -s (number of state) -f (file name)\n";
}   
if(argc==5){

      if((argv[1][0]=='-'&& argv[1][1]=='s')&&(argv[3][0]=='-'&& argv[3][1]=='f')){
        int x;
         x=argv[2][0]-48;
         state *array;
         array= create_states(x);

         ifstream File;
         File.open(argv[4][0]);

在这一行：

file.open(argc[4][0]);

我收到错误。

Answer 1

应该是：

  ifstream File;
  File.open(argv[4]);

此外，如果文件路径包含空格，请使用双引号。