为什么我的代码打印两次相同的命令行参数？

Question

我通过三个参数传递给我的程序，这是所有的文本文件： ARG 1：一个 ARG 2：二 ARG 3：三

为什么ARG 1被打印了两次？

#include <iostream> 
#include <fstream> 
using namespace std; 

int main(int argc, char *argv[]) 
{ 
if(argc < 2)  //check if files exist 1st 
{ 
    cout << "usage: " << argv[0] << " <filename>\n"; 
} 
else  //proceed if files exist 
{ 
    for(int x = 1; x < argc; x++)  //while x < the argument count, read each argument 
    { 
     ifstream infile; 
     infile.open(argv[x]); 
     if(!infile.is_open())  //check is file opens 
     { 
      cout << "Could not open file." << endl; 
     } 
     else  //if it opens, proceed 
     { 
      string s; 
      while(infile.good()) 
      { 
       infile >> s;  //declare string called s 
       if(s[s.length()-1] == ',')  //if the end of the arg string has a ',' replace it with a null 
       { 
        s[s.length()-1] = '\0'; 
       } 
       cout << s; 
       if(x != (argc -1)) 
       { 
        cout << ", "; 
       } 
      } 
     } 
    } 
    cout << endl; 
} 
return 0; 
} 

此代码输出：

一个，一个，两个，三个

Answer 1

你的错误

  cout << s; // here 
      if(x != (argc -1)) 
      { 
       cout << ", "; 
      } 

如何解决

  cout << s; 
      s = ""; // fix 
      if(x != (argc -1)) 
      { 
       cout << ", "; 
      } 

您只需将流两次放入的s字符串。而已。

---

短代码为您的目的：

std::ostringstream oss; 
    for(std::size_t index = 1; index < argc; ++ index){ 
     oss << std::ifstream(argv[ index ]).rdbuf() ? assert(1==1) : assert(1==0); 
    } 
    std::cout << oss.str(); 

输出

one 
two 
three