C++ 将命令行参数转换为字符串

Convert command line argument to string

提问by theta

I have a program that reads hard-coded file-path and I want to make it read file-path from command line instead. For that purpose I changed the code like this:

我有一个读取硬编码文件路径的程序，我想让它从命令行读取文件路径。为此，我更改了这样的代码：

#include <iostream>

int main(char *argv[])
{
...
}

but, argv[1]variable exposed this way seems to be of type pointer, and I need it as a string. What should I do to convert this command line argument to string?

但是，argv[1]以这种方式公开的变量似乎是指针类型，我需要它作为字符串。我应该怎么做才能将此命令行参数转换为字符串？

回答by masoud

It's already an array of C-style strings:

它已经是一个 C 风格的字符串数组：

#include <iostream>
#include <string>
#include <vector>


int main(int argc, char *argv[]) // Don't forget first integral argument 'argc'
{
  std::string current_exec_name = argv[0]; // Name of the current exec program
  std::vector<std::string> all_args;

  if (argc > 1) {
    all_args.assign(argv + 1, argv + argc);
  }
}

Argument argcis count of arguments plus the current exec file.

参数argc是参数的计数加上当前的 exec 文件。

回答by juanchopanza

You can create an std::string

您可以创建一个 std::string

#include <string>
#include <vector>
int main(int argc, char *argv[])
{
  // check if there is more than one argument and use the second one
  //  (the first argument is the executable)
  if (argc > 1)
  {
    std::string arg1(argv[1]);
    // do stuff with arg1
  }

  // Or, copy all arguments into a container of strings
  std::vector<std::string> allArgs(argv, argv + argc);
}

回答by kayleeFrye_onDeck

No need to upvote this. It would have been cool if Benjamin Lindleymade his one-liner comment an answer, but since he hasn't, here goes:

无需为此点赞。如果本杰明·林德利 (Benjamin Lindley)将他的单行评论作为答案，那会很酷，但既然他没有，那么这里是：

std::vector<std::string> argList(argv, argv + argc);

std::vector<std::string> argList(argv, argv + argc);

If you don't want to include argv[0]so you don't need to deal with the executable's location, just increment the pointer by one:

如果你不想包含argv[0]所以你不需要处理可执行文件的位置，只需将指针加一：

std::vector<std::string> argList(argv + 1, argv + argc);

std::vector<std::string> argList(argv + 1, argv + argc);

回答by kayleeFrye_onDeck

I'm not sure if this is 100% portable but the way the OS SHOULDparse the args is to scan through the console command string and insert a nil-term char at the end of each token, and int main(int,char**)doesn't use const char**so we can just iterate through the args starting from the third argument (@notethe first arg is the working directory) and scan backward to the nil-term char and turn it into a space rather than start from beginning of the second argument and scanning forward to the nil-term char. Here is the function with test script, and if you do need to un-nil-ify more than one nil-term char then please comment so I can fix it; thanks.

我不确定这是否是 100% 可移植的，但操作系统应该解析 args 的方式是扫描控制台命令字符串并在每个标记的末尾插入一个 nil-term 字符，并且int main(int,char**)不使用，const char**所以我们可以只需从第三个参数开始遍历 args（@note第一个 arg 是工作目录）并向后扫描到零项字符并将其转换为空格，而不是从第二个参数的开头开始向前扫描到空字符。这是带有测试脚本的函数，如果您确实需要取消零化多个零项字符，请发表评论，以便我可以修复它；谢谢。

#include <cstdio>
#include <iostream>

using namespace std;

namespace _ {
/* Converts int main(int,char**) arguments back into a string.
@return false if there are no args to convert.
@param arg_count The number of arguments.
@param args      The arguments. */
bool ArgsToString(int args_count, char** args) {
  if (args_count <= 1) return false;
  if (args_count == 2) return true;
  for (int i = 2; i < args_count; ++i) {
    char* cursor = args[i];
    while (*cursor) --cursor;
    *cursor = ' ';
  }
  return true;
}
}  // namespace _

int main(int args_count, char** args) {
  cout << "\n\nTesting ArgsToString...\n";

  if (args_count <= 1) return 1;
  cout << "\nArguments:\n";
  for (int i = 0; i < args_count; ++i) {
    char* arg = args[i];
    printf("\ni:%i\"%s\" 0x%p", i, arg, arg);
  }
  cout << "\n\nContiguous Args:\n";
  char* end = args[args_count - 1];
  while (*end) ++end;
  cout << "\n\nContiguous Args:\n";
  char* cursor = args[0];
  while (cursor != end) {
    char c = *cursor++;
    if (c == 0)
      cout << '`';
    else if (c < ' ')
      cout << '~';
    else
      cout << c;
  }
  cout << "\n\nPrinting argument string...\n";
  _::ArgsToString(args_count, args);
  cout << "\n" << args[1];
  return 0;
}

回答by clockw0rk

Because all attempts to print the argument I placed in my variable failed, here my 2 bytes for this question:

因为打印我放在变量中的参数的所有尝试都失败了，这里是我这个问题的 2 个字节：

std::string dump_name;

(stuff..)

if(argc>2)
{
    dump_name.assign(argv[2]);
    fprintf(stdout, "ARGUMENT %s", dump_name.c_str());
}

Note the use of assign, and also the need for the c_str() function call.

注意assign的使用，以及c_str()函数调用的需要。

回答by Careal Manic

It's simple. Just do this:

这很简单。只需这样做：

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

int main(int argc, char *argv[])
{
    std::vector<std::string> argList;
    for(int i=0;i<argc;i++)
        argList.push_back(argv[i]);
    //now you can access argList[n]
}

@Benjamin Lindley You are right. This is not a good solution. Please read the one answered by juanchopanza.

@Benjamin Lindley 你是对的。这不是一个好的解决方案。请阅读 juanchopanza 的回答。

回答by brkeyal

#include <iostream>

std::string commandLineStr= "";
for (int i=1;i<argc;i++) commandLineStr.append(std::string(argv[i]).append(" "));