如何将 char 指针数组拆分为两个带有分隔符的新 char 指针数组？

Question

我目前在 main 中有一个 char *command[SIZE] 数组，它通过接受用户输入来填充。可以填写的示例是 {"ls", "-1", "|" “种类”}。我想将此作为函数的参数，并使用分隔符“|”将其拆分为两个数组（char *command1[SIZE]、char *command2[SIZE]）。所以 char *command1[SIZE] 包含 {"ls" 和 "-l"}，而 char *command2[SIZE] 包含 {"sort"}。 Command1 和 command2 不应包含分隔符。

下面是我的部分代码...

** void executePipeCommand(char *command) {

  char *command1[SIZE];
  char *command2[SIZE];


 //split command array between the delimiter for further processing. (the delimiter 
   is not needed in the two new array)

}

int main(void) {

  char *command[SIZE];

  //take in user input...

  executePipeCommand(command);

}

**

Answer 1

适用于任意数量的拆分令牌，您可以选择拆分令牌。

std::vector<std::vector<std::string>> SplitCommands(const std::vector<std::string>& commands, const std::string& split_token)
{
    std::vector<std::vector<std::string>> ret;
    if (! commands.empty())
    {
        ret.push_back(std::vector<std::string>());
        for (std::vector<std::string>::const_iterator it = commands.begin(), end_it = commands.end(); it != end_it; ++it)
        {
            if (*it == split_token)
            {
                ret.push_back(std::vector<std::string>());
            }
            else
            {
                ret.back().push_back(*it);
            }
        }
    }
    return ret;
}

转换成需要的格式

std::vector<std::string> commands;
char ** commands_unix;
commands_unix = new char*[commands.size() + 1]; // execvp requires last pointer to be null
commands_unix[commands.size()] = NULL;
for (std::vector<std::string>::const_iterator begin_it = commands.begin(), it = begin_it, end_it = commands.end(); it != end_it; ++it)
{
    commands_unix[it - begin_it] = new char[it->size() + 1]; // +1 for null terminator
    strcpy(commands_unix[it - begin_it], it->c_str());
}


// code to delete (I don't know when you should delete it as I've never used execvp)
for (size_t i = 0; i < commands_unix_size; i++)
{
    delete[] commands_unix[i];
}
delete[] commands_unix;