C ++：从stringstream到char ** [英] C++: From stringstream to char**

问题描述

我有一个类 parse（int argc，char * argv []）函数，我必须使用它来设置一个对象的期望状态。我使用 stringstream 从gui参数，然后我试图将它们转换为char **将它们传递给函数。这是我得到的：

  std :: stringstream sstream; 
 
 sstream<< -clip< < min_x_entry.get_text（）
<< < max_x_entry.get_text（）; //等等。
 
 std :: cout<< sstream.str（）; // All looks good here 
 
 std :: vector< std :: string> args; 
 std :: vector< char *> argv; 
 std :: string arg; 
 
 while（sstream>> arg）
 {
 args.push_back（arg）; 
 argv.push_back（const_cast< char *>（args.back（）。c_str（）））; 
} 
 argv.push_back（0）; 
 
 int argc = args.size（）; 
 
 for（int i = 0; i  std :: cout< & argv [0] [i]; //这会输出垃圾
 
 my_object.parse（argc，& argv [0]）//这会失败
  

我缺少什么？

解决方案

一个问题是重新分配 args 向量作为 push_back（） 将根据需要增加向量的大小：

如果新size（）不大于capacity （），没有迭代器或引用无效。 否则所有的迭代器和引用都会失效。

argv vector存储指向 args 中元素内部的指针，因此这些指针将无效。

解决方案是先创建 args 向量，然后再创建 argv 向量：

  while（sstream>> arg）args.push_back（arg）; 
 
 for（auto i = args.begin（）; i！= args.end（）; i ++）
 {
 argv.push_back（const_cast< char *> i-> c_str（）））; 
} 
 argv.push_back（0）; 
   循环打印出 argv 字符串不正确。这：
 & argv [0] [i] 
  
是 char * ，但从 i  th元素的 argv 中的第一个条目。例如，如果 argv 中的第一个c字符串string：
 & argv [0] [1]是tring
& argv [0] [2]是ring
  
更改为：
  for（int i = 0; i  std :: cout < argv [i]<< std :: endl; //添加'endl'来刷新'cout'。 
  
 
I have a class with parse(int argc, char* argv[]) function which I have to use to set a desired state of an object. I'm taking the parameters from the gui using stringstream and then I'm trying to convert them to char** to pass them to the function. Here's what I've got:
std::stringstream sstream;

sstream << "-clip" << " " << min_x_entry.get_text()
        << " " << max_x_entry.get_text(); // etc.

std::cout << sstream.str();    // All looks good here

std::vector<std::string> args;
std::vector<char*> argv;
std::string arg;

while (sstream >> arg)
{
    args.push_back(arg);
    argv.push_back(const_cast<char*>(args.back().c_str()));
}
argv.push_back(0);

int argc = args.size();

for (int i = 0; i < argc; ++i)
    std::cout << &argv[0][i];    // This outputs garbage

my_object.parse(argc, &argv[0])  // And this fails
What am I missing? Is there a better way of achieving this?
解决方案
A problem would be reallocation of the args vector as push_back() will grow the size of the vector if required:

  
If new size() is not larger than capacity(), no iterators or references are invalidated. Otherwise all iterators and references are invalidated.
The argv vector is storing pointers to the internals of elements in args, so these would be invalidated.

A solution would be to create the args vector first then create the argv vector afterwards:
while (sstream >> arg) args.push_back(arg);

for (auto i = args.begin(); i != args.end(); i++)
{
    argv.push_back(const_cast<char*>(i->c_str()));
}
argv.push_back(0);
The for loop that prints out the argv strings is incorrect. This:
&argv[0][i]
is a char* but starts from ith element of the first entry in argv. For example, if the first c-string in argv was "string":
&argv[0][1] is "tring"
&argv[0][2] is "ring"
change to:
for (int i = 0; i < argc; i++)
    std::cout << argv[i] << std::endl; // Added 'endl' to flush 'cout'.


                        
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