将未知长度的char **（c）转换为vector （c ++）[duplicate]

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converting an array of null terminated const char* strings to a std::vector< std::string > 4 answers

将一个以null结尾的const char *字符串数组转换为std :: vector 4个答案

copying c array of strings into vector of std::string 4 answers

将c数组的字符串复制到std :: string 4答案的向量中

How would one go about converting a C char** to a C++ vector? Is there some built-in functionality one can utilize to do this, or is it better to accomplish it through a series of iterative steps?

如何将C char **转换为C ++向量?是否有一些内置功能可以用来做到这一点,或者通过一系列迭代步骤完成它是否更好?

EDIT: For various reasons, the number of elements in the C array is unknown. It is possible I could pass that as another parameter, but is that absolutely necessary?

编辑:由于各种原因,C数组中的元素数量未知。我可以将其作为另一个参数传递,但这是绝对必要的吗?

4 个解决方案

#1

You can simply use the constructor of std::vector that takes two iterators:

你可以简单地使用带有两个迭代器的std :: vector的构造函数:

const char* arr[] = {"Hello", "Friend", "Monkey", "Face"};
std::vector<std::string> v(std::begin(arr), std::end(arr));

Or if you really have a const char**:

或者,如果你真的有一个const char **:

const char** p = arr;
std::vector<std::string> v(p, p + 4);

Which will also work with directly using arr instead of p due to array-to-pointer conversion.

由于数组到指针的转换,这也可以直接使用arr而不是p。

#2

char** c;
vector<string> v(c, c + 10);

Will construct elements from element of given range. 10 is number of elements here

将从给定范围的元素构造元素。 10是这里的元素数量

#3

You can use the constructor of std::vector that takes two iterators, a.k.a. the range constructor:

你可以使用带有两个迭代器的std :: vector的构造函数,a.k.a。范围构造函数:

char* strings[] = {"aaa", "bbb", "ccc", "ddd"};
std::vector<std::string> v(strings, strings + 4);

where 4 is the size of your array. In this concrete example, the calculation of the size of the strings array would be also possible by using sizeof operator:

其中4是数组的大小。在这个具体的例子中,使用sizeof运算符也可以计算字符串数组的大小:

int len = sizeof(strings)/sizeof(char*);
std::vector<std::string> v2(strings, strings + len);

which wouldn't be possible with pure char** though since the size of the array can not be directly retrieved from a pointer in any way (also worth to read something about array decaying).

虽然因为无法以任何方式直接从指针中检索数组的大小(这也值得阅读有关数组衰减的内容),因此使用纯char **是不可能的。

#4

This one-liner is useful for capturing command line arguments...

这个单行程序对捕获命令行参数很有用......

int
main(int argc, char ** argv)
{
  std::vector<std::string> arguments(argv, argv + argc);
}

#1