将非空终止的无符号字符数组复制到 std::string [英] Copying non null-terminated unsigned char array to std::string
问题描述
如果数组是空终止,这将非常简单:
If the array was null-terminated this would be pretty straight forward:
unsigned char u_array[4] = { 'a', 's', 'd', '\0' };
std::string str = reinterpret_cast<char*>(u_array);
std::cout << "-> " << str << std::endl;
但是,我想知道复制非空终止无符号字符数组的最合适方法是什么,如下所示:
However, I wonder what is the most appropriate way to copy a non null-terminated unsigned char array, like the following:
unsigned char u_array[4] = { 'a', 's', 'd', 'f' };
变成std::string
.
有没有什么方法可以不用迭代无符号字符数组?
Is there any way to do it without iterating over the unsigned char array?
谢谢大家.
推荐答案
std::string
有一个 constructor 接受一对迭代器和 unsigned char
可以(以实现定义的方式)转换为 char
> 所以这是有效的.不需要 reinterpret_cast
.
std::string
has a constructor that takes a pair of iterators and unsigned char
can be converted (in an implementation defined manner) to char
so this works. There is no need for a reinterpret_cast
.
unsigned char u_array[4] = { 'a', 's', 'd', 'f' };
#include <string>
#include <iostream>
#include <ostream>
int main()
{
std::string str( u_array, u_array + sizeof u_array / sizeof u_array[0] );
std::cout << str << std::endl;
return 0;
}
当然,数组大小"模板函数比sizeof
计算更健壮.
Of course an "array size" template function is more robust than the sizeof
calculation.
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