C++ 如何将 char* 转换为 std::vector？
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How to convert char* to std::vector?
提问by Richard Knop
I have a char* variable:
我有一个 char* 变量：
// char* buffer;
// ...
fread (buffer, 1, lSize, pFile);
// ...
How can I convert it to std::vector? Casting will get me an error.
如何将其转换为 std::vector？铸造会让我出错。
采纳答案by James McNellis
Why not use a std::vectorto begin with:
为什么不使用 astd::vector开头：
std::vector<char> buffer(lSize);
std::fread(&buffer[0], 1, buffer.size(), pFile);
回答by ronag
std::vector<char> data(buffer, buffer + size);
James McNellis answer is better if you can do it like that.
如果你能这样做，詹姆斯麦克内利斯的回答会更好。
回答by mgiuca
You can't cast between a char* and a vector; pointer casting causes the result to have the exact same bytes as the input, so it's generally a bad idea unless you're doing low-level bit manipulation.
你不能在 char* 和 vector 之间进行转换；指针转换会导致结果与输入具有完全相同的字节，因此除非您进行低级位操作，否则这通常是一个坏主意。
Assuming you wanted to build a vector of chars, you can create a new vector object of type std::vector<char>(note that you need to supply the type of the vector's elements), and initialise it from the array by passing the begin and end pointers to the constructor:
假设您想构建一个字符向量，您可以创建一个新的向量类型对象std::vector<char>（请注意，您需要提供向量元素的类型），并通过将开始和结束指针传递给构造函数来从数组中初始化它：
std::vector<char> vec(buffer, buffer+lSize);
Note that the second argument is an "end pointer", very common in C++. Importantly, it is a pointer to the character afterthe end of the buffer, not the last character in the buffer. In other words, the start is inclusive but the end is exclusive.
请注意，第二个参数是“结束指针”，在 C++ 中很常见。重要的是，它是指向缓冲区末尾之后的字符的指针，而不是缓冲区中的最后一个字符。换句话说，开始是包容性的，但结束是排斥性的。
Another possibility (since you're dealing with chars) is to use a std::stringinstead. This may or may not be what you want (depending on whether you're thinking about the char* as a string or an array of bytes). You can construct it in the same way:
另一种可能性（因为您正在处理字符）是使用 astd::string代替。这可能是也可能不是您想要的（取决于您是将 char* 视为字符串还是字节数组）。你可以用同样的方式构造它：
std::string str(buffer, buffer+lSize);
or with a different constructor, taking the size as the second argument:
或者使用不同的构造函数，将大小作为第二个参数：
std::string str(buffer, lSize);
