在 C++ 中将字符串转换为 uint8_t 数组
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Converting a string to uint8_t array in C++
提问by Carlo
I want an std::string object (such as a name) to a uint8_t array in C++. The
function reinterpret_cast<const uint8_t*>rejects my string. And since I'm coding using NS-3, some warnings are being interpreted as errors.
我想要一个 std::string 对象（例如名称）到 C++ 中的 uint8_t 数组。该函数reinterpret_cast<const uint8_t*>拒绝我的字符串。由于我使用 NS-3 进行编码，因此一些警告被解释为错误。
回答by Rob?
If you want a pointer to the string's data:
如果你想要一个指向string's 数据的指针：
reinterpret_cast<const uint8_t*>(&myString[0])
If you want a copy of the string's data:
如果您想要string's 数据的副本：
std::vector<uint8_t> myVector(myString.begin(), myString.end());
uint8_t *p = &myVector[0];
回答by sth
String objects have a .c_str()member function that returns a const char*. This pointer can be cast to a const uint8_t*:
字符串对象有一个.c_str()成员函数，它返回一个const char*. 此指针可以转换为 a const uint8_t*：
std::string name("sth");
const uint8_t* p = reinterpret_cast<const uint8_t*>(name.c_str());
Note that this pointer will only be valid as long as the original string object isn't modified or destroyed.
请注意，此指针仅在原始字符串对象未被修改或销毁时才有效。
