c++ - 不存在合适的构造函数来从 "uint8_t *"转换为 "std::vector>"

Question

no suitable constructor exists to convert from "uint8_t *" to "std::vector<uint8_t, std::allocator<uint8_t>>"

类型转换不工作

编辑:

const std::vector<uint8_t> Test (const std::vector<uint8_t> buffer) const;

uint8_t* buffer="...";

//so i can use Test() function
Test(buffer);

Error
no suitable constructor exists to convert from "uint8_t *" to "std::vector<uint8_t, std::allocator<uint8_t>>"

Answer 1

您不能将array 转换为std::vector，您需要显式构造一个。一种方法是像这样使用 vector 的范围构造函数:

uint8_t* buffer="...";
// +1 for the terminating \0
std::vector<uint8_t> vector( buffer, buffer + strlen( buffer ) + 1 ); 
Test( vector );

---

作为旁注，如果您的缓冲区嵌入了 \0，则 strlen 将返回不正确的值。作为解决方法，您可以执行以下操作:

uint8_t[] buffer="...";
std::vector<uint8_t> vector( buffer, buffer + sizeof( buffer ) ); 
Test( vector );