[英]no suitable constructor exists to convert from “uint8_t *” to “std::vector<uint8_t, std::allocator<uint8_t>>”
no suitable constructor exists to convert from "uint8_t *" to "std::vector<uint8_t, std::allocator<uint8_t>>"
和演員不起作用
編輯:
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>>"
您不能將array
轉換為std::vector
,您需要顯式構造一個array
。 一種方法是使用vector
的range構造函數,如下所示:
uint8_t* buffer="...";
// +1 for the terminating \0
std::vector<uint8_t> vector( buffer, buffer + strlen( buffer ) + 1 );
Test( vector );
附帶說明一下,如果您的緩沖區已嵌入\\0
則strlen
將返回錯誤的值。 作為解決方法,您可以執行以下操作:
uint8_t[] buffer="...";
std::vector<uint8_t> vector( buffer, buffer + sizeof( buffer ) );
Test( vector );
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