轉換和分離2字節的十六進制兩個chararacters，上面的兩個十六進制進入一個字符和兩個十六進制下進入到在C++第二系統字符

Question

我在

u_int_variable = 43981; // This is equal to ABCD in HEX 

一個unsigned int值I可以打印此安慰使用

printf("Value in Hex is %X", u_int_variable); 

// Output : Value in Hex is ABCD 

我如何轉換和分離這使得

unsigned char uhex, lhex; 

uhex = 0xAB; and lhex = 0xCD; 

Answer 1

uhex = static_cast<unsigned char>(u_int_variable >> 8); 
lhex = static_cast<unsigned char>(u_int_variable & 0xFF); 

（十六進制）ABCD

或者，成爲> 32位無符號整數絕對安全：

uhex = static_cast<unsigned char>((u_int_variable >> 8) & 0xFF); 

Answer 2

爲Variant：關於位運算

unsigned char uhex, lhex; 
lhex = static_cast<unsigned char>(u_int_variable & 0xFF); 
uhex = static_cast<unsigned char>((u_int_variable >> 8) & 0xFF); 

良好的第一深度文章是here

Answer 3

我給你使用位字段和工會，解決方案看起來像這樣：

struct hex 
    { 
     unsigned int lhex : 8; // upper and lower part alignement are system dependent 
     unsigned int uhex : 8; 
    }; 

    union number 
    { 
     number(int val): m_val(val){}; 
     void lhex(void) {printf("lhex %X", this->num.lhex);}; 
     void uhex(void) {printf("uhex %X", this->num.uhex);}; 
     int m_val; 
     hex num; 
    }; 

    number n(43981); 
    n.lhex(); 
    n.uhex();