在32位整數中對4個值進行編碼

Question

我正在嘗試使用一個32位整數對4個值進行編碼的類。到目前爲止，它保存並輸出所有的值而不會丟失任何東西，但由於某種原因，按錯誤的順序。

struct encoder { 
    uint32_t val; 

    encoder(uint32_t _val = 0) : val(_val) {} 

    uint32_t first(uint32_t v = 0) { 
     if (!v) return (val << (8*3)) >> (8*3); 

     val |= v; 
    } 

    uint32_t second(uint32_t v = 0) { 
     if (!v) return (val << (8*2)) >> (8*3); 

     encoder _backupval(val); 

     val = (val >> (8*1)); 
     val |= v; 
     val = (val << (8*1)); 
     val |= _backupval.first(); 
    } 

    uint32_t third(uint32_t v = 0) { 
     if (!v) return (val << (8*1)) >> (8*3); 

     encoder _backupval(val); 

     val = (val >> (8*2)); 
     val |= v; 
     //now restore 
     val = val << 8; 
     val |= _backupval.second(); 
     val = val << 8; 
     val |= _backupval.first(); 
    } 

    uint32_t fourth(uint32_t v = 0) { 
     if (!v) return (val << (8*0)) >> (8*3); 

     encoder _backupval(val); 

     val = (val >> (8*3)); 
     val |= v; 
     //now restore 
     val = val << 8; 
     val |= _backupval.second(); 
     val = val << 8; 
     val |= _backupval.first(); 
     val = val << 8; 
     val |= _backupval.third(); 
    } 
}; 

int main() { 
    encoder t; 
    t.first(6); 
    t.second(42); 
    t.third(212); 
    t.fourth(23); 

    cout << "first number: " << t.first() 
     << "\nsecond number: " << t.second() 
     << "\nthird number: " << t.third() 
     << "\nfourth number: " << t.fourth() << endl; 

} 

請注意，我正在做這個練習，而且這個程序不會在真正的程序中使用。因此，不建議替代方法來做到這一點，並且不要指出缺陷（例如，如果數字的大小超過256，會損壞另一個整數等）。

Answer 1

您混合了函數的順序在fourth：

uint32_t fourth(uint32_t v = 0) { 
    if (!v) return (val << (8*0)) >> (8*3); 

    encoder _backupval(val); 

    val = (val >> (8*3)); 
    val |= v; 
    //now restore 
    val = val << 8; 
    val |= _backupval.second(); 
    val = val << 8; 
    val |= _backupval.first(); 
    val = val << 8; 
    val |= _backupval.third(); // this should be called first! 
} 

這將導致加擾的返回值。先致電third，然後it will work：

uint32_t fourth(uint32_t v = 0) { 
    if (!v) return (val << (8*0)) >> (8*3); 

    encoder _backupval(val); 

    val = (val >> (8*3)); 
    val |= v; 
    //now restore 
    val = val << 8; 
    val |= _backupval.third(); 
    val = val << 8; 
    val |= _backupval.second(); 
    val = val << 8; 
    val |= _backupval.first(); 
}