C++ - 顯示來自ARGB值的參數不起作用？

Question

我正在開發一個簡單的控制檯應用程序，您可以在其中輸入ARGB值的十六進制代碼，並且應用程序會告訴您255個值中的紅色，綠色，藍色和Alpha內容。

例如：

Enter a 32-bit RGBA color value in hexadecimal (e.g. FF7F3300): FF7F3300 
Your color contains: 
255 of 255 red 
127 of 255 green 
51 of 255 blue 
0 of 255 alpha 

從理論上講，它是應用程序應該如何工作。但事實並非如此。它爲每種顏色顯示0。

我在C＃應用程序中使用了與此應用程序相同的代碼，除了經過語法調整以適應語言，並且它正常工作。

但是，由於我缺乏C++知識，即使經過半小時的思考，我也無法正確調試此應用程序。我用於此

我的C++代碼如下：

#include "stdafx.h" //Always imported for VisualStudio 

#include <iostream> //For input/output operations 

using namespace std; //To simplify code - no conflicting namespaces are being used, and yet the std namespace is used often 

void introduce() 
{ 
    cout << "Welcome to 'ARGB to Decimal'! This program will take a hex ARGB colour value (such as FF7F3300) and output the colour parameters that this value stores."; 
} 

uint32_t getValue() 
{ 
    cout << "\n\nPlease enter an ARGB value: "; 

    uint32_t value; 

    cin >> hex >> value; 

    cout << hex << "You have selected an ARGB value of " << value; 

    return value; //Converted to hex due to bitwise operations that will be performed on this value 
} 

int main() 
{ 
    introduce(); 

    uint32_t ARGBValue{ getValue() }; 

    uint32_t bitComparison{ 0xFF000000 }; //Used as the right operand of a bitwise AND operator to single out the bits for each byte of the ARGB value (with its bits being shifted 8 bits to the right before the 2nd, 3rd, and 4th comparison, and so display the appropriate byte value for that parameter 

    cout << "\n\nThe selected value has the following parameter values (out of 255):\n- Alpha:\t" 
     << ((ARGBValue & bitComparison) >> 24) 
     << "\n- Red:\t\t" << ((ARGBValue & (bitComparison >>= 8)) >> 16)   
     << "\n- Green:\t" << ((ARGBValue & (bitComparison >>= 8)) >> 8) 
     << "\n- Blue:\t\t" << (ARGBValue & (bitComparison >>= 8)) 
     << "\n\n"; 

    system("pause"); 

    return 0; 
} 

這挑出在ARGB值中的每個參數的比特通過使用與該參數值的按位與運算符和一個bitComparison值，其具有在相應位置處開啓的一組8位的位，在該位置當前參數的位被單出。

bitComparison位在cout語句內移位，因爲它正在執行。

如果有人能告訴我如何解決這個問題，我將不勝感激。

下面是示例輸出，它不工作：

Welcome to 'ARGB to Decimal'! This program will take a hex ARGB colour value (such as FF7F3300) and output the colour parameters that this value stores. 

Please enter an ARGB value: FF7F3300 
You have selected an ARGB value of ff7f3300 

The selected value has the following parameter values (out of 255): 
- Alpha:  0 
- Red:   0 
- Green:  0 
- Blue:   0 

Press any key to continue . . . 

Answer 1

此編譯精細和得到預期的輸出（編譯時的g ++ C++ 11支持）：轉換

#include <iostream> //For input/output operations 
#include <cstdint> 

void introduce() 
{ 
    std::cout << "Welcome to 'ARGB to Decimal'! This program will take a hex ARGB colour value (such as FF7F3300) and output the colour parameters that this value stores."; 
} 

uint32_t getValue() 
{ 
    std::cout << "\n\nPlease enter an ARGB value: "; 
    std::uint32_t value; 
    std::cin >> std::hex >> value; 
    std::cout << std::hex << "You have selected an ARGB value of " << value; 
    return value; //Converted to hex due to bitwise operations that will be performed on this value 
} 

int main() 
{ 
    introduce(); 

    std::uint32_t ARGBValue{ getValue() }; 

    std::cout << std::dec << "\n\nThe selected value has the following parameter values (out of 255):\n- Alpha:\t" 
       << ((ARGBValue >> 24) & 0xFF) 
       << "\n- Red:\t\t" 
       << ((ARGBValue >> 16) & 0xFF) 
       << "\n- Green:\t" 
       << ((ARGBValue >> 8) & 0xFF) 
       << "\n- Blue:\t\t" 
       << (ARGBValue & 0xFF) 
       << "\n\n"; 

    return 0; 
}