什么是| 运算符在函数调用中的意思？ [C ++]

Question

I usually see this when looking at Win32 gui code. My assumption is that it is a standard bitwise or, but I also occasionaly see it in C#, and it seems like there would be a better (well higher level) way to do the same thing there. Anyway, here's an example:

MessageBox(NULL, "Window Creation Failed!", "Error!", MB_ICONEXCLAMATION | MB_OK);

Thanks,

Seamus

Answer 1

The | is a bitwise OR. MB_OK and MB_ICONEXCLAMATION are defined constants which are a power of 2 (such as 32 or 128), so that the bitwise OR can combine them (128 | 32 would be 160, which has two bits set). This is normal when the bits are used as flags.

Answer 2

Its for bitmasking. Lets use this incredibly trivally example. You have a binary value for colors, with the following values.

100 = BLUE

010 = RED

001 = GREEN

When you say SomeFunction ( BLUE | RED | GREEN ); you are infact passing the value 111, which can then be decoded to mean BLUE and RED and GREEN.

Google Bitwise operators for more details.

Answer 3

It's the bitwise or operator like other places. Basically, this technique is used when you want to set some attributes that are not mutually exclusive.

The function can easily check them with some code like this:

if (arg & MB_ICONEXCLAMATION) { // Show an exclamation icon...

}

// ...

if (arg & MB_OK) { // Show an OK button

}

Answer 4

Think of MB_ICONEXCLAMATION and MB_OK as "options" that aren't anything fancier than ints. What you care about is the bit representation of those ints.

Say:

//MessageBox.cs or whatever
public static int MB_ICONEXCLAMATION = 0x1 // 0001 in binary
public static int MB_OK = 0x2 // 0010 in binary

When you OR them together, you get 0011 in binary. So you are requesting both options for the MessageBox using just one argument instead of having to have more arguments, one for each option you want to specify.

Answer 5

这是一个按位OR。

Answer 6

It does a bitwise or.

int x = 5 | 3; // x == 7 now