如何检查比特值等于1? [英] How to check is bit value equals to 1?

问题描述

我正在运行OpenCL内核.在某个时刻，我想检查变量中所选位置的位是否等于1?例如，我发现在C#中，我们可以使用代码

I have running OpenCL kernel. At some moment I want to check is bit at selected position in variable equals to one or not? For example, I found that in C# we can convert uint to string containing bits using code like this:

// Here we will store our bit

string bit;

// Unsigned integer that will be converted

uint hex = 0xfffffffe;

// Perform conversion and store our string containing bits

string str = uint2bits(hex);

// Print all bits "11111111111111111111111111111110"

Console.WriteLine(str);

// Now get latest bit from string

bit = str.Substring(31, 1);

// And print it to console

Console.WriteLine(bit);

// This function converts uint to string containing bits

public static string uint2bits(uint x) {
    char[] bits = new char[64];
    int i = 0;
    while (x != 0) {
        bits[i++] = (x & 1) == 1 ? '1' : '0';
        x >>= 1;
    }
    Array.Reverse(bits, 0, i);
    return new string(bits);
}

如何像上面的代码那样在内核内部制作?

How to make inside kernel like in code above?

__kernel void hello(global read_only uint* param) {

/*

Do something with "param"

...

*/

int bit;
int index = 31;

/*

Store bit value ("0" or "1") at variable "bit"

...

*/

if (bit == 1) {

// ...

} else {

// ...

}

// Or maybe something more easy to check a bit value at "index" in "param" ... ?

}

在哪里可以了解更多信息?

Where I can read more about that?

推荐答案

通过用1<<b进行掩码并将结果与​​零进行比较，可以检查数字中的b位是设置为1还是0. /p>

You can check if a bit b in a number is set to 1 or 0 by masking with 1<<b, and comparing the result to zero:

static bool isBitSet(uint n, uint b) {
    return (n & (1U << b)) != 0;
}

二进制文件中的1 << b的值由第b个位置中的单个1从背面开始，从零开始计数，在所有其他位置为零.当使用按位AND运算符&将此掩码应用于n时，将得到一个数字n的单个位，而在所有其他位置为零.因此，当n的第b位设置为零时，总体操作的结果为零，否则为非零.

The value of 1 << b in binary is composed of a single 1 in b-th position counting from the back starting at zero, and zeros in all other positions. When you apply this mask to n with the bitwise AND operator & you get a single bit of number n, and zeros in all other positions. Therefore, the result of the overall operation is zero when b-th bit of n is set to zero, and a non-zero otherwise.

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