从简单函数中删除 if 条件

Question

我需要从以下两个函数中删除尽可能多的 if 条件：

inline int inc_with_1bit_saturation(int counter)
{
    if (counter == 1)
        return --counter;
    return ++counter;
}

void branch_prediction_1bit_saturation(int* input, int* output, int size)
{
    int counter = 0;

    for (int i = 0; i < size; ++i)
    {
        if (input[i] != counter)
        {
            counter = inc_with_1bit_saturation(counter);
            output[i] = 0;
        }
        else output[i] = 1;
    }
}

我该怎么做？if 分支是绝对必要的，不能删除，哪个分支可以用简单的按位运算或类似的东西替换？

更新 1

根据用户 JSF 的重要提示，代码现在看起来像这样：

void branch_prediction_1bit_saturation(int* input, int* output, int size)
{
    int counter = 0;

    for (int i = 0; i < size; ++i)
    {
        if (input[i] != counter)
        {
            counter = 1 - counter;
            output[i] = 0;
        }
        else output[i] = 1;
    }
}

更新 2

感谢Cantfindname，代码变成了这样：

void branch_prediction_1bit_saturation(int* input, int* output, int size)
{
    int counter = 0;

    for (int i = 0; i < size; ++i)
    {
        output[i] = counter == input[i];
        counter = output[i] * counter + (1 - output[i])*(1 - counter);
    }
}

这就完全解决了这个问题。

Answer 1

函数inc_with_1bit_saturation等价于模2。所以你可以替换

counter = inc_with_1bit_saturation(counter);

有

counter = (counter+1) % 2;

Answer 2

对于循环内的if语句：

output[i] = (int)(input[i]==counter);
counter = output[i]*counter + (1-output[i])*(1-counter) //used JSF's trick

True转为1，false转为0，依此：bool to int conversion

Answer 3

void branch_prediction_1bit_saturation(int* input, int* output, int size) {

    int counter = 0;

    for (int i = 0; i < size; ++i)
    {
        output[i] = (int)!((!!input[i]) ^ counter);
        counter = (int)((!!input[i]) & counter) | ((!!input[i]) & !counter);
    }
}

A为逻辑输入[i];

B为逻辑计数器；

输入[i] != 计数器的真相table是：

A B

0 0 | 0 --> (0 & 0) | (0 &!0) = 0 | 0 = 0

0 1 | 0 --> (0 & 1) | (0 &!1) = 0 | 0 = 0

1 0 | 1 --> (1 & 0) | (1 &!0) = 0 | 1 = 1

1 1 | 1 --> (1 & 1) | (1 & !1) = 1 | 0 = 1

输出的真相table[i]

A B

0 0 | 1 --> !(0 ^ 0) = !(0) = 1

0 1 | 0 --> !(0 ^ 1) = !(1) = 0

1 0 | 0 --> !(1 ^ 0) = !(1) = 0

1 1 | 1 --> !(1 ^ 1) = !(0) = 1

:)

从简单函数中删除 if 条件

Remove if conditions from simple function

c++

optimization

micro-optimization

更新 1

更新 2