如果条件来自简单函数

Question

I need to remove as many if conditions as possible from the two functions below:

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;
    }
}

How can I do that and what if branch is absolutely necessary and cannot be removed and which one can be replaced by simple bitwise operations or something like that?

Update 1

According to User JSF 's great tip, the code is now looking like this:

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;
    }
}

Update 2

Thanks to Cantfindname , the code became like this:

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);
    }
}

And this completely solves the question.

Answer 1

For the if statement inside the loop:

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

True converts to 1 and false to 0, according to this: bool to int conversion

Answer 2

function inc_with_1bit_saturation is equivalent of modulo 2 . So you can replace

counter = inc_with_1bit_saturation(counter);

With

counter = (counter+1) % 2;

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 is logic input[i];

B is logic counter;

The truth table for input[i] != counter is:

AB

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

The truth table for output[i]

AB

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

:)