我需要从以下两个函数中删除尽可能多的条件:
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,不能被删除,哪一个可以被简单的按位操作或类似的东西取代?
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;
}
}
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);
}
}
这完全解决了这个问题。
答案 0 :(得分:6)
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转换为1,false转换为0,根据:bool to int conversion
答案 1 :(得分:1)
函数inc_with_1bit_saturation
相当于 modulo 2 。所以你可以替换
counter = inc_with_1bit_saturation(counter);
使用
counter = (counter+1) % 2;
答案 2 :(得分:1)
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是逻辑输入[i];
B是逻辑计数器;
输入[i]!=计数器的真值表是:
A B
0 0 | 0 - &gt; (0&amp; 0)| (0&amp;!0)= 0 | 0 = 0
0 1 | 0 - &gt; (0&amp; 1)| (0&amp;!1)= 0 | 0 = 0
1 0 | 1 - &gt; (1&amp; 0)| (1&amp;!0)= 0 | 1 = 1
1 1 | 1 - &gt; (1&amp; 1)| (1&amp;!1)= 1 | 0 = 1
输出[i]
的真值表A B
0 0 | 1 - &gt; !(0 ^ 0)=!(0)= 1
0 1 | 0 - &gt; !(0 ^ 1)=!(1)= 0
1 0 | 0 - &gt; !(1 ^ 0)=!(1)= 0
1 1 | 1 - &gt; !(1 ^ 1)=!(0)= 1
:)