我正在努力摆脱函数中的 float 。
我有一个函数对 16 位整数值进行操作，该值是一个放大的 8 位值。然后将缩小的 8 位发送到输出。
我确定我解释得不好。像这样:
int8 spot_value = 21; //arbitrary. just need a starting point
int16 running_value;
running_value = spot_value << 8; //multiply by 256 which is 5376
running_value += 154; //my upscaled value is now 5530
spot_value = running_value >> 8; //downscale again
如果我们使用 float ，我缩小后的值将是 21.6，我可以轻松地将其四舍五入为 22 并转换为 int8。但它会截断为我不想要的 21。
如果它是一个 float 但不将任何内容转换为 float (即使是暂时的)，如果这样做是合适的，是否有一种方法可以将整数“四舍五入”？
这可能比我想象的要简单得多。
最佳答案
您可以将 1/2 添加到该值。在您的缩放系统中为 128。
int8 spot_value = 21; //arbitrary. just need a starting point
int16 running_value;
running_value = spot_value << 8; //multiply by 256 which is 5376
running_value += 154; //my upscaled value is now 5530
spot_value = running_value + 128; //add an additional 1/2 for rounding
spot_value = spot_value >> 8; //downscale again
或者...
scale_shift = 8;
scale = 1 << scale_shift
round = scale >> 1;
int8 spot_value = 21; //arbitrary. just need a starting point
int16 running_value;
running_value = spot_value << scale_shift; //multiply by 256 which is 5376
running_value += 154; //my upscaled value is now 5530
spot_value = (running_value + round) >> scale_shift; //downscale again
或者基于 Dav 的回答的小技巧......
running_value = spot_value << scale_shift; //multiply by 256 which is 5376
running_value += 154; //my upscaled value is now 5530
spot_value = (running_value >> 8) + ((running_value >> 7) & 1)
.
关于c++ - "rounding"整数，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/1384348/