I found these conversions from uint16_t to int32_t here https://gist.github.com/mux2000/5a0db5be900ca8b7f2d528fb37968b38.
int32_t ac6 = (int32_t)(int16_t)bmp180_cal_data[AC6],
ac5 = (int32_t)(int16_t)bmp180_cal_data[AC5],
mc = (int32_t)(int16_t)bmp180_cal_data[MC],
md = (int32_t)(int16_t)bmp180_cal_data[MD];
<代码>bmp180_cal_data is a range of uint16_t。 因此,在从不同大小的未签名和副手转换时,首先应投向原始价值的大小(bmp180_cal_data[x],然后投向新的尺寸。
这是否正确? 如果是,为什么?
我认为,根据具体执行行为,如果<代码>bmp180_cal_data的数值高于INT16_>,就试图做“正确”的事情。 将这一编号改为int16_t在此情况下取决于执行,但共同行为是仅以两种补充形式解释借方,这样就可归纳为负数(在将签名改为未签名时如何处理负数)。 然后,它利用第二版扩大至32个轨道,从而保持负数。
如果它们直接投到<条码>int32_t,那么16倍的数值就只是保留其原有价值,因为它们符合32条轨道,而不是消极的。
因此,如果原始数值为65535,则该数值为-1。 当使用双重投射物时,将使用单一投射物的655。
The single cast of (int16_t) is useful to map the values above INT16_MAX to negative values (in a common implementation defined manner): [32,768 ... 65,535] to [-32,768 ... -1].
<代码>(int32_t)的投放没有用途,使代码WET。 功能性的定义是明确的,与(没有)<代码>(int32_t)相同。
也许,作者们根本不认识到不需要<代码>(int32_t)?
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