Mismatch between C and Java bitwise operation on hex values

Question

I have the following line in a C code that claims to convert a signed int8 into unsigned int16.

float x = (float) (((int16_t) ((temp[0] <<8) & 0xff00) | (temp[1] & 0x00ff)));

I converted this for Java as

float x = (((temp[0] << 8) & 0xff00) | (temp[1] & 0x00ff));

For the same input array,

temp[] = {0xFC, 0x10}

x= -1008 // In C
x= 64528 // In Java

I referred on SO and google for various posts on this but could not identify what is missing.

Tried the other data types short, int, float etc but to no avail

How could we get the same value of -1008 in java? Please help.

Thanks in advance

Answer 1

In C you have one additional cast: (int16_t) . If you did the same in Java - (short) , you would get the same result.

The issue is that in both C and Java the temporary results are cast at least to int (or higher type which can hold the value). With cast to int16_t or short you force the compiler to lose the higher bits again.