So i am trying to subtract two uint_16 values from one another and every time i do it i get an error.
conversion from 'int' to 'uint16_t', signed/unsigned mismatch
uint16_t a = 432;
uint16_t d= 60000;
uint16_t total = d -a;
I know you would need to cast it but how so,and wouldn't their be an loss of data when doing so.
uint16_t most likely has a lower integer rank than int. It is probably an alias for unsigned short.
If that is the case, then integral promotion will be applied to d and a before they are subtracted.
Integral promotion will try to convert unsigned short to int if int can hold all possible values of unsigned short. Only if that is not the case will integral promotion to unsigned int be performed.
Therefore the subtraction is most likely done in the type int, not uint16_t.
The warning is then telling you that you are casting that signed int result to an unsigned type (uint16_t) when you initialize total with it, which is normally not what you want to do, because unsigned types cannot store negative values that a signed type might hold.
If it is possible that the subtraction yields a negative value, then you should not be using uint16_t at all. Instead you should manually cast the operands to a suitable signed integral type and store the result as a suitable signed integer type (i.e. int or int32_t):
int32_t total = static_cast<int32_t>(d)-static_cast<int32_t>(a);
or
auto total = static_cast<int32_t>(d)-static_cast<int32_t>(a);
for a bit less repetition.
If you guarantee that the subtraction doesn't yield any negative values, then you can add a static_cast to tell the compiler that you really know what you are doing:
uint16_t total = static_cast<uint16_t>(d-a);
or
auto total = static_cast<uint16_t>(d-a);
if you don't want to repeat the type name.
