不需要的舍入C ++

Question

I have the following equations:

//get thermistor resistor value
temp=(THERMISTOR_R0)/((temp2/temp)-1);

//get temperature value in Kelvins and convert to Celsiuis
temp=(THERMISTOR_BETA)/log(temp/(THERMISTOR_R0*exp((-THERMISTOR_BETA)/298)));
temp-=273;

desiredVoltage =((15700-(25*temp))/10);

THERMISTOR_R0 and THERMISTOR _BETA are constant.

temp , temp2 and desiredVoltage are unsigned int and are defined before calculations.

The problem is, for example, when the term ((temp2/temp)-1) falls below 1, it rounds down to 0. I want to get rid of this rounding as it is causing huge problems with my calculations.

How do I do this?

Answer 1

It's not rounding, it's integer division. If both operands of the / operator are of integer types the behavior of C++ is to perform an integer division, which keeps only the integer part of the result (this is often needed in some algorithms because it's faster).

To get a "regular" division make sure that at least one of the operands involved is of a floating point type ( float , double or long double ); you can do this either declaring the variables involved as FP types

double temp2, temp;

either sticking a cast in front of one of the operands.

temp=(THERMISTOR_R0)/((double(temp2)/temp)-1);

(notice that here you'll incur in truncation if temp is still of integral type).

Most probably, here you'll simply want to declare temp and temp2 as double (or float if you are working in a really resource-tight environment).

Also, when dividing by a numeric literal, keep in mind that if you don't write the decimal point it will be an int literal, if you write it it will be a double . Eg, 298 is an int , 298. is a double , so 1/2 is 0 , but 1/2. is 0.5 .

Answer 2

如果需要浮点除法行为，请确保使用浮点类型。

Answer 3

The compiler rounds floating point values down.

Generally this problem is solved by adding 0.5 to the value before the conversion to an integer. You may want to explicitly use floating point values (or cast), then add 0.5 to the result before you cast back to integers.