Important changes to forums and questions
All forums and questions are now archived. To start a new conversation or read the latest updates go to forums.mbed.com.
12 years, 5 months ago.
what is pow?
hello i found this code:
altitude = (float)44330 * (1 - pow(((float) pressure/p0), 0.190295));
but mbed doesnt want to accept this, i get:
"more than one instance of overloaded function "pow" matches the argument list:" in file "/main.cpp", Line: 67, Col: 42
" function "std::pow(float, int)"" in file "/"
" argument types are: (float, double)" in file "/"
anbody has an idea? thanx in advance!
3 Answers
12 years, 5 months ago.
Try this:
altitude = (float)44330 * (1 - pow( (float)(pressure/p0), 0.190295) );
It's subtle, but this says the entire expression, "pressure/p0," is float, rather than just "pressure."
Did you see this one?
http://mbed.org/forum/mbed/topic/476/
Particularly Simon's comment?
posted by 17 Mar 2013
12 years, 5 months ago.
the problem is the function pow needs an integer, but i want my exponent is 0.190295 is there an other function then power??
12 years, 5 months ago.
Instead of casting pressure/p0 as a float, cast it as a double.
float altitude = (float)44330 * (1 - pow(((double) (pressure/p0)), 0.190295));
Tim
You can also declare that your exponent is a float:
float altitude = (float)44330 * (1 - pow(((float) (pressure/p0)), (float)0.190295))
http://www.cplusplus.com/reference/cmath/pow/
The standard library function pow does not have a mixed float/double or double/float input overloading. What you originally typed caused pressure/p0 to be a float while your exponent was interpreted as a double by the compiler. Explicitly declaring either the base as a double and letting the exponent be interpreted as a double, or declaring both as floats clears up the problem.
posted by 17 Mar 2013