问题
Let's say I have the following code snippet:
int i; double value;
for(i = 0; i < CONSTANT; i++) {
value = (double)pow(2, i);
}
Trying to compile this code yields an "undefined reference to `pow'" error.
Including or excluding math.h makes no difference, since it ends up being included anyway.
Raising 2.0 to a hardcoded power works okay, but everything fails if I substitute the exponent by an expression that contains i.
What am I doing wrong? Thanks.
回答1:
It's a very interesting behavior, and a good learning example.
To solve your problem, add
-lm
to your gcc command line (provided you're using gcc). This tells the compiler to link against the math library.
What seems to be going on, is that if you're using
pow(2.0, 3);
the compiler realizes this expression evaluates to a constant, and does mere substitution.
Thus, no library function has to be called.
回答2:
You need to link with -lm to actually include the math library.
It worked for a hardcoded value because the compiler optimized the pow call away.
回答3:
You must link against the math library:
gcc program.c -lm
The reason is that GCC (and some other compilers) have a built-in pow() function for literal constants. So if you call pow() with 2.0 manually, the compiler will actually figure-out what the answer is and substitute that for you. With a variable input, the compiler must rely on the math library, which you must link against.
回答4:
The code for pow is part of the math library. You need to link in that library (in addition to the C library that is linked in by default).
To do that, with gcc, specify -lm on the compiler invocation
gcc ... -lm
回答5:
http://www.cplusplus.com/reference/clibrary/cmath/pow/
In C, only the version taking two double parameters exists with this name. The other overloads are only available in C++.
It looks like you can't pass an int, so just make i a double and that should work.
来源：https://stackoverflow.com/questions/4431053/cs-pow-function-refuses-to-work-with-variable-exponent