std::pow gives a wrong approximation for fractional exponents

问题

Here is what I mean trying to do

 double x=1.1402
 double pow=1/3;
 std::pow(x,pow) -1;

result is 0 but I expect 0.4465

the equation is (1 + x) ^3= 1.1402, find x.

回答1:

1/3 is done as integer arithmetic, so you're assigning 0 to pow. Try pow(x, 1.0/3.0);

回答2:

1/3 is 0. That's integer division.

Try:

double pow = 1.0 / 3.0;

For:

#include <iostream>
#include <cmath>

int main(void)
{
 double x = 1.1402;
 double pow = 1.0/3.0;
 std::cout << std::pow(x, pow) - 1;

}

回答3:

Many have stated that 1/3 = 0, but have not explained why this is so.

C and C++ will perform the operation based on the the types of the operands. Since both operands are integers, it performs an integer division creating an integer result. When it is forced to assign that integer result to a double variable, it converts the integer 0 to a double 0.0.

It is not necessary to make both operands double, if either one is double the compiler will convert the other to double as well before performing the operation. 1.0/3 or 1/3.0 will both return the result you expected, as will 1.0/3.0.

回答4:

your 1/3 is integer division, the result of the integer division is 0.

来源：https://stackoverflow.com/questions/1580332/stdpow-gives-a-wrong-approximation-for-fractional-exponents