Why pow(10,5) = 9,999 in C++

Question

Recently i write a block of code:

const int sections = 10;

for(int t= 0; t < 5; t++){
   int i = pow(sections, 5- t -1);  
   cout << i << en         


        

Answer 1

From Here

Looking at the pow() function: double pow (double base, double exponent); we know the parameters and return value are all double type. But the variable num, i and res are all int type in code above, when tranforming int to double or double to int, it may cause precision loss. For example (maybe not rigorous), the floating point unit (FPU) calculate pow(10, 4)=9999.99999999, then int(9999.9999999)=9999 by type transform in C++.

How to solve it?

Solution1

Change the code:

    const int num = 10;

    for(int i = 0; i < 5; ++i){
       double res = pow(num, i);
       cout << res << endl;
    }

Solution2

Replace floating point unit (FPU) having higher calculation precision in double type. For example, we use SSE in Windows CPU. In Code::Block 13.12, we can do this steps to reach the goal: Setting -> Compiler setting -> GNU GCC Compile -> Other options, add

-mfpmath=sse -msse3

The picture is as follows:

_{(source: qiniudn.com)}