Logarithm computing without math.h

问题

I'm trying to compute ln(x) by Taylor series. Here is my code:

#define N 10

float ln(float x){ 
    int i;
    float result;
    float xt; 
    float xtpow;
    int sign;
    if(x > 0 && x <= 1){ 
        xt = x - 1.0;
        sign = -1; 
        xtpow = 1.0;
        result = 0;
        for(i = 1 ; i < N + 1; i++ );
        {   
            // Problem here
            printf("%d\n", i); 
            sign = sign * (-1);
            xtpow *= xt; 
            result += xtpow * sign / i;
        }   
    }else if(x >= 1)
    {   
        return -1 * ln(1.0 / x); 
    }   
    return result; 
}

The problem is with my series cycle(see above). It seems like after 1 cycle variable i becomes equal N + 1, and nothing going on after it. Have you any ideas why it is so?

回答1:

It seems like after 1 cycle variable i becomes equal N + 1

remove ; after for loop:

for(i = 1 ; i < N + 1; i++ );
                            ^

Your loop continue execute without executing code in block you putted in braces { } after for loop and for loop just increments i till for loop breaks. After the loop code block (where you commented "problem is here") get executes with i = N + 1 value.

I am not sure but I have an additional doubt about conditional expressions in if(). You code pattern is something like:

 if(x > 0 && x <= 1){ <-- "True for x == 1"
   // loop code
 }
 else if(x >= 1){ <-- "True for x == 1"  
      // expression code here 
 }

So for x == 1 else code never execute. Check this code too.

来源：https://stackoverflow.com/questions/18961889/logarithm-computing-without-math-h