C：具有幂函数将二进制转换为十进制

Question

I am trying to convert decimal number to binary number using recursion. But I am getting wrong output...I tried to debug the program and got stuck with "pow()". I inserted many printf statements to see the status of the variables... Here is my program...

#include<stdio.h>
#include<conio.h>   
#include<math.h>

void main(){
int n,k;
int z;
long b,j;
long binary(int,int);
printf("Enter a Number : ");
scanf("%d",&n);
printf("Binary Equivalent is : ");
k=log2(n);
b=binary(n,k);
z=pow(10,k); 
printf("\n\nb=%d ,z=%d",b,z); 
}



long binary(int n,int c)
{
    int a,np;
    static long b=0;
    if(n<=1){
        b=(n%2)*(int)pow(10,c)+b;
        printf("\n nmod2=%d\nb=%ld\nc=%d\nn=%d ",(n%2),b,c,n);
        return b;
    }
    else{
        a=n%2;  
        np=pow(10,c);
        b=a*np+b;
        printf("\n a=%d,np=%d \n nmod2=%d \n b=%ld \n c=%d \n n=%d ",a,np,(n%2),b,c,n);
        binary(n/2,--c);
    }
}


Output is:
Enter a Number : 5
Binary Equivalent is :
 a=1,np=99
nmod2=1
b=99    
c=2    
n=5
 a=0,np=10
nmod2=0
b=99
c=1
n=2
 nmod2=1
b=100
c=0
n=1

b=100 ,z=100

why the "pow(10,c)" in binary() when c=2 equals 99 why not 100 as in main()?

Answer 1

The problem is here:

(int)pow(10,c)

The result is around 100, but rounding it down with (int) can get you either 99 or 100 depending on the roundoff error of pow .

This is a handy function lround in math.h that should help you. It rounds a number towards the closest integer, rounding 0.5 towards zero.

Alternatively, you could define:

int my_round(double f) {
    return (int)(f+0.5);
}

Adding 0.5 and rounding down looks like rounding to the closest integer.

Answer 2

The pow() function is a floating-point function, which tends to screw up such stuff. You shouldn't use it for integers without proper rounding to compensate.

Answer 3

Multiple problems.

1. As others have point out that the result of pow(10,c) is the problem. A good pow() would return exactly 100.0 for pow(10,2) . It is reasonable to expect that behavior from pow() , but a cautious programmer would do as others suggest and round-to-nearest a floating point number to an int .

2. Round to nearest is best handle with a library function. Should one want to roll your own:

   int IRound(double f) { return (f > 0.0) ? (int)(f + 0.5) : (int)(f - 0.5); }

3. long binary(int n,int c) does not return a value when n > 1 . Suspect return binary(n/2,--c) is wanted.

4. The static long b=0; is a bad way to deal with recursion. long binary(int n,int c) would nominally correctly run only once. A second call, b would not necessarily start at 0 .

5. Given pow() is flaky, log2() may need special care too.