Commit ecbf5106 by Uwe Schulzweida

### Declare find_element() static.

parent 1673358d
 ... ... @@ -20,6 +20,7 @@ * @param nelem length of the sorted list * @param x the element to find a position for */ static long find_element(double x, long nelem, const double *restrict array) { long ii; ... ... @@ -29,10 +30,10 @@ long find_element(double x, long nelem, const double *restrict array) if ( array[0] < array[nelem-1] ) // ascending order { /* return the length of the array if x is out of bounds */ // return the length of the array if x is out of bounds if ( x < array[0] || x > array[nelem-1] ) return nelem; /* search for the interval in which x fits */ // search for the interval in which x fits // implementation: binary search algorithm for ( ii = 1; ii < nelem; ++ii ) { ... ... @@ -41,7 +42,7 @@ long find_element(double x, long nelem, const double *restrict array) // faster! mid = (first + last) >> 1; /* return the bigger interval border of the interval in which x fits */ // return the bigger interval border of the interval in which x fits // if ( x >= array[mid-1] && x <= array[mid] ) break; // faster! if ( !(x < array[mid-1] || x > array[mid]) ) break; ... ... @@ -55,10 +56,10 @@ long find_element(double x, long nelem, const double *restrict array) } else { /* return the length of the array if x is out of bounds */ // return the length of the array if x is out of bounds if ( x < array[nelem-1] || x > array[0] ) return nelem; /* search for the interval in which x fits */ // search for the interval in which x fits // implementation: binary search algorithm for ( ii = 1; ii < nelem; ++ii ) { ... ... @@ -67,7 +68,7 @@ long find_element(double x, long nelem, const double *restrict array) // faster! mid = (first + last) >> 1; /* return the bigger interval border of the interval in which x fits */ // return the bigger interval border of the interval in which x fits // if ( x >= array[mid] && x <= array[mid-1] ) break; // faster! if ( !(x < array[mid] || x > array[mid-1]) ) break; ... ...
Markdown is supported
0% or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!