| 1 | /* Range object implementation */
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| 2 |
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| 3 | #include "Python.h"
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| 4 |
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| 5 | typedef struct {
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| 6 | PyObject_HEAD
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| 7 | long start;
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| 8 | long step;
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| 9 | long len;
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| 10 | } rangeobject;
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| 11 |
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| 12 | /* Return number of items in range/xrange (lo, hi, step). step > 0
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| 13 | * required. Return a value < 0 if & only if the true value is too
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| 14 | * large to fit in a signed long.
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| 15 | */
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| 16 | static long
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| 17 | get_len_of_range(long lo, long hi, long step)
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| 18 | {
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| 19 | /* -------------------------------------------------------------
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| 20 | If lo >= hi, the range is empty.
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| 21 | Else if n values are in the range, the last one is
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| 22 | lo + (n-1)*step, which must be <= hi-1. Rearranging,
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| 23 | n <= (hi - lo - 1)/step + 1, so taking the floor of the RHS gives
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| 24 | the proper value. Since lo < hi in this case, hi-lo-1 >= 0, so
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| 25 | the RHS is non-negative and so truncation is the same as the
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| 26 | floor. Letting M be the largest positive long, the worst case
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| 27 | for the RHS numerator is hi=M, lo=-M-1, and then
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| 28 | hi-lo-1 = M-(-M-1)-1 = 2*M. Therefore unsigned long has enough
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| 29 | precision to compute the RHS exactly.
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| 30 | ---------------------------------------------------------------*/
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| 31 | long n = 0;
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| 32 | if (lo < hi) {
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| 33 | unsigned long uhi = (unsigned long)hi;
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| 34 | unsigned long ulo = (unsigned long)lo;
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| 35 | unsigned long diff = uhi - ulo - 1;
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| 36 | n = (long)(diff / (unsigned long)step + 1);
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| 37 | }
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| 38 | return n;
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| 39 | }
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| 40 |
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| 41 | static PyObject *
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| 42 | range_new(PyTypeObject *type, PyObject *args, PyObject *kw)
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| 43 | {
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| 44 | rangeobject *obj;
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| 45 | long ilow = 0, ihigh = 0, istep = 1;
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| 46 | long n;
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| 47 |
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| 48 | if (!_PyArg_NoKeywords("xrange()", kw))
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| 49 | return NULL;
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| 50 |
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| 51 | if (PyTuple_Size(args) <= 1) {
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