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py: In string.count, handle case of zero-length needle.

genexit-inst
Damien George 2014-03-13 22:07:55 +00:00
parent de4d7aecc8
commit 536dde254b
2 changed files with 33 additions and 7 deletions
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14
py/objstr.c
View File

@ -495,8 +495,8 @@ STATIC mp_obj_t str_count(uint n_args, const mp_obj_t *args) {
GET_STR_DATA_LEN(args[0], haystack, haystack_len);
GET_STR_DATA_LEN(args[1], needle, needle_len);
size_t start = 0;
size_t end = haystack_len;
machine_uint_t start = 0;
machine_uint_t end = haystack_len;
/* TODO use a non-exception-throwing mp_get_index */
if (n_args >= 3 && args[2] != mp_const_none) {
start = mp_get_index(&str_type, haystack_len, args[2], true);
@ -505,13 +505,13 @@ STATIC mp_obj_t str_count(uint n_args, const mp_obj_t *args) {
end = mp_get_index(&str_type, haystack_len, args[3], true);
}
machine_int_t num_occurrences = 0;
// needle won't exist in haystack if it's longer, so nothing to count
if (needle_len > haystack_len) {
MP_OBJ_NEW_SMALL_INT(0);
// if needle_len is zero then we count each gap between characters as an occurrence
if (needle_len == 0) {
return MP_OBJ_NEW_SMALL_INT(end - start + 1);
}
// count the occurrences
machine_int_t num_occurrences = 0;
for (machine_uint_t haystack_index = start; haystack_index + needle_len <= end; haystack_index++) {
if (memcmp(&haystack[haystack_index], needle, needle_len) == 0) {
num_occurrences++;

26
tests/basics/string_count.py
View File

@ -1,3 +1,29 @@
print("".count(""))
print("".count("a"))
print("a".count(""))
print("a".count("a"))
print("a".count("b"))
print("b".count("a"))
print("aaa".count(""))
print("aaa".count("a"))
print("aaa".count("aa"))
print("aaa".count("aaa"))
print("aaa".count("aaaa"))
print("aaaa".count(""))
print("aaaa".count("a"))
print("aaaa".count("aa"))
print("aaaa".count("aaa"))
print("aaaa".count("aaaa"))
print("aaaa".count("aaaaa"))
print("aaa".count("", 1))
print("aaa".count("", 2))
print("aaa".count("", 3))
print("aaa".count("", 1, 2))
print("asdfasdfaaa".count("asdf", -100))
print("asdfasdfaaa".count("asdf", -8))
print("asdf".count('s', True))