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builtin.string: optimize string.count where substr.len == 1 (#9337)
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@ -834,7 +834,21 @@ pub fn (s string) count(substr string) int {
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if substr.len > s.len {
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return 0
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}
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mut n := 0
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if substr.len == 1 {
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target := substr[0]
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for letter in s {
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if letter == target {
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n++
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}
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}
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return n
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}
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mut i := 0
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for {
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i = s.index_after(substr, i)
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