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Function std.algorithm.searching.count
The first version counts the number of elements x in r for
which pred(x, value) is true. pred defaults to
equality. Performs Ο(haystack.length) evaluations of pred.
The second version returns the number of times needlehaystackneedle.empty, as the count
of the empty range in any range would be infinite. Overlapped counts
are not considered, for example count("aaa", "aa")1, not
2.
The third version counts the elements for which pred(x) is true. Performs Ο(haystack.length) evaluations of pred.
Prototypes
size_t count(alias pred, Range, E)( Range haystack, E needle ) if (isInputRange!Range && !isInfinite!Range && is(typeof(binaryFun!pred(haystack.front, needle)) : bool)); size_t count(alias pred, R1, R2)( R1 haystack, R2 needle ) if (isForwardRange!R1 && !isInfinite!R1 && isForwardRange!R2 && is(typeof(binaryFun!pred(haystack.front, needle.front)) : bool)); size_t count(alias pred, R)( R haystack ) if (isInputRange!R && !isInfinite!R && is(typeof(unaryFun!pred(haystack.front)) : bool));
Note
Regardless of the overload, counthaystack
Parameters
| Name | Description |
|---|---|
| pred | The predicate to evaluate. |
| haystack | The range to count. |
| needle | The element or sub-range to count in the |
Returns
The number of positions in the haystackpred returned true.
Example
import std.uni : toLower; // count elements in range int[] a = [ 1, 2, 4, 3, 2, 5, 3, 2, 4 ]; assert(count(a, 2) == 3); assert(count!("a > b")(a, 2) == 5); // count range in range assert(count("abcadfabf", "ab") == 2); assert(count("ababab", "abab") == 1); assert(count("ababab", "abx") == 0); // fuzzy count range in range assert(count!((a, b) => std.uni.toLower(a) == std.uni.toLower(b))("AbcAdFaBf", "ab") == 2); // count predicate in range assert(count!("a > 1")(a) == 8);