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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 needle occurs in haystack. Throws an exception if needle.empty, as the count of the empty range in any range would be infinite. Overlapped counts are not considered, for example count("aaa", "aa") is 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, count will not accept infinite ranges for haystack.

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);

Authors

Andrei Alexandrescu

License

Boost License 1.0.

Comments