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						Function std.algorithm.searching.countUntil
Counts elements in the given
    forward range
    until the given predicate is true for one of the given needles.
						
				ptrdiff_t countUntil(alias pred, R, Rs...)
				(
				
				  R haystack,
				
				  Rs needles
				
				)
				
				if (isForwardRange!R && (Rs
				
				ptrdiff_t countUntil(alias pred, R, N)
				(
				
				  R haystack,
				
				  N needle
				
				)
				
				if (isInputRange!R && is(typeof(binaryFun!pred(haystack
				
				ptrdiff_t countUntil(alias pred, R)
				(
				
				  R haystack
				
				)
				
				if (isInputRange!R && is(typeof(unaryFun!pred(haystack
				Parameters
| Name | Description | 
|---|---|
| pred | The predicate for determining when to stop counting. | 
| haystack | The input range to be counted. | 
| needles | Either a single element, or a
            forward range
            of elements, to be evaluated in turn against each
            element in haystackunder the given predicate. | 
Returns
The number of elements which must be popped from the front of
    haystack before reaching an element for which
    startsWith!pred(haystack, needles) is true. If
    startsWith!pred(haystack, needles) is not true for any element in
    haystack, then -1 is returned. If only pred is provided,
    pred(haystack) is tested for each element.
See Also
Example
writeln(countUntil("hello world", "world")); // 6
writeln(countUntil("hello world", 'r')); // 8
writeln(countUntil("hello world", "programming")); // -1
writeln(countUntil("日本語", "本語")); // 1
writeln(countUntil("日本語", '語')); // 2
writeln(countUntil("日本語", "五")); // -1
writeln(countUntil("日本語", '五')); // -1
writeln(countUntil([0, 7, 12, 22, 9], [12, 22])); // 2
writeln(countUntil([0, 7, 12, 22, 9], 9)); // 4
writeln(countUntil!"a > b"([0, 7, 12, 22, 9], 20)); // 3
Example
import stdAuthors
License
					Copyright © 1999-2024 by the D Language Foundation | Page generated by ddox.