std::mismatch
Template:ddcl list begin <tr class="t-dsc-header">
<td><algorithm>
<td></td> <td></td> </tr> <tr class="t-dcl ">
<td >std::pair<InputIt1,InputIt2>
mismatch( InputIt1 first1,
InputIt1 last1,
<td > (1) </td> <td class="t-dcl-nopad"> </td> </tr> <tr class="t-dcl ">
<td >std::pair<InputIt1,InputIt2>
mismatch( InputIt1 first1,
InputIt1 last1,
InputIt2 first2,
<td > (2) </td> <td class="t-dcl-nopad"> </td> </tr> Template:ddcl list end
Returns the first mismatching pair of elements from two ranges: one defined by [first1, last1)
and another starting at first2
. The first version of the function uses operator==
to compare the elements, the second version uses the given binary predicate p
.
Contents |
Parameters
first1, last1 | - | the first range of the elements |
first2 | - | the beginning of the second range of the elements |
p | - | binary predicate which returns true if the elements should be treated as equal. The signature of the predicate function should be equivalent to the following: bool pred(const Type1 &a, const Type2 &b); While the signature does not need to have const &, the function must not modify the objects passed to it and must be able to accept all values of type (possibly const) |
Return value
std::pair with iterators to the first two non-equivalent elements, or, if no different elements found, pair with last1
and the corresponding iterator from the second range.
Complexity
At most last1
- first1
applications of the predicate
Possible implementation
First version |
---|
template<class InputIt1, class InputIt2> std::pair<InputIt1, InputIt2> mismatch(InputIt1 first1, InputIt1 last1, InputIt2 first2) { while (first1 != last1 && *first1 == *first2) { ++first1, ++first2; } return std::make_pair(first1, first2); } |
Second version |
template<class InputIt1, class InputIt2, class BinaryPredicate> std::pair<InputIt1, InputIt2> mismatch(InputIt1 first1, InputIt1 last1, InputIt2 first2, BinaryPredicate p) { while (first1 != last1 && p(*first1, *first2)) { ++first1, ++first2; } return std::make_pair(first1, first2); } |
Example
This program determines the the longest substring that is simultaneously found at the very beginning of the given string and at the very end of it, in reverse order (possibly overlapping)
#include <iostream> #include <string> #include <algorithm> std::string mirror_ends(const std::string& in) { return std::string(in.begin(), std::mismatch(in.begin(), in.end(), in.rbegin()).first); } int main() { std::cout << mirror_ends("abXYZba") << '\n' << mirror_ends("abca") << '\n' << mirror_ends("aba") << '\n'; }
Output:
ab a aba