mismatch

Search two sequences for a mismatched item.

Use a pair of iterators i and j to traverse two sequences, starting at first1 and first2 , respectively. Return the iterator pair ( i, j ) when either their respective elements mismatch or i reaches last1 . The first version uses operator== to perform the comparisons, whereas the second version uses binary_pred .

Library

Standards<ToolKit>

Declaration


#include <algorithm>

template< class InputIterator1, class InputIterator2 >
pair< InputIterator1, InputIterator2 > mismatch
  (
  InputIterator1 first1,
  InputIterator1 last1,
  InputIterator2 first2
  );

template
  <
  class InputIterator1,
  class InputIterator2,
  class BinaryPredicate
  >
pair< InputIterator1, InputIterator2 > mismatch
  (
  InputIterator1 first1,
  InputIterator1 last1,
  InputIterator2 first2,
  BinaryPredicate binary_pred
  );

Complexity

Time complexity is linear, as at most ( last1 - first1 ) comparisons are performed. Space complexity is constant.

Example <ospace/osstd/examples/mismtch0.cpp>
#include <iostream>
#include <algorithm>

int n1[ 5 ] = { 1, 2, 3, 4, 5 };
int n2[ 5 ] = { 1, 2, 3, 4, 5 };
int n3[ 5 ] = { 1, 2, 3, 2, 1 };

void
main()
  {
  pair< int*, int* > result;

  result = mismatch( n1, n1 + 5, n2 );

  if ( result.first ==( n1 + 5 ) && result.second ==( n2 + 5 ) )
    cout << "n1 and n2 are the same\n";
  else
    cout << "Mismatch at offset: " << ( result.first - n1 ) << "\n";

  result = mismatch( n1, n1 + 5, n3 );

  if ( result.first ==( n1 + 5 ) && result.second ==( n3 + 5 ) )
    cout << "n1 and n3 are the same\n";
  else
    cout << "Mismatch at offset: " << ( result.first - n1 ) << "\n";
  }

n1 and n2 are the same
Mismatch at offset: 3
Example <ospace/osstd/examples/mismtch1.cpp>
#include <iostream>
#include <algorithm>
#include <utility>
#include <vector>

void
main()
  {
  vector< int > v1( 10 );
  vector< int > v2( v1.size() );
  for ( size_t i = 0; i < v1.size(); ++i )
    {
    v1[ i ] = i;
    v2[ i ] = i;
    }

  pair< vector< int >::iterator, vector< int >::iterator > result =
    mismatch
      (
      v1.begin(),
      v1.end(),
      v2.begin()
      );

  if ( result.first == v1.end() && result.second == v2.end() )
    cout << "v1 and v2 are the same\n";
  else
    cout << "mismatch at index: " << ( result.first - v1.begin() ) << "\n";
  v2[ v2.size() / 2 ] = 42;

  result = mismatch( v1.begin(), v1.end(), v2.begin() );

  if ( result.first == v1.end() && result.second == v2.end() )
    cout << "v1 and v2 are the same\n";
  else
    cout << "mismatch at index: " << ( result.first - v1.begin() ) << "\n";
  }

v1 and v2 are the same
mismatch at index: 5
Example <ospace/osstd/examples/mismtch2.cpp>
#include <iostream>
#include <string.h>
#include <algorithm>

const unsigned size = 5;
char* n1[ size ] = { "Brett", "Graham", "Jack", "Mike", "Todd" };

bool
str_equal( const char* a_, const char* b_ )
  {
  return ::strcmp( a_, b_ ) == 0 ? 1 : 0;
  }

void
main()
  {
  char* n2[ size ];
  copy( n1, n1 + 5, n2 );

  pair< char**, char** > result
    (
    mismatch( n1, n1+ size, n2, str_equal )
    );

  if ( result.first == n1 + size && result.second == n2 + size )
    cout << "n1 and n2 are the same\n";
  else
    cout << "mismatch at index: " << ( result.first - n1 ) << "\n";
  n2[ 2 ] = "QED";

  result = mismatch( n1, n1 + size, n2, str_equal );

  if ( result.first == n2 + size && result.second == n2 + size )
    cout << "n1 and n2 are the same\n";
  else
    cout << "mismatch at index: " << ( result.first - n1 ) << "\n";
  }

n1 and n2 are the same
mismatch at index: 2

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