Compress the array, removing from it all elements whose value is in the interval [a, b]. C ++

And it will not do? :)

 int arr[10] = { 1, 2, 5, 8, 12, 15, 45, 3, 4, 7 }; int main(int argc, const char * argv[]) { int a = 6, b = 20; fill(remove_if(arr,arr+10,[a,b](int x){ return (x >= a && x <= b); }),arr+10,0); for(auto x: arr) cout << x << " "; cout << endl; } 

Especially for @pavel: :-)

 void removeArr(int * arr, int size, int a, int b) { struct pred { pred(int a, int b):a(a),b(b){} bool operator()(int x) { return (x >= a && x <= b); } int a, b; }; fill(remove_if(arr,arr+size,pred(a,b)),arr+size,0); } 

In this cycle

 for (i = 0; i < n; i++){ if ( arr[i] >= a && arr[i] <= b ){ arr[i] = arr[i+1]; arr[i] = 0; } 

out-of-bounds occurs when i equals n-1 in a sentence

  arr[i] = arr[i+1]; 

and besides, the body of the loop does not make sense, since first the assignment for the element with index i value arr[i+1] , and then this value is overwritten with zero.

  arr[i] = arr[i+1]; arr[i] = 0; 

Below is a demo program showing how you can "remove" elements from an array that lie in the [a, b] segment, use loops

 #include <iostream> int main() { int arr[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }; const size_t N = sizeof( arr ) / sizeof( *arr ); for ( size_t i = 0; i < N; i++ ) std::cout << arr[i] << ' '; std::cout << std::endl; int a = 3, b = 6; size_t k = 0; for ( size_t i = 0; i < N; i++ ) { if ( not ( a <= arr[i] && arr[i] <= b ) ) { if ( i != k ) arr[k] = arr[i]; ++k; } } for ( size_t i = k; i < N; i++ ) arr[i] = 0; for ( size_t i = 0; i < N; i++ ) std::cout << arr[i] << ' '; std::cout << std::endl; return 0; } 

Output of the program to the console

 0 1 2 3 4 5 6 7 8 9 0 1 2 7 8 9 0 0 0 0 

The same can be done using standard algorithms. For example,

 #include <iostream> #include <algorithm> int main() { int arr[] = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 }; const size_t N = sizeof( arr ) / sizeof( *arr ); for ( size_t i = 0; i < N; i++ ) std::cout << arr[i] << ' '; std::cout << std::endl; int a = 3, b = 6; size_t k = 0; std::fill( std::remove_if( arr, arr + N, [&]( int x ) { return a <= x && x <= b; } ), arr + N, 0 ); for ( size_t i = 0; i < N; i++ ) std::cout << arr[i] << ' '; std::cout << std::endl; return 0; } 

The output of this program is similar to the one shown above.

 0 1 2 3 4 5 6 7 8 9 0 1 2 7 8 9 0 0 0 0 

In addition to the answers already given. You can do just one cycle with two indices. As soon as the first index reaches the end of the array, all elements between the second index and the end can be replaced with 0.

 for(int i=0, j=0;i<N;){ if(j<N){ if(!(arr[j] >= a && arr[j] <=b)){ i++; } arr[i]=++j<N? arr[j]:0; } else{ arr[i++]=0; } } 

Example on ideone

What do you think will happen here?

arr[i] = arr[i+1];

with i = n - 1 ?

Here is the code for you. Optimization at your discretion

 int i = 0; int lastIdx = n - 1; while (i <= lastIdx) { if (arr[i] >= a && arr[i] <= b) { for (int j = i; j < lastIdx; j++) arr[j] = arr[j + 1]; arr[lastIdx--] = 0; } else i++; }