#include <iostream>
#include <list>
using namespace std;


int main () {
    list<int> mylist;
    list<int>::iterator it;

    for(int i=1;i<6;i  ){
        mylist.push_back(i);
    }
    for (it=mylist.begin(); it!=mylist.end();   it)
    cout << ' ' << *it;
    cout<<endl;

    for(it=mylist.begin(); it!=mylist.end();it  ){
        if((*it)==2){
            mylist.erase(it);
            mylist.insert(it,9);
            break;
        }
    }
    
    for (it=mylist.begin(); it!=mylist.end();   it)
    cout << ' ' << *it;
    cout<<endl;
    return 0;
}

It seems that mylist.erase(it) is not working because when I delete it, the program works.

The output for the above program is

1 2 3 4 5 bash: line 1: 88370 Segmentation fault: 11 "/Users/alimtleuliyev/Desktop/quiadratic" [Finished in 441ms with exit code 139]

CodePudding user response：

You need to write

   if((*it)==2){
        it = mylist.erase(it);
        mylist.insert(it,9);
        break;
    }

To output the list it is better to use the range-based for loop.

Also to find an element in the list with the given value it is better to use the standard algorithm std::find instead of using a for loop.

Here is a demonstration program.

#include <iostream>
#include <list>
#include <iterator>
#include <algorithm>

int main()
{
    std::list<int> mylist;

    for ( int i = 1; i < 6; i   ) 
    {
        mylist.push_back( i );
    }

    for (const auto &item : mylist)
    {
        std::cout << item << ' ';
    }
    std::cout << std::endl;

    auto it = std::find( std::begin( mylist ), std::end( mylist ), 2 );

    if ( it != std::end( mylist) )
    {
        it = mylist.erase( it );
        mylist.insert( it, 9 );
    }

    for (const auto &item : mylist)
    {
        std::cout << item << ' ';
    }
    std::cout << std::endl;
        
    return 0;
}

The program output is

1 2 3 4 5
1 9 3 4 5

Pay attention to that instead of calling the member functions erase and insert you could just write

    if ( it != std::end( mylist) ) *it = 9;