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stl issue

Started by denethor Jun 13, 2002 at 6:09 PM 4 replies 400+ views

Original Post

Jun 13, 2002 at 6:09 PM

  
list<string> lst;
lst.push_back("hello");
lst.push_back("hello");
lst.push_back("hello");
lst.push_back("hello");

list<string>::iterator iter;
for( iter=lst.begin; iter!=lst.end(); iter++)
{
    (*iter).erase();    //empty the string

    lst.erase(iter);    //removed the node from the list

}

In the above code, do I need the [(*iter).erase();] line, or will the string's destructor be called and erase itself when the node is destoyed? *the list is of type string [edited by - denethor on June 13, 2002 7:21:11 PM] [edit: added source tags] [edited by - Magmai Kai Holmlor on June 13, 2002 10:43:54 PM]

felonius

Jun 13, 2002 at 6:15 PM

Yes, the strings destructor will automatically be called for each element in the list. You do not need that first line.

In fact, you don''t even have to erase all elements of the list manually. When the list goes out of scope its elements will automatically be deleted and their destructors will be called.

Fruny

203

Fruny

Jun 13, 2002 at 6:17 PM

When you remove elements from a STL container, their destructor automatically gets called. Therefore, a std::string would automatically get cleaned up.

Note however, that a pointer doesn''t have a destructor (i.e. the element doesn''t get destroyed when the pointer gets out of scope, you have to manually delete them).

Your list seems to be a list of char pointers that point to statically allocated string constants (i.e. you didn''t allocate memory for the "hello" strings). Therefore there is no cleanup to be done (no pointers to delete).

Therefore, in your case, you do not have to call erase on the strings contained in you list. You would have to call delete if the strings were dynamically allocated (list of pointer to string or pointer to char*), though.

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Shannon Barber

Mod

368

Shannon Barber

Jun 13, 2002 at 9:47 PM

It''s a list of std::string, I added source tags so that it shows up. By default < and > are used to indicate html tags, so stuff between them is not displayed.

IndirectX

Jun 13, 2002 at 11:39 PM

When you use erase on an iterator, the iterator becomes invalid. Use

  
for( iter=lst.begin; iter!=lst.end(); ){
    (*iter).erase();    //empty the string

    iter = lst.erase(iter);    //removed the node from the list

}

Stoffel

Jun 14, 2002 at 11:21 AM

or just lst.clear ().

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