c++ - The string erase() function is giving different results for similar calls -

i tried using string::erase() function 2 parameters start , end point giving me different results same type of call.

here code:

#include<iostream> using namespace std;  int main(){     string s = "azxxzy";    s.erase(2,2);    cout << s;    s.erase(1,1);    cout << endl << s; } 

it deletes 2 characters i.e xx first call second call deleting 1 z.

can please explain why happening?

corrected-

the question wrong used overloaded version i.e

'string& erase (size_t pos = 0, size_t len = npos); '

but expected output

'iterator erase(iterator first,iterator last)'

you're using std::string::erase() following synoptic:

string& erase (size_t pos = 0, size_t len = npos); 

it erase part of string specified @ position pos length len. note position index pos begins 0. default parameter len = npos indicates characters until end.

in example means:

string s = "azxxzy"; s.erase(2,2);        /* azzy: deleting 2 characters position 2 */ s.erase(1,1);        /* azy:  deleting 1 character position 1  */ 

in textbook written 2 parameters of std::string::erase iterator first , iterator last. why assumed giving strange result.

you mean overloaded version:

iterator erase (iterator first, iterator last); 

but didn't provide iterator. you're passing int literals implicitly converted size_t. in link i've posted above can see overloaded version + example.

i expected output 'ay'

to output iterators following std::string::begin() , std::string::end():

string s = "azxxzy"; s.erase(s.begin() + 1, s.end() - 1); /* ay: deleting characters        */                                      /* between 1st , last character */