type conversion - Calling unexisting constructor in C++ -

suppose have class string:

class string {    public:     string (int n);    string(const char *p); } 

what happen if try:

string mystring='x'; 

here written char 'x' converted int , call string(int) constructor. however, not understand it.

first, how 'x' can converted int? can imagine "3" converted 3 "x" converted to? second, have 2 constructors in class. first constructor takes 1 argument of int type , constructor takes pointer char variable argument. try call not existing constructor takes char argument. so, convert char integer, why not try convert char pointer char , use second constructor?

first, how 'x' can converted int? can imagine "3" converted 3 "x" converted to?

don't make mistake of confusing 'x' , "x".

• "x" string literal and, you're right, meaningless try convert int;
• 'x' character literal, number (equal underlying value of whatever character x maps in basic source character set; ascii, 120); converting integer trivial.

so, convert char integer, why not try convert char pointer char , use second constructor?

there rules in c++ govern "best match". in case, it's int overload. in case both equal match, compiler present "ambiguous overload" error , forced convert 'x' explicitly (i.e. manually, cast) in order dictate wanted.

in case not problem, since non-pointers cannot convert pointers.