string - how to define a var in C++ -

this code compiles perfect:

if ( args.length() > 0 ) {     if ( args[0]->isstring() ) {         string::utf8value szqmn( args[0]->tostring() ) ;         printf( "(cc)>>>> qmn [%s].\n", (const char*)(* szqmn) ) ;     } ; } ; 

but 1 not :

if ( args.length() > 0 ) {     if ( args[0]->isstring() ) {         string::utf8value szqmn( args[0]->tostring() ) ; // <<<< (a)     } ; } ; printf( "(cc)>>>> qmn [%s].\n", (const char*)(* szqmn) ) ; // <<<< (b) 

error says : "error c2065: 'szqmn' : undeclared identifier" on line (b)

this means me sentence marked (a) definition @ same time assignement, right ?

and compiler decides "conditionally" defined within 2 "if's" ?

my question : how move declaration out of 2 "if's" ?

in way can give defalut value ... in case if fails.

if write line out of 2 "if's"

string::utf8value szqmn ("") ; 

... error :

cannot convert argument 1 'const char [1]' 'v8::handle<v8::value>' 

any ideas?

this means me sentence marked (a) definition @ same time assignement, right?

technically constructor call creates variable , initializes it.

also note automatic variables exist until end of scope (usually block inside {} brackets). why second code example not compile.

if (condition) {     int x = 5; } x = 6; // error, x not exist anymore 

my question : how move declaration out of 2 "if's"?

string::utf8value szqmn (""); 

this constructor call of class string::utf8value class. error message takes parameter of type v8::handle<v8::value>. without knowing cannot give answer how call it. wanted pass "" of type const char* or const char[1] , compiler telling not take parameter.

edit:

from link deepblackdwarf provided in comment, how create utf8value string:

std::string something("hello world");  handle<value> something_else = string::new( something.c_str() ); 

so in case do:

string::utf8value szqmn (string::new(""));