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1. (http://group.jobbole.com/13838/)

Sogou 2016 research and development engineer Pen test

1 intMainintargcChar*argv[])2 {3     stringA="Hello World";4     stringb=A;5     if(A.C_STR () = =b.c_str ())6     {7cout<<"true"<<Endl;8     }9     Elsecout<<"false"<<Endl;Ten     stringC=b; OneC=""; A     if(A.C_STR () = =b.c_str ()) -     { -cout<<"true"<<Endl; the     } -     Elsecout<<"false"<<Endl; -A=""; -     if(A.C_STR () = =b.c_str ()) +     { -cout<<"true"<<Endl; +     } A     Elsecout<<"false"<<Endl; at     return 0; -}

The correct result about the code output is ()

1. False false
2. True false False
3. True True True
4. True True False

Run the result on my WINDOWS10+TDM gcc 5.x is all false

As you can see from the debug results in GdB, C_STR returns a character pointer, which means that every string object that applies the function results in a character array, the first address pointer of the array is returned, and according to subsequent debugs, such as a= "", The position of the first address of this array is fixed, that is, the first time the pointer returns to the A.C_STR is

(GDB) P A.c_str ()

$ = 0x24fe20 "Hello World"

After A= ""

(GDB) P A.c_str ()

$6 = 0x24fe20 ""

The address has not changed.

Therefore, all the c_str results of A and b above are not equal in comparison.

In addition to the knowledge points in C + + primer 5, the use of B=a and B (a) effect is similar, is the copy initialization rather than direct initialization, so two objects are not the same, but for the overloaded = =, is based on the same characters are returned true.

Some pen questions on the internet