c++ - Difference between a static and dynamic array -


Friends I was just playing around with some pointer programs and felt that the GCC (and probably standard standard) Differentiates and dynamic arrays.

In an array there is a place holder for the address of elements, while for the stable array, there is no memory place where the compiler element stores the initial address of the array.

I have an example program for displaying my confusions. 

  #include & lt; Iostream & gt; # Key (zero) {int_static [10]; Int * _dynamic; _dynamic = new int [10]; Std :: Court & lt; & Lt; "_static =" & lt; & Lt; _static & lt; & Lt; "& Amp; _static =" & lt; & Lt; & Amp; _static & lt; & Lt; "& Amp; _static [0] =" & lt; & Lt; & Amp; _static [0] to be & lt; & Lt; Std :: endl; Std :: Court & lt; & Lt; "_dynamic =" & lt; & Lt; _dynamic & lt; & Lt; "& Amp; _dynamic =" & lt; & Lt; & Amp; _dynamic & lt; & Lt; "& Amp; _dynamic [0] =" & lt; & Lt; & Amp; _dynamic [0] is followed by & lt; & Lt; Std :: endl; Return 0; }   
 For the above program,  _static  and  & amp; _static [0]  Return the same address to the expected lines However,  & amp; _static  also the other two returns as a single address  
 then,  _static  and  & amp; _static  Refer to the same number (or address we want to call it) as expected,  _dynamic  and  & amp; _dynamic  indicate different locations.  
 So, why did the C standard say that  _static  and  & amp; _static  should mention the same place. It seems misleading. One reason I think is that  & amp; _static  does not matter much  
 Can anyone help me with this illusion?   
 
 
 Actually  _static  and  & amp; _static  is not the reason for not mentioning the same place because you use  _static  in that context, in which it is decayed in an indicator, this is the reason that you Use, you  create  refer to them in the same place. But before you did not do that - there was one array and the other was an indicator they could not be equal because they were basically different things.

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