Question

Sometimes i see different array-access styles in C++ and thought it could be relative to an assembly addressing mode:

C++:

int * aa=new int[2];
0[aa]=15; //a little different than aa[0]
1[aa]=15;
aa[0]=15;
aa[1]=15;
printf("%d %d \n",aa[0],aa[1]);

Assembly:

__asm
{
    mov aa[0],ebx
    mov aa[1],eax
    mov 0[aa],ebx
    mov 1[aa],eax
}

Is this C++ array-access notation a standard and if yes, was it derived from an assembly addressing mode?

When i try [aa]1=5; , compiler gives

• “aa attribute not found”,
• “missing ‘;’ before ‘constant'”

• “left operand must be l-value”.

  //When i try pointer arithmetic,

  *(aa+1)=0//gives no error

  *(aa+0)=0//gives no error 🙂

Is this rule same for the operator [] overloading?

MSVC++ 2010

Thank you.

Answer 1

This is a well-known quirk of C and C++, they don’t really care which half of an expression goes in the square brackets []. They both evaluate to the same equivalent expression:

*(0+aa)=15;