This is the output from the Microsoft VS 2009 compiler with default
optimizations. the increment operator is only evaluated once in the
equation.. It's evaluated the second time after the compiler is done with
the math. And the output in both cases will be 12. Because you surrounded
m++ in parens it will be evaluated before it's used. (with this particular
compiler).
; 12 : v=(m++)+(++m);
mov eax, DWORD PTR _m$[ebp] ; gets m from storage into a
useful register.
add eax, 1 ; adds one to
it
mov DWORD PTR _m$[ebp], eax ; put's it back in storage
mov ecx, DWORD PTR _m$[ebp] ; move it again into a useful
register
add ecx, DWORD PTR _m$[ebp] ; add storage value to the
register
mov DWORD PTR _v$[ebp], ecx ; move the register into where
v is stored.
mov edx, DWORD PTR _m$[ebp] ; NOW increment again..
add edx, 1
mov DWORD PTR _m$[ebp], edx
; 11 : v=v+(m++)+(++m);
mov eax, DWORD PTR _m$[ebp]
add eax, 1
mov DWORD PTR _m$[ebp], eax
mov ecx, DWORD PTR _v$[ebp] ; the only difference here is
getting v into a useful register and then adding the m storage twice..
add ecx, DWORD PTR _m$[ebp]
add ecx, DWORD PTR _m$[ebp]
mov DWORD PTR _v$[ebp], ecx
mov edx, DWORD PTR _m$[ebp]
add edx, 1
mov DWORD PTR _m$[ebp], edx
Michael
_____
From: [email protected] [mailto:[email protected]] On Behalf Of
Aliya Awais
Sent: Wednesday, May 27, 2009 5:07 AM
To: [email protected]
Subject: [c-prog] confusing output
hi ! can someone please tell me the difference between output of given two
expression in C++ programs , how expressions are evaluated in these two
cases?
#include<iostream.h>
void main()
{
int m=5,v=0,u=0;
v=v+(m++)+(++m);
cout<<v<<endl; //output is 11
}
#include<iostream.h>
void main()
{
int m=5,v=0,u=0;
v=(m++)+(++m);
cout<<v<<endl; //output is 12 (why confusing)
}
what is difference between v=v+(m++)+(++m); and v=(m++)+(++m); .
thanks
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[Non-text portions of this message have been removed]
