How is it calculated???

If we have
a=5;
c=++a + ++a + ++a;

What would be value of c after this statement and how does it come?

If we have the code
a=5;
c=a++ + a++ + a++;

What would be the value of c then?
And also please explain how it is calculated by the compiler???

[QUOTE=ashwini_nith;21579]If we have
a=5;
c=++a + ++a + ++a;

What would be value of c after this statement and how does it come?

If we have the code
a=5;
c=a++ + a++ + a++;

What would be the value of c then?
And also please explain how it is calculated by the compiler???[/QUOTE]
left to right assigment
c= ++a + ++a + ++a;
<- <- <-
(5) + (5) +(5);
c=15;

Hi!

In this equatin c=++a+ ++a+ ++a;
The "a " value is changing every step so, ++a is consider as a=1+a;
In this case, "a" value is changing it's memory location(&a).
so final value of "a" is a=8. you should put the "a" value as 8 in this equation.
c=a+a+a; c=8+8+8;c=24;

you should not do this problem as like this c=++a+ ++a+ ++a;
c=6+7+8 --->false



Following procedure may help you

int a=5;
c=++a+ ++a+ ++a; ---step(1)

(((((((((((This case output is 24

c=(++a)+( )+( );
c=(++5)+( )+( ); ---Step(2)

After this step a=6; &a-->6

c=a+ ++a+( );
c=a+ ++6+( ); ---(3)

After This step a=7; &a-->7

c=a+a+ ++a;
c= a+a+ ++7; ---(4)

After This step a=8;&a-->8

c=a+a+a;
c=8+8+8;

Finally c=24;


In the next case, You will follow above procedure. It will work well.

Even after any confusion ask me that i will help you

Hi!

Your first case output is 24
c=++a+ ++a+ ++a;--->(1)

In every step "a" value is change it's memory location.

Finally a=8; i.e &a=8;

so, c=(++a)+ (++a)+ (++a);
c= 8+8+8;c=24;

This step may help you

c=(++a)+ (++a)+ (++a);---->(2)

c=(++5)+()+();

c=(&a=6)+()+();--->(3)

c=()+(++6)+();

c=()+(&a=7)+();--->(4)

c=()+()+(++7);

c=()+()+(&a=8);--->(5)

Now &a=8

Finally, c=8+8+8;

c=24;

You will follow this steps to your second case.

Every steps "a " variable value is changing in it's memory location.


If any dought you will ask me


Regards
venkat.G

Hi!

your second case output is 15
c=a++ +a++ +a++;

The "+" "=" operators are first precedence then post increment operator "++"
********************************************************************************************
NOTE:
(But, not in preincrement eg:++i.Hence, preincrement operator is first precedence then "+" and "=" ).
*******************************************************************************************

so, first addition(+) will be done and after it's value assign(=) in variable "c".

i.e c=a++ +a++ +a++;
c=5++ +5++ +5++;
c=15 and a=5-----step(1)
variable "a" is increment 3 times after step(1)
so, a=a+3
a=8;


Finall output is c=15 and a=8



EXAMPLES;

int b=5,d;
d=b++;
what is the value of d?

you know d=5 as well as b is incremented by 1. so, b=6;

Ques1: int d,b=5;
d=b++; ------>(a)
Ans: d=b; ------>Step1
b=b+1; ------>step2
d=5; and b=6;

Ques2: int t=3,s;
s=t++ +t++;
Ans:
s=t+t; ----->step(1)
t=t+2;

s=6; and t=5;

Ques3:
int l=5,m;
m=l++ +l++ +l++;
Ans:
m=l+l+l; ---->step1
l=l+3; ------>step2
m=15; and l=8;

Reply your suggestion .

Regards
venkat.G

[QUOTE=ashwini_nith;21579]If we have
a=5;
c=++a + ++a + ++a;

What would be value of c after this statement and how does it come?

If we have the code
a=5;
c=a++ + a++ + a++;

What would be the value of c then?
And also please explain how it is calculated by the compiler???[/QUOTE]

Dear ashwini_nith,

In the first case, value of 'a' will be incremented first and then it will be used
resulting in the following:

O/P in first case will be 6+7+8=21

In the second case, value of 'a' will be used first and then incremented
resulting in the following:

O/P in second case will be 5+6+7=18

I hope this will help you.

Have a pleasant time.

[QUOTE=venki_2007;21710]Hi!

Your first case output is 24
c=++a+ ++a+ ++a;--->(1)

In every step "a" value is change it's memory location.

Finally a=8; i.e &a=8;

so, c=(++a)+ (++a)+ (++a);
c= 8+8+8;c=24;

This step may help you

c=(++a)+ (++a)+ (++a);---->(2)

c=(++5)+()+();

c=(&a=6)+()+();--->(3)

c=()+(++6)+();

c=()+(&a=7)+();--->(4)

c=()+()+(++7);

c=()+()+(&a=8);--->(5)

Now &a=8

Finally, c=8+8+8;

c=24;

You will follow this steps to your second case.

Every steps "a " variable value is changing in it's memory location.


If any dought you will ask me


Regards
venkat.G[/QUOTE]
you are wrong c=15 is the answer

[QUOTE=ashwini_nith;21579]If we have
a=5;
c=++a + ++a + ++a;

What would be value of c after this statement and how does it come?

If we have the code
a=5;
c=a++ + a++ + a++;

What would be the value of c then?
And also please explain how it is calculated by the compiler???[/QUOTE]

Dear ashwini_nith,

In the first case, value of 'a' will be incremented first and then it will be used
resulting in the following:

O/P in first case will be 6+7+8=21

In the second case, value of 'a' will be used first and then incremented
resulting in the following:

O/P in second case will be 5+6+7=18

I hope this will help you.

Have a pleasant time.

[QUOTE=sarathi trichy;21825]you are wrong c=15 is the answer[/QUOTE]


Hi!

you will check the program at TURBO C/C++ compiler
To first case
c=++a+ ++a+ ++a;
if a=5
c=24;
To second case

c=a++ +a++ +a++;
if a=5;
c=15;
-----------------> you will check in LINUX2.6.X ( In my version give this output)

To first case
c=22;
To second case
c=15;

BUT, TURBO C/C++ compiler only correct.




Regards
venkat.G

[QUOTE=sreekumar_nair_it;21780]Dear ashwini_nith,

In the first case, value of 'a' will be incremented first and then it will be used
resulting in the following:

O/P in first case will be 6+7+8=21

In the second case, value of 'a' will be used first and then incremented
resulting in the following:

O/P in second case will be 5+6+7=18

I hope this will help you.

Have a pleasant time.[/QUOTE]
you are wrong tne answer is 15 only

[QUOTE=ashwini_nith;21579]If we have
a=5;
c=++a + ++a + ++a;

What would be value of c after this statement and how does it come?

If we have the code
a=5;
c=a++ + a++ + a++;

What would be the value of c then?
And also please explain how it is calculated by the compiler???[/QUOTE]

hi aswani, does that syntax exists ? i think increment operator lone can exist to the right of assignment operator. check the syntax.

Hi all,

The answer of Venki is ONLY correct answer. As well as the explanation given by Venki is absolutely correct as per C++ language is concern. Inded i validated the result on VS2005.

Others please do give it a thought before posting a message it may misguide freshers who are new to such a beautiful language. We are here keep our loving c++ at the top and guide others to right path and solution inplace of misguiding them.

Guys sarathi trichy , sreekumar_nair_it and nagarjun_p002 please do understand the issue and work on it by validating it on a tool before givign your opinion.

no offence to anyone, m just another C++ fan joined this community recently.

++a + ++a + ++a = undefined
++a + ++a + ++a + ++a + ++a = undefined

Because C expressions can contain side effects, issues of sequencing are important in expression evaluation. Most operators impose no sequencing requirements, but a few operators impose sequence points upon their evaluation: comma, logical-AND, logical-OR and conditional.
Thus in the expressions ++a + ++a + ++a and similar expressions are said to have side effects or undefined behavior. :(
Therefore we see different values being output by different compilers as there is no rule specifying the evaluation of such expressions. These compilers assume the sequence points to be at different places.

Undefined Behavior/Side Effects :(

The standards define "Between two sequence points, an object is modified more than once, or is modified and the prior value is accessed other than to determine the value to be stored"

Different compilers will evaluate the expression differently and give different output.
e.g int a[5]={1,2,3,4,5};
printf("\\n%d %d",a[3],++3[a]);

would print the value 5 5 since the sequence point in the expression is [B]","[/B]

[U][B]The following are the sequence points described:
[/B][/U]
The call to a function, after the arguments have been evaluated
The end of the first operand of the following operators:
[LIST][*]logical AND && [*]logical OR || [*]conditional ? [*]comma ,[/LIST]
The end of a full declarator: declarators
The end of a full expression: an initializerthe expression in an expression statement
The controlling expression of a selection statement (if or switch)
The controlling expression of a while or do statement
Each of the three expressions of a for statement
The expression in a return statement
Immediately before a library function returns
After the actions associated with each formatted input/output function conversion
Immediately before and immediately after each call to a comparison function, Between any call to a comparison function and any movement of the objects
passed as arguments to that call

refer the wiki source [url=http://en.wikipedia.org/wiki/Sequence_point]Sequence point - Wikipedia, the free encyclopedia[/url]

If asked in an interview tell the output with the compiler you used and explain about the side effects in brief. :)

A = ++c + ++c + ++c; the ans is ......21..bcoz ++a means 1st increment and then assign the value a=++a + ++a + 6(++a) (right to left value readed by compiler[<-----]) a=++a + 7(++a) + 6 a=8 (++a) + 7 +6
a=21.... Thats all.

A = ++c + ++c + ++c; the ans is ......21..bcoz ++a means 1st increment and then assign the value a=++a + ++a + 6(++a) (right to left value readed by compiler[<-----]) a=++a + 7(++a) + 6 a=8 (++a) + 7 +6 a=21.... Thats all. Bst wses, sankar