how ++n and n++ differ ?

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C++

what is the basic difference between i++ and ++i ? which is better?

Posted 13-Jul-11 1:53am

consim

Updated 13-Jul-11 2:49am

ThatsAlok

v2

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ThatsAlok 13-Jul-11 8:49am

editor comment: changed second i++ to ++i typo mistake

Sergey Alexandrovich Kryukov 13-Jul-11 16:43pm

Why not simply reading C++ reference?
--SA

4 solutions

Solution 3

Try to find a difference :)

// 1.
{
  int a[2] = {0};
  int i = 0;

  do {
    a[i++] = 1; // the index 0 is touched as well (0 and 1 are touched)
  } while (i < 2)
}

// 2.
{
  int a[2] = {0};
  int i = 2;

  do {
    a[--i] = 1; // the index 2 is not touched (0 and 1 are touched)
  } while (i)
}

Posted 13-Jul-11 3:17am

Eugen Podsypalnikov

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John R. Shaw · Accepted Answer · 2011-07-13T10:48:00

n = ++i; // #1 : assigns i to n after adding 1 to i
n = i++; // #2 : assigns i to n before adding 1 to i

which is better?

++i should be preferred over i++; I believe Stroustrup, Myers and others have stated this in various books and/or post.

The reason we preferred this in C (especially for graphics), was that the compiler created a temporary (at assembly level) to hold the previous value of i; the temporary was then assigned to n after adding 1 to i. Therefore, it was less efficient.

For built-in types a modern compiler will normally optimize out the temporary; although they are not required to do so.

In C++ it is even more important to get in the habit of preferring ++i. The reason for this is operator overloading; which, unless the class is very simple, the compiler cannot optimize out the temporary; which can be very costly. You do not want to accidentally create unneeded overhead because you are in the habit of writing i++.

Just create a simple class and override the pre and post increments, then step through it to see the problem.

Most modern books consistently use i++ when presenting code examples. This is bad, because 99% of the time ++i could have been used instead.

for( i=0; i<10; ++i ) {} // Good habit
for( i=0; i<10; i++ ) {} // Bad habit

వేంకటనారాయణ(venkatmakam) · Accepted Answer · 2011-07-13T02:02:00

Solution 1

1. i++ will increment i and returns the i value.
2. i+1 is rvalue that wont effect i and returns i+1 value.
3. ++i, increments i and returns i+1 value;

example:

int i = 9;
int j;
j = i++; //After this statement j value will be 9, i value will be 10;
j = i+1; //After this statement j value will be 10, i value will be 9;
j = ++i; //After this statement j value will be 10, i value will be 10;

so normally in stand alone expressions like i++; and ++i; have no difference.but the difference comes when you use them in assign statements(as shown in example);

Posted 13-Jul-11 2:02am

వేంకటనారాయణ(venkatmakam)

Updated 13-Jul-11 2:05am

v2

Comments

Joan M 13-Jul-11 11:34am

Good answer that covers all the possibilities... 5ed!

వేంకటనారాయణ(venkatmakam) 14-Jul-11 2:08am

Thank you Joan Murt

Philippe Mori 16-Jul-11 19:04pm

This is a good explanation for primitive types but it does not covers user defined type (like STL iterators).

CPallini · Accepted Answer · 2011-07-13T02:04:00

Solution 2

try:

C

int i = 5;
int a = i++;
printf("%d\n", a);

vs

C

int i = 5;
int a = ++i;
printf("%d\n", a);

Hint: first snippet will output 5, second one will output 6.

Or you may even try reading the documentation (it won't hurt you).

Posted 13-Jul-11 2:04am

CPallini

Updated 13-Jul-11 2:49am

v3

Comments

వేంకటనారాయణ(venkatmakam) 13-Jul-11 8:08am

Good home work :-).My 5.

Guyverthree 13-Jul-11 8:44am

don#'t forget that ++ is an operator and + 1 is not.

therefore if i is a pointer + 1 might only move the pointer a single byte.
where as ++i or i++ will move it the size of the data that the pointer points to :)

CPallini 13-Jul-11 8:48am

You are wrong: the pointer math is consistent with the 'integers' one, hence (p+1) and, for instance, (++p) point to the same memory address.

Guyverthree 18-Jul-11 8:57am

LPVOID does not have a size on Linux so + 1 will move it a single byte. Windows for instance it is 4 bytes.
and ++ will not compile.

CPallini 18-Jul-11 11:26am

That's true just for the very special case of (void *) pointers.

Guyverthree 18-Jul-11 11:30am

Glad we agree.

CPallini 18-Jul-11 11:33am

:-)

ThatsAlok 13-Jul-11 8:47am

right you say.... there is one instruction less in i++ or ++i , i forget!

CPallini 13-Jul-11 8:51am

Do you mean it is a language gift to the optimizer? Well, maybe.

ThatsAlok 14-Jul-11 2:32am

yeap!

Joan M 13-Jul-11 11:35am

I love when I read those jewels: "Or you may even try reading the documentation (it won't hurt you)." good answer and great comment and 5ed, of course!

CPallini 13-Jul-11 11:46am

Moltes gràcies.

Joan M 13-Jul-11 11:53am

Prego!

how ++n and n++ differ ?

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