difference between a += b and a = a + b

Question

In Java, I really wonder is there a difference between using a += b; or a = a + b; . Which one should I use principally? I know that first one is something like shortcut but does the compiler get those two indications differently?

Solution 1

See the Java language specification, 15.26.2 Compound assignment operators

To quote the relevant parts:

A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T)((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

short x = 3;
x += 4.6;

and results in x having the value 7 because it is equivalent to:

short x = 3;
x = (short)(x + 4.6);

So it is more than syntactic sugar, as

int x = 1;
long a = 2l;
x+= a;

compiles, where

int x =1;
long a =2l;
x = x+a;

gives you a compile error, as was discussed here on StackOverflow quite recently

Solution 2

it does depend on the language, but in c# it is very slightly more efficient to use a += b;.

a is only evaluated once.

in a = a + b, a is evaluated twice.

http://msdn.microsoft.com/en-us/library/sa7629ew.aspx

Solution 3

Just syntactic sugar in most languages that I know that would include c, c++, C#, java, javascript..

notable difference noted by Cicada in regards to c++:

On numeric types (int and friends) there is no difference. On user-defined classes there may be a difference. A notable one would be D3DXVECTOR3 from DirectX, for example. Using + would construct a temporary object while += would not. The latter is about 30% faster.

Solution 4

In most languages that support this notation, a = a + b is the same as a += b, but it's not always the case.

Here is an example in Python (using Numpy):

>>> import numpy as np
>>> a = np.array([1])
>>> b = np.array([2])
>>> c = a
>>> a = a + b
>>> print a
[3]
>>> print c
[1]

Here a = a + b creates a new array for a + b and stores it into a. c, which was using the same reference as the initial a still holds the initial array (with value 1).

>>> a = np.array([1])
>>> b = np.array([2])
>>> c = a
>>> a += b
>>> print a
[3]
>>> print c
[3]

Here a += b re-uses the initial array a. As a result, since both a and c refer to the same array, both a and c are modified.

Solution 5

In addition to everything said above, you can also use the following shortcuts:

Operator (+=)

x+=y;

same as:

x=x+y;

Operator (-=)

x-=y;

same as:

x=x-y;

Operator (*=)

x*=y;

same as:

x=x*y;

Operator (/=)

x/=y;

same as:

x=x/y;

Operator (%=)

x%=y;

same as :

x=x%y;

Operator (&=)

x&=y;

same as :

x=x&y;

Operator (|=)

x|=y;

same as :

x=x|y;

Operator (^=)

x^=y; 

same as :

x=x^y;

Operator (>>=)

x>>=y;

same as

result=x>>y;

The same operation to operator (<<=) and operator (>>>=) .

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Comments

[Mark Rolich]

Please specify language.

[Bruno]

Which language? In which context? Depending on the language, it may also depend on the type of a and b.

[Hovercraft Full Of Eels]

look up "sytactic sugar".

[Mysticial]

See: stackoverflow.com/questions/8710619/java-operator

[Jeremy D]

The a+=b thing is useful to make more concise instruction but not necessarily the easiest to understand! That depends only of you I guess.

[El3ctr0n1c4]

@Mysticial I've just checked the link. "wow". those were really different. Thanks

[user703016]

It's not syntactic sugar in C++.

[Rogel Garcia]

Can an overridden operator in C++ be different? That is, override + in one way and += other way?

[Ry-]

@rogelware: Yes. (And I wish it couldn't be.)

[Bassam Mehanni]

@Cicada it has been a while since I used C++, could you please tell me what the difference is between a += b; and a = a + b;?

[Bassam Mehanni]

yes if it's overridden but in its default state, isn't it essentially the same, or am I missing something?

[user703016]

@BassamMehanni On numeric types (int and friends) there is no difference. On user-defined classes there may be a difference. A notable one would be D3DXVECTOR3 from DirectX, for example. Using + would construct a temporary object while += would not. The latter is about 30% faster.

[Bassam Mehanni]

@Cicada Thanks, do you mind if I added this to the answer for completeness?

[user703016]

@BassamMehanni Feel free to do so.

[Hovercraft Full Of Eels]

How is this different from the link provided previously by Mystical in the comments to the original question?

[TacticalCoder]

@HovercraftFullOfEels: I'd say it's different in that this answer specifies, before any edit, that there's more to it than syntactic sugar as has been discussed extensively here a few days ago. And he did provide the link to one these recent questions.

[Rob W]

See the answer below ( stackoverflow.com/questions/8773349/… ) for the explanation.