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What does the comma operator , do?
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Is comma operator free from side effect?
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Closed 2 years ago.
I am trying to grok the comma operator. The reference says:
In a comma expression E1, E2, the expression E1 is evaluated, its result is discarded (...), and its side effects are completed before evaluation of the expression E2 begins (...).
I am unable to grok the clause - "... the expression E1 is evaluated, its result is discarded (...), and its side effects are completed before...". Specifically, what is retained and what is discarded?
For e.g., in the example from this SO answer:
int x = 0;
int y = some_number;
for(; x < y; ++x, --y)
{
// Do something which uses a converging x and y
}
When we do a ++x, what is the 'result' (that is discarded) and what is the 'side-effect' (that is 'completed' and perhaps 'retained')? Shouldn't the result of ++x be discarded and the value of x remain unchanged? Working example showing incrementation of x is here.
In other words, given a certain expression, how do I deduce if it will be evaluated and its results would be discarded, or if it is a side-effect and its results would perhaps be computed and retained?
In C++ an expression can result in a value and can cause side effects. In the expression ++x, --y you have two sub expressions forming the whole expression. The left, ++x returns x after increment, and the right returns y after decrement. The expression will return the right side of the comma (y) rather than the left side x.
The side effects of the left hand side are preserved, so x is still incremented.
This might make more sense if you were looking to perform assignment.
For example
int x = 1;
int y = 1;
int& z = (++x, --y);
std::cout << z << std::endl;
z becomes a reference to y and thus we will print 0
Reading through the Bjarne's CPP book I've encountered such a statement: "Where logically feasible, the result of an operator that takes an lvalue operand is an lvalue denoting that lvalue operand", and I really can't wrap my head around it. These are the examples included with this statement:
void f(int x, int y)
{
int j = x = y; // the value of x=y is the value of x after the assignment
int∗ p = &++x; // p points to x
int∗ q = &(x++); // error : x++ is not an lvalue (it is not the value stored in x)
int∗ p2 = &(x>y?x:y); // address of the int with the larger value
int& r = (x<y)?x:1; // error : 1 is not an lvalue
}
Code itself makes sense to me, but it makes sense from my personal understanding of how these operators work. But I can't really apply the statement here, for example for the first line. OK, = is an operator which can take both lvalue and rvalue operands (and from my understanding lvalue is implicitly converted to rvalue in this case) , then the result of x = y is an lvalue denoting y? If I follow this correctly, then I could write int* j = x = y, but this would be a compile-time error, as the result of x = y is clearly an rvalue. So I am really confused here.
Can anyone clarify what is this statement is semantically about and how it relates to the given examples step by step?
x = y returns lvalue referring to left-hand operand.
int* j = x = y, this is erroneous because of invalid types of the operands.
But,
int* j = &(x = y) :
works, as x = y returns l-value so this boils down to int *j = &x (value of x is y)
and similarly, this int& j = x = y also works
Additional links:
For conditional expressions: Return type of '?:' (ternary conditional operator)
I am new to c++ and haven't seen such a declaration in programs anywhere. Please help me
For integral type
i >>= 3;
is equal to:
i = i >> 3;
ie bitwise shift variable i right (in his case) in place.
For user defined classes it can be overloaded to whatever functionality author wants this operator to implement
Use of compound assignment operators is encouraged since it evaluates the LHS only once.
From the C++11 Standard:
5.17 Assignment and compound assignment operators
...
7 The behavior of an expression of the form E1 op = E2 is equivalent to E1 = E1 op E2 except that E1 is evaluated only once.
If you have a function that returns a reference to an object, it is more efficient and less error prone when you use a compound assignment operator.
E.g.
std::vector<int> v(10, 1);
v[4] <<= 2;
is better than
std::vector<int> v(10);
v[4] = v[4] << 2;
those are called compound assignment operators. There are almost 10 of them in C/C++ language. ex += -= *= <<= >>=. When one of the operand is same as the variable to which final result to be assigned is same, in a binary operation this can be used as a short hand syntax. Ex a=a+b can be written as a+=b. in the same a=a<<2 can be written as a<<=2. This type of syntactic sugar is also supported by other languages too.
Left shift assignment <<=
x <<= y Shift x left by y bits
Right shift assignment >>=
x >>= y Shift x right by y bits
Read This to know more about bitwise operators in c++.
In c in general, an operator with an equals following it is shorthand for "do the operation, and put the result back in the same variable."
x = x >> 4; // shift x right by 4 bits
x >>= 4; // exactly the same as above
x <<= 3; // shift x left by 3 bits
This assumes that x is a numeric type like long, unsigned short, etc.
If x is not numeric, then the original programmer probably defined the operation as an overloaded operator in c++, for example, lopping off n characters from a string.
Recently I came across this piece of code. I don't know why I never saw this kind of syntax in all my "coding life".
int main()
{
int b;
int a = (b=5, b + 5);
std::cout << a << std::endl;
}
a has value of 10. What exactly is this way of initialization called? How does it work?
This statement:
int a = (b=5, b + 5);
Makes use of the comma operator. Per Paragraph 5.18/1 of the C++11 Standard:
[...] A pair of expressions separated by a comma is evaluated left-to-right; the left expression is a discarded value
expression (Clause 5).83 Every value computation and side effect associated with the left expression
is sequenced before every value computation and side effect associated with the right expression. The type
and value of the result are the type and value of the right operand; the result is of the same value category
as its right operand, and is a bit-field if its right operand is a glvalue and a bit-field. If the value of the right
operand is a temporary (12.2), the result is that temporary.
Therefore, your statement is equivalent to:
b = 5;
int a = b + 5;
Personally, I do not see a reason for using the comma operator here. Just initialize your variable the easily readable way, unless you have a good reason for doing otherwise.
operator , evaluates arguments one after another and return the last value
It may be used not only in initialization
The comma , operator allows you to separate expressions. The compount statement made by
exp1, exp2, ..., expn
evaluates to expn.
So what happens is that first b is set to 5 and then a is set to b + 5.
A side note: since , has the lowest precedence in the table of operators the semantics of
int a = b = 5, b+5;
is different from
int a = (b = 5, b+5);
because the first is parsed as (int a = b = 5), b + 5
When used in an expression the comma operator will evaluate all of its operands (left-to-right) and return the last.
The initialization is called copy initialization. If you ignore the complex expression on the right, it's the same as in:
int a = 10;
This is to be contrasted with direct initialization, which looks like this:
int a(10);
(It's possible that you were separately confused about how to evalue a comma expression. Please indicate if that's the case.)
I don't understand the concept of postfix and prefix increment or decrement. Can anyone give a better explanation?
All four answers so far are incorrect, in that they assert a specific order of events.
Believing that "urban legend" has led many a novice (and professional) astray, to wit, the endless stream of questions about Undefined Behavior in expressions.
So.
For the built-in C++ prefix operator,
++x
increments x and produces (as the expression's result) x as an lvalue, while
x++
increments x and produces (as the expression's result) the original value of x.
In particular, for x++ there is no no time ordering implied for the increment and production of original value of x. The compiler is free to emit machine code that produces the original value of x, e.g. it might be present in some register, and that delays the increment until the end of the expression (next sequence point).
Folks who incorrectly believe the increment must come first, and they are many, often conclude from that certain expressions must have well defined effect, when they actually have Undefined Behavior.
int i, x;
i = 2;
x = ++i;
// now i = 3, x = 3
i = 2;
x = i++;
// now i = 3, x = 2
'Post' means after - that is, the increment is done after the variable is read. 'Pre' means before - so the variable value is incremented first, then used in the expression.
The difference between the postfix increment, x++, and the prefix increment, ++x, is precisely in how the two operators evaluate their operands. The postfix increment conceptually copies the operand in memory, increments the original operand and finally yields the value of the copy. I think this is best illustrated by implementing the operator in code:
int operator ++ (int& n) // postfix increment
{
int tmp = n;
n = n + 1;
return tmp;
}
The above code will not compile because you can't re-define operators for primitive types. The compiler also can't tell here we're defining a postfix operator rather than prefix, but let's pretend this is correct and valid C++. You can see that the postfix operator indeed acts on its operand, but it returns the old value prior to the increment, so the result of the expression x++ is the value prior to the increment. x, however, is incremented.
The prefix increment increments its operand as well, but it yields the value of the operand after the increment:
int& operator ++ (int& n)
{
n = n + 1;
return n;
}
This means that the expression ++x evaluates to the value of x after the increment.
It's easy to think that the expression ++x is therefore equivalent to the assignmnet (x=x+1). This is not precisely so, however, because an increment is an operation that can mean different things in different contexts. In the case of a simple primitive integer, indeed ++x is substitutable for (x=x+1). But in the case of a class-type, such as an iterator of a linked list, a prefix increment of the iterator most definitely does not mean "adding one to the object".
No one has answered the question:
Why is this concept confusing?
As an undergrad Computer Science major it took me awhile to understand this because of the way I read the code.
The following is not correct!
x = y++
X is equal to y post increment. Which would logically seem to mean X is equal to the value of Y after the increment operation is done. Post meaning after.
or
x = ++y
X is equal to y pre-increment. Which would logically seem to mean X is equal to the value of Y before the increment operation is done. Pre meaning before.
The way it works is actually the opposite. This concept is confusing because the language is misleading. In this case we cannot use the words to define the behavior.
x=++y is actually read as X is equal to the value of Y after the increment.
x=y++ is actually read as X is equal to the value of Y before the increment.
The words pre and post are backwards with respect to semantics of English. They only mean where the ++ is in relation Y. Nothing more.
Personally, if I had the choice I would switch the meanings of ++y and y++. This is just an example of a idiom that I had to learn.
If there is a method to this madness I'd like to know in simple terms.
Thanks for reading.
It's pretty simple. Both will increment the value of a variable. The following two lines are equal:
x++;
++x;
The difference is if you are using the value of a variable being incremented:
x = y++;
x = ++y;
Here, both lines increment the value of y by one. However, the first one assigns the value of y before the increment to x, and the second one assigns the value of y after the increment to x.
So there's only a difference when the increment is also being used as an expression. The post-increment increments after returning the value. The pre-increment increments before.
int i = 1;
int j = 1;
int k = i++; // post increment
int l = ++j; // pre increment
std::cout << k; // prints 1
std::cout << l; // prints 2
Post increment implies the value i is incremented after it has been assigned to k. However, pre increment implies the value j is incremented before it is assigned to l.
The same applies for decrement.
Post-increment:
int x, y, z;
x = 1;
y = x++; //this means: y is assigned the x value first, then increase the value of x by 1. Thus y is 1;
z = x; //the value of x in this line and the rest is 2 because it was increased by 1 in the above line. Thus z is 2.
Pre-increment:
int x, y, z;
x = 1;
y = ++x; //this means: increase the value of x by 1 first, then assign the value of x to y. The value of x in this line and the rest is 2. Thus y is 2.
z = x; //the value of x in this line is 2 as stated above. Thus z is 2.
Since we now have inline javascript snippets I might as well add an interactive example of pre and pos increment. It's not C++ but the concept stays the same.
let A = 1;
let B = 1;
console.log('A++ === 2', A++ === 2);
console.log('++B === 2', ++B === 2);
From the C99 standard (C++ should be the same, barring strange overloading)
6.5.2.4 Postfix increment and decrement operators
Constraints
1 The operand of the postfix increment
or decrement operator shall have
qualified or unqualified real or
pointer type and shall be a modifiable
lvalue.
Semantics
2 The result of the postfix ++
operator is the value of the operand.
After the result is obtained, the
value of the operand is incremented.
(That is, the value 1 of the
appropriate type is added to it.) See
the discussions of additive operators
and compound assignment for
information on constraints, types, and
conversions and the effects of
operations on pointers. The side
effect of updating the stored value of
the operand shall occur between the
previous and the next sequence point.
3 The postfix -- operator is analogous
to the postfix ++ operator, except
that the value of the operand is
decremented (that is, the value 1 of
the appropriate type is subtracted
from it).
6.5.3.1 Prefix increment and decrement operators
Constraints
1 The operand of the prefix increment
or decrement operator shall have
qualified or unqualified real or
pointer type and shall be a modifiable
lvalue.
Semantics
2 The value of the operand of the
prefix ++ operator is incremented. The
result is the new value of the operand
after incrementation. The expression
++E is equivalent to (E+=1). See the discussions of additive operators and
compound assignment for information on
constraints, types, side effects, and
conversions and the effects of
operations on pointers.
3 The prefix -- operator is analogous
to the prefix ++ operator, except that
the value of the operand is
decremented.
Post increment(a++)
If
int b = a++,then this means
int b = a;
a = a+1;
Here we add 1 to the value. The value is returned before the increment is made,
For eg a = 1; b = a++;
Then b=1 and a=2
Pre-increment (++a)
If int b = ++a; then this means
a=a+1;
int b=a ;
Pre-increment: This will add 1 to the main value. The value will be returned after the increment is made, For a = 1; b = ++a;
Then b=2 and a=2.
Already good answers here, but as usual there seems to be some general lack of clarity in simply remembering which way round these work. I suppose this arises because semantically resolving the nomenclature is not entirely straightforward. For example, you may be aware that "pre-" means "before". But does the pre-increment ++i return the value of i before the increment, or does it increment i before returning a value?
I find it much easier to visually follow the expression through from left to right:
++ i
-------------------------------------------------->
Increment i Then supply the value of i
i ++
-------------------------------------------------->
Supply the value of i Then increment i
Of course, as Alf points out in the accepted answer, this may not reflect when the 'real i' is updated, but it is a convenient way of thinking about what gets supplied to the expression.
#include<stdio.h>
void main(){
char arr[] ="abcd";
char *p=arr,*q=arr;
char k,temp;
temp = *p++; /* here first it assigns value present in address which
is hold by p and then p points to next address.*/
k = ++*q;/*here increments the value present in address which is
hold by q and assigns to k and also stores the incremented value in the same
address location. that why *q will get 'h'.*/
printf("k is %c\n",k); //output: k is h
printf("temp is %c\n",temp);//output: temp is g
printf("*p is %c\n",*p);//output: *p is e
printf("*q is %c",*q);//output: *q is h
}
Post and Pre Increment with Pointers
The pre increment is before increment value ++ e.g.:
(++v) or 1 + v
The post increment is after increment the value ++ e.g.:
(rmv++) or rmv + 1
Program:
int rmv = 10, vivek = 10;
cout << "rmv++ = " << rmv++ << endl; // the value is 10
cout << "++vivek = " << ++vivek; // the value is 11
You should also be aware that the behaviour of postincrement/decrement operators is different in C/C++ and Java.
Given
int a=1;
in C/C++ the expression
a++ + a++ + a++
evaluates to 3, while in Java it evaluates to 6. Guess why...
This example is even more confusing:
cout << a++ + a++ + a++ << "<->" << a++ + a++ ;
prints 9<->2 !! This is because the above expression is equivalent to:
operator<<(
operator<<(
operator<<( cout, a++ + a++ ),
"<->"
),
a++ + a++ + a++
)