While reading through a C++ book, I came across an expression which was not explained properly (or maybe I just didn't understand the explanation). This is the expression:
c = a+++b;
Which of these does it mean?
c = a + (++b); // 1
c = (a++) + b; // 2
Thanks.
Its interpreted as:
c = a++ + b; //which is same as you're ve written : (a++) + b
Its following the Maximal munch rule.
Related
Does spaces have any meaning in these expressions:
assume:
int a = 1;
int b = 2;
1)
int c = a++ +b;
Or,
2)
int c = a+ ++b;
When I run these two in visual studio, I get different results. Is that the correct behavior, and what does the spec says?
In general, what should be evaluated first, post-increment or pre-increment?
Edit: I should say that
c =a+++b;
Does not compile on visual studio. But I think it should. The postfix++ seems to be evaluated first.
Is that the correct behavior
Yes, it is.
Postfix ++ first returns the current value, then increments it. so int c = a++ +b means compute the value of c as the sum between current a(take the current a value, and only after taking it, increment a) and b;
Prefix ++ first increments the current value, then returns the value already incremented, so in this case, int c = a+ ++b means compute c as the sum between a and the return of the next expression, ++b, which means b is first incremented, then returned.
In general, what should be evaluated first, post-increment or
pre-increment?
In this example, it is not about which gets evaluated first, it is about what each does - postfix first returns the value, then increments it; prefix first increments the value, then returns it.
Hth
Maybe it helps to understand the general architecture of how programs are parsed.
In a nutshell, there are two stages to parsing a program (C++ or others): lexer and parser.
The lexer takes the text input and maps it to a sequence of symbols. This is when spaces are handled because they tell where the symbols are. Spaces really matter at some places (like between int and c, to not confuse with the symbol intc) but not others (like between a and ++ because there is no ambiguity to separate them).
The first example:
int c = a++ +b;
gives the following symbols, each on its own row (implementations may do this in slightly different ways of course):
int
c
=
a
++
+
b
;
While in the other case:
int c = a+ ++b;
the symbols are instead:
int
c
=
a
+
++
b
;
The parser then builds a tree (Abstract Syntax Tree, AST) out of the symbols and according to some grammar. In particular, according to the C++ grammar, + as an addition has a lower precedence than the unary ++ operator (regardless of postfix or prefix). This means that the first example is semantically the same as (a++) + b while the second is like a+ (++b).
For your examples, the ASTs will be different, because the spaces already lead to a different output at the lexer phase.
Note that spaces are not required between ++ and +, so a+++b would theoretically be fine, but this is not recommended for readability. So, some spaces are important for technical reasons while others are important for us users to read the code.
Yes they should be different; the behaviour is correct.
There are a few possible sources for your confusion.
This question is not about "spaces in operators". You have different operators. If you were to remove the space, you would have a different question. See What is i+++ increment in c++
It's also not about "what should be evaluated first, post-increment or pre-increment". It's about understanding the difference between post-increment and pre-increment.
Both increment the variable to which they apply.
But the post-increment expression returns the value from before the increment.
Whereas the pre-increment expression returns the value after the increment.
I.e.
//Given:
int a = 1;
int b = 2;
//Post-increment
int c = a++ +b; =>
1 + 2; (and a == 2) =>
3;
//Pre-increment
int c = a+ ++b; =>
1 + 3; (and b == 3) =>
4;
Another thing that might be causing confusion. You wrote: a++ +b;. And you may be assuming that +b is the unary + operator. This would be an incorrect assumption because you have both left and right operands making that + a binary additive operator (as in x + y).
Final possible confusion. You may be wondering why:
in a++ +b the ++ is a post-increment operator applied to a.
whereas in a+ ++b it's a pre-increment operator applied to b.
The reason is that ++ has higher precedence than the binary additive +. And in both cases it would be impossible to apply ++ to +.
Consider the following snippets:
C++:
#include <iostream>
using namespace std;
int main()
{
int x = 10, y = 20;
y = x + (x=y)*0;
cout << y;
return 0;
}
which gives a result of 20, because the value of y is assigned to x since the bracket is executed first according to the Operator Precedence Table.
VB.NET:
Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
Dim x As Integer = 10
Dim y As Integer = 20
y = x + (x = y) * 0
MsgBox(y)
End Sub
which instead gives a result of 10.
What is the reason for this difference?
What is the order of execution of operators in VB.NET?
Unlike in C++, VB.NET's = is not always an assignment. It can also be the equality comparison operator (== in C++) if it appears inside an expression. Therefore your two expressions are not the same. They are not even equivalent. The VB.NET code does not do what you might think it does.
First to your C++ code: Like you're saying, the assignment x=y happens first; thus your code is roughly equivalent to this: (It seems that was incorrect; see Jens' answer.) Since you end up with y being 20, it is likely that your C++ compiler evaluated your code as if you had written this:
int x = 10, y = 20;
x = y;
y = x + x*0; // which is equivalent to `y = 20 + 20*0;`, or `y = 20 + 0;`, or `y = 20;`
In VB.NET however, because the = in your subexpression (x=y) is not actually interpreted as an assignment, but as a comparison, the code is equivalent to this:
Dim x As Integer = 10
Dim y As Integer = 20
y = 10 + False*0 ' which is equivalent to `y = 10 + 0*0`, or `y = 10` '
Here, operator precedence doesn't even come into play, but an implicit type conversion of the boolean value False to numeric 0.
(Just in case you were wondering: In VB.NET, assignment inside an expression is impossible. Assignments must always be full statements of their own, they cannot happen "inline". Otherwise it would be impossible to tell whether a = inside an expression meant assignment or comparison, since the same operator is used for both.)
Your C++ snippet is undefined behavior. There is no sequence point between using x as the first argument and assigning y to x, so the compiler can evaluate the sub-expressions in any order. Both
First evaluate the left side: y = 10 + (x=y) * 0 -> y = 10 + (x=20) * 0 -> y = 10 + 20*0
First evaluate the right side: y = x + (x=20) * 0 -> y = 20 + 20 * 0
It is also generally a very bad style to put assignments inside expressions.
This answer was intended as a comment, but its length quickly exceeded the limit. Sorry :)
You are confusing operator precedence with evaluation order. (This is a very common phenomenon, so don't feel bad). Let me try to explain with a simpler example involving more familiar operators:
If you have an expression like a + b * c then yes, the multiplication will always happen before the addition, because the * operator binds tighter than + operator. So far so good? The important point is that C++ is allowed to evaluate the operands a, b and c in any order it pleases. If one of those operands has a side effect which affects another operand, this is bad for two reasons:
It may cause undefined behavior (which in your code is indeed the case), and more importantly
It is guaranteed to give future readers of your code serious headaches. So please don't do it!
By the way, Java will always evaluate a first, then b, then c, "despite" the fact that multiplication happens before addition. The pseudo-bytecode will look like push a; push b; push c; mul; add;
(You did not ask about Java, but I wanted to mention Java to give an example where evaluating a is not only feasible, but guaranteed by the language specification. C# behaves the same way here.)
see simple example:
int a = 0;
int b = (a ++ , a + 1); // result of b is UB or well defined ? (c++03).
This was changed in c++11/c++14 ?
The result is well defined and has been since C++98. The comma operator introduces a sequence point (or a "sequenced before" relationship in later C++s) between the the write and the second read of a and I don't see any other potential reasons for undefined behavior.
See the code
#include<iostream.h>
#include<conio.h>
class A{
private:
int i;
public:
A()
{
i=10;
}
A operator++(int)
{
A tmp=*this;
i +=1;
return tmp;
}
display()
{
cout<<i;
}
};
int main()
{
A a,b;
b=a++ + a++;
cout<<endl<<b<<"\t"<<a;
return 0;
}
For the statement b = a++ + a++, the expected value we think will be 20,
But the above statement resulted into 21.
How?
Kindly help me out.
According to cppreference, your code is equals to b = (a++) + (a++)
So, computing it, we have:
a = 10;
tmp1 = a++;//tmp1 = 10, a = 11
tmp2 = a++;//tmp2 = 11, a = 12
b = tmp1 + tmp2 // 10 + 11 = 21
Also remember that constructions like b = a++ + a++; are very poor readable, so you should always use brackets, also it's always a good idea to avoid using increments and decrements like a++ in complex expressions. Readability is much better than showing that you know operator priorities.
As Charles pointed out ++ called on A object are function calls. Thus you first increment i from 10 to 11 and return 10, then in the second call you increment i from 11 to 12 and return 11. The you add 10 and 11 ending up with 21.
The first call increments a to 11 and returns 10. The second call increments a to 12 and returns 11. Sounds like 21 is correct.
That said, the order of evaluation (which ++ is "the first call") is unspecified (thanks JD), so using it twice in the same expression is generally not a good idea.
I shall render my answer in the form of a simple comparison.
Your code is:
b = a++ + a++;
I think you rather confused it with:
b = a + (a++)++;
Overloaded operators are just functions with funny names. They don't neccesarily behave the same as built-in ones. If you're tempted to do the same expression with an object of built-in type - don't, the behaviour would be undefined.
You haven't shown the definition of operator+ for A - I'll asume it's a free function. The expression b = a++ + a++; can then be rewritten as
b = operator+( a.operator++(0), a.operator++(0) );
Hope that helps make things clearer.
The two calls to postfix increment are indeterminately sequenced - that means we can't know which one will be called first. It doesn't matter in your case, since they're both called on the same object, but don't rely on any particular order - it need not be consistent, even during the same execution of the program.
The following code has me confused
int a=2,b=5,c;
c=a+++b;
printf("%d,%d,%d",a,b,c);
I expected the output to be 3,5,8, mainly because a++ means 2 +1 which equals 3, and 3 + 5 equals 8, so I expected 3,5,8. It turns out that the result is 3,5,7. Can someone explain why this is the case?
It's parsed as c = a++ + b, and a++ means post-increment, i.e. increment after taking the value of a to compute a + b == 2 + 5.
Please, never write code like this.
Maximal Munch Rule applies to such expression, according to which, the expression is parsed as:
c = a++ + b;
That is, a is post-incremented (a++) and so the current value of a (before post-increment) is taken for + operation with b.
a++ is post incrementing, i.e. the expression takes the value of a and then adds 1.
c = ++a + b would do what you expect.
This is an example of bad programming style.
It is quite unreadable, however it post increments a so it sums the current value of a to b and afterwards increments a!
a++ gets evaluated after the expression.
c = ++a + b; would give you what you thought.
The post increment operator, a++, changes tge value of a after the value of a is evaluated in the expression. Since the original value of a is 2, that's what's used to compute c; the value of a is changed to reflect the new value after the ++ is evaluated.
a++ + b ..it gives the result 7 and after the expression value of a is update to 3 because of the post increment operator
According to Longest Match rule it is parsed as a++ + +b during lexical analysis phase of compiler. Hence the resultant output.
Here c= a+++b; means c= (a++) +b; i.e post increment.
In a++, changes will occur in the next step in which it is printing a, b and c.
In ++a, i.e prefix-increment the changes will occur in the same step and it will give an output of 8.