The shorthand for declaration and initialization in go is
var a, b, c = 1 , 2, 3
Equivalent to following way of declaration and initialization (as per specs)
a:=1
b:=2
c:=3
var a int
var b int
var c int
a=1
b=2
c=3
But I am not getting the answer for the problem found in following code:
package main
import "fmt"
func main() {
var a int = 0
var b int = 1
fmt.Println("init a ",a)
fmt.Println("init b ",b)
a, b = b, a+b
fmt.Println("printing a after `a, b = b, a+b`",a)
fmt.Println("printing b after `a, b = b, a+b`",b)
}
Output should be:
printing a after 'a, b = b, a+b' 1
printing b after 'a, b = b, a+b' 2
Since the value of b is evaluated with a + b i.e 1+1 = 2. But its giving 1.
Here is the playground links of both the working code where you can observe the difference.
a,b = b, a+b
a=b, b=a+b
I know I am missing something to understand, basically how the shorthand expression are evaluated especially when the same variable is involved in the expression.
But where is the proper documentation to refer. Could anyone help on this?
See here
The assignment proceeds in two phases. First, the operands of index
expressions and pointer indirections (including implicit pointer
indirections in selectors) on the left and the expressions on the
right are all evaluated in the usual order. Second, the assignments
are carried out in left-to-right order.
Based on that a+b (0+1) is evaluated first. Then it's assigned. Thus you get the result of a = 1 and b = 1
Related
I am given 2 DFAs. * denotes final states and -> denotes the initial state, defined over the alphabet {a, b}.
1) ->A with a goes to A. -> A with b goes to *B. *B with a goes to *B. *B with b goes to ->A.
The regular expression for this is clearly:
E = a* b(a* + (a* ba* ba*)*)
And the language that it accepts is L1= {w over {a,b} | w is b preceeded by any number of a's followed by any number of a's or w is b preceeded by any number of a's followed by any number of bb with any number of a's in middle of(middle of bb), end or beginning.}
2) ->* A with b goes to ->* A. ->*A with a goes to *B. B with b goes to -> A. *B with a goes to C. C with a goes to C. C with b goes to C.
Note: A is both final and initial state. B is final state.
Now the regular expression that I get for this is:
E = b* ((ab) * + a(b b* a)*)
Finally the language that this DFA accepts is:
L2 = {w over {a, b} | w is n 1's followed by either k 01's or a followed by m 11^r0' s where n,km,r >= 0}
Now the question is, is there a cleaner way to represent the languages L1 and L2 because it does seem ugly. Thanks in advance.
E = a* b(a* + (a* ba* ba*)*)
= a*ba* + a*b(a* ba* ba*)*
= a*ba* + a*b(a*ba*ba*)*a*
= a*b(a*ba*ba*)*a*
= a*b(a*ba*b)*a*
This is the language of all strings of a and b containing an odd number of bs. This might be most compactly denoted symbolically as {w in {a,b}* | #b(w) = 1 (mod 2)}.
For the second one: the only way to get to state B is to see an a in A, and the only way to get to C from outside C is to see an a in B. C is a dead state and the only way to get to it is to see aa starting in A. That is: if you ever see two as in a row, the string is not in the language; the language is the set of all strings over a and b not containing the substring aa. This might be most compactly denoted symbolically as {(a+b)*aa(a+b)*}^c where ^c means "complement".
I'm trying to understand closures, but literally every definition of a closure that I can find uses the same cryptic and vague phrase: "closes over".
What's a closure? "Oh, it's a function that closes over another function."
But nowhere can I find a definition of what "closes over" means. Can someone explain what it means for Thing A to "close over" Thing B?
A closure is a pair consisting of a code pointer and an environment pointer. The environment pointer contains all of the free variables of a given function. For example:
fun f(a, b) =
let fun g(c, d) = a + b + c + d
in g end
val g = f(1, 2)
val result = g(3, 4) (*should be 10*)
The function g contains two free variables: a and b. If you are not familiar with the term free variable, it is a variable that is not defined within the scope of a function. In this context, to close over something, means to remove any occurrences of a free variable from a function. The above example provides good motivation for closures. When the function f returns, we need to be able to remember what the values of a and b are for later. The way this is compiled, is to treat function g as a code pointer and a record containing all the free variables, such as:
fun g(c, d, env) = env.a + env.b + c + d
fun f(a, b, env) = (g, {a = a, b = b})
val (g, gEnv) = f(1, 2)
val result = g(3, 4, gEnv)
When we apply the function g, we supply the environment that was returned when calling function f. Note that now function g no longer has any occurrences of a variable that is not defined in its scope. We typically call a term that doesn't have any free variables as closed. If you are still unclear, Matt Might has an excellent in depth explanation of closure conversion at http://matt.might.net/articles/closure-conversion/
Same example in Javascript
Before closure conversion
function f(a, b){
function g(c, d) {
return a + b + c + d
}
return g
}
var g = f(1, 2)
var result = g(3, 4)
After closure conversion:
function g(c, d, env) {
return env.a + env.b + c + d
}
function f(a, b, env) {
return [g, {"a": a, "b": b}]
}
var [g, gEnv] = f(1, 2)
var result = g(3, 4, gEnv)
From apple documentation
Closures are self-contained blocks of functionality that can be passed
around and used in your code. Closures in Swift are similar to blocks
in C and Objective-C and to lambdas in other programming languages.
But what that means?
It means that a closure captures the variables and constants of the context in which it is defined, referred to as closing over those variables and constants.
I hope that helps!
I read that three variables a,b, and c can be swapped using the single statement below:
c = a ^ b ^ c ^ (a=b) ^ (b=c)
Similarly, two variables a and b can be swapped as:
a = a ^ b ^ (b=a)
Can somebody please explain how does this work?
P.S. Here is the link saying so.
http://p--np.blogspot.ro/2011/04/reverse-linked-list-using-only-2.html
You only need to make in the original statement all assignments (except the one you want to change to) are countered:
a = a ^ b ^ c ^ (b=c) ^ (c=a);
You assign c=a, b=c. The return value of those is a and c respectevely, so, you want to "counter" a and c, and you use the fact that a^a == 0 and c^c == 0, and just add a,c at the beginning.
In addition, you add a b, to assign a to the value of b.
Here is a demo on ideone
Note that as discussed in comments, it is language dependent if the above is guaranteed to succeed or not. In Java, for example - it is: The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right., while in others, such as C, it is not.
Can anyone say how ruby evaluates this:
a = 1
b = 2
a, b = b, a + b
a will be 2 and b will be 3, not 4 as you might expect
It seems that instead of working from left to right it does both sides in parallel somehow?
It is expressed as :-
a = 1
b = 2
a, b = b, (a + b)
a # => 2
b # => 3
This is called parallel assignment. Here all RHS expressions will be evaluated first (left to right). After that assignment will be happened from left to right.
It means, the calculation as follows :
a, b = b, a + b
a, b = 2, (2 + 1)
a, b = 2, 3 # now the real assignment will be happened here.
This is called parallel association, and, like name suggests, it works like all the assignments are done in parallel. You can for example write:
a = 1
b = 2
a, b = b, a
a #=> 2
b #=> 1
a = 1
b = 2
a, b = b, a + b
a
#=> 2
b
#=> 3
Here first rvalue is assigned to first lvalue and the result of second rexp is assigned to second lvalue. These assignments are parallel in nature not sequential.
a, b = b, a is a swap operation using parallel assignments. This makes me think Ruby might be using temporary variables to perform parallel assignments. I invite for corrections here.
I saw the following code: a += b += c - please, does anyone know what does it mean?
a = a + b
b = b + c
or:
b = b + c
a = a + b
It depends on associativity of += operator. Usually it's right to left, so it's the second answer: a+=b is evaluated after b+=c.
It can only be evaluated as:
a += (b += c)
If it would be evaluated as this:
(a += b) += c
then it would evaluate a += b to a numeric value, and then try to change the numeric value, not a variable, which is not possible.
In most high-level languages, expressions like this are evaluated left to right because += is right-to-left associative. In this case, the value of c is added to the value of b, then the value of b (after addition by c) is added to a.
This is equivalent to your second block.