getting an error when I try to run this line of code and I can't figure out why
let validCol column value : bool =
for i in 0..8 do
if sudokuBoard.[i,column] = value then
false
else true
As Tyler Hartwig says a for loop cannot return a value except unit.
On the other hand, inside a list comprehension or a seq Computation Expression you can use for to yield the values and then test if the one you are looking for exists:
let validCol column value : bool =
seq { for i in 0..8 do yield sudokuBoard.[i,column] }
|> Seq.exists value
|> not
In F#, the last call made is what is returned, you have explicitly declared you are returning a bool.
The for loop is unable to return or aggregate multiple values, bun instead, returns unit.
let validCol column value : bool =
for i in 0..8 do
if sudokuBoard.[i,column] = value then
false
else
true
Here, you'll need to figure out how to aggregate all the bool to get your final result. I'm not quite sure what this is supposed to return, or I'd give an example.
It looks like you are looking for a short-cut out of the loop like in C# you can use continue, break or return to exit a loop.
In F# the way to accomplish that with performance is to use tail-recursion. You could achieve it with while loops but that requires mutable variables which tail-recursion doesn't need (although we sometimes uses it).
A tail-recursive function is one that calls itself at the very end and doesn't look at the result:
So this is tail-recursive
let rec loop acc i = if i > 0 then loop (acc + i) (i - 1) else acc
Where this isn't
let rec loop fib i = if i < 1 then 1 else fib (i - 1) + fib (i - 2)
If F# compiler determines a function is tail-recursive the compiler applies tail-recursion optimization (TCO) on the function, basically it unrolls it into an efficient for loop that looks a lot like the loop would like in C#.
So here is one way to write validCol using tail-recursion:
let validCol column value : bool =
// loops is tail-recursive
let rec loop column value i =
if i < 9 then
if sudokuBoard.[i,column] = value then
false // The value already exists in the column, not valid
else
loop column value (i + 1) // Check next row.
else
true // Reach the end, the value is valid
loop column value 0
Unfortunately; F# compiler doesn't have an attribute to force TCO (like Scala or kotlin does) and therefore if you make a slight mistake you might end up with a function that isn't TCO. I think I saw GitHub issue about adding such an attribute.
PS. seq comprehensions are nice in many cases but for a sudoku solver I assume you are looking for something that is as fast as possible. seq comprehensions (and LINQ) I think adds too much overhead for a sudoku solver whereas tail-recursion is about as quick as you can get in F#.
PS. In .NET 2D arrays are slower than 1D arrays, just FYI. Unsure if it has improved with dotnet core.
What are the benefits and drawbacks of the ?: operator as opposed to the standard if-else statement. The obvious ones being:
Conditional ?: Operator
Shorter and more concise when dealing with direct value comparisons and assignments
Doesn't seem to be as flexible as the if/else construct
Standard If/Else
Can be applied to more situations (such as function calls)
Often are unnecessarily long
Readability seems to vary for each depending on the statement. For a little while after first being exposed to the ?: operator, it took me some time to digest exactly how it worked. Would you recommend using it wherever possible, or sticking to if/else given that I work with many non-programmers?
I would basically recommend using it only when the resulting statement is extremely short and represents a significant increase in conciseness over the if/else equivalent without sacrificing readability.
Good example:
int result = Check() ? 1 : 0;
Bad example:
int result = FirstCheck() ? 1 : SecondCheck() ? 1 : ThirdCheck() ? 1 : 0;
This is pretty much covered by the other answers, but "it's an expression" doesn't really explain why that is so useful...
In languages like C++ and C#, you can define local readonly fields (within a method body) using them. This is not possible with a conventional if/then statement because the value of a readonly field has to be assigned within that single statement:
readonly int speed = (shiftKeyDown) ? 10 : 1;
is not the same as:
readonly int speed;
if (shifKeyDown)
speed = 10; // error - can't assign to a readonly
else
speed = 1; // error
In a similar way you can embed a tertiary expression in other code. As well as making the source code more compact (and in some cases more readable as a result) it can also make the generated machine code more compact and efficient:
MoveCar((shiftKeyDown) ? 10 : 1);
...may generate less code than having to call the same method twice:
if (shiftKeyDown)
MoveCar(10);
else
MoveCar(1);
Of course, it's also a more convenient and concise form (less typing, less repetition, and can reduce the chance of errors if you have to duplicate chunks of code in an if/else). In clean "common pattern" cases like this:
object thing = (reference == null) ? null : reference.Thing;
... it is simply faster to read/parse/understand (once you're used to it) than the long-winded if/else equivalent, so it can help you to 'grok' code faster.
Of course, just because it is useful does not mean it is the best thing to use in every case. I'd advise only using it for short bits of code where the meaning is clear (or made more clear) by using ?: - if you use it in more complex code, or nest ternary operators within each other it can make code horribly difficult to read.
I usually choose a ternary operator when I'd have a lot of duplicate code otherwise.
if (a > 0)
answer = compute(a, b, c, d, e);
else
answer = compute(-a, b, c, d, e);
With a ternary operator, this could be accomplished with the following.
answer = compute(a > 0 ? a : -a, b, c, d, e);
I find it particularly helpful when doing web development if I want to set a variable to a value sent in the request if it is defined or to some default value if it is not.
A really cool usage is:
x = foo ? 1 :
bar ? 2 :
baz ? 3 :
4;
Sometimes it can make the assignment of a bool value easier to read at first glance:
// With
button.IsEnabled = someControl.HasError ? false : true;
// Without
button.IsEnabled = !someControl.HasError;
I'd recommend limiting the use of the ternary(?:) operator to simple single line assignment if/else logic. Something resembling this pattern:
if(<boolCondition>) {
<variable> = <value>;
}
else {
<variable> = <anotherValue>;
}
Could be easily converted to:
<variable> = <boolCondition> ? <value> : <anotherValue>;
I would avoid using the ternary operator in situations that require if/else if/else, nested if/else, or if/else branch logic that results in the evaluation of multiple lines. Applying the ternary operator in these situations would likely result in unreadable, confusing, and unmanageable code. Hope this helps.
The conditional operator is great for short conditions, like this:
varA = boolB ? valC : valD;
I use it occasionally because it takes less time to write something that way... unfortunately, this branching can sometimes be missed by another developer browsing over your code. Plus, code isn't usually that short, so I usually help readability by putting the ? and : on separate lines, like this:
doSomeStuffToSomething(shouldSomethingBeDone()
? getTheThingThatNeedsStuffDone()
: getTheOtherThingThatNeedsStuffDone());
However, the big advantage to using if/else blocks (and why I prefer them) is that it's easier to come in later and add some additional logic to the branch,
if (shouldSomethingBeDone()) {
doSomeStuffToSomething(getTheThingThatNeedsStuffDone());
doSomeAdditionalStuff();
} else {
doSomeStuffToSomething(getTheOtherThingThatNeedsStuffDone());
}
or add another condition:
if (shouldSomethingBeDone()) {
doSomeStuffToSomething(getTheThingThatNeedsStuffDone());
doSomeAdditionalStuff();
} else if (shouldThisOtherThingBeDone()){
doSomeStuffToSomething(getTheOtherThingThatNeedsStuffDone());
}
So, in the end, it's about convenience for you now (shorter to use :?) vs. convenience for you (and others) later. It's a judgment call... but like all other code-formatting issues, the only real rule is to be consistent, and be visually courteous to those who have to maintain (or grade!) your code.
(all code eye-compiled)
One thing to recognize when using the ternary operator that it is an expression not a statement.
In functional languages like scheme the distinction doesn't exists:
(if (> a b) a b)
Conditional ?: Operator
"Doesn't seem to be as flexible as the if/else construct"
In functional languages it is.
When programming in imperative languages I apply the ternary operator in situations where I typically would use expressions (assignment, conditional statements, etc).
While the above answers are valid, and I agree with readability being important, there are 2 further points to consider:
In C#6, you can have expression-bodied methods.
This makes it particularly concise to use the ternary:
string GetDrink(DayOfWeek day)
=> day == DayOfWeek.Friday
? "Beer" : "Tea";
Behaviour differs when it comes to implicit type conversion.
If you have types T1 and T2 that can both be implicitly converted to T, then the below does not work:
T GetT() => true ? new T1() : new T2();
(because the compiler tries to determine the type of the ternary expression, and there is no conversion between T1 and T2.)
On the other hand, the if/else version below does work:
T GetT()
{
if (true) return new T1();
return new T2();
}
because T1 is converted to T and so is T2
If I'm setting a value and I know it will always be one line of code to do so, I typically use the ternary (conditional) operator. If there's a chance my code and logic will change in the future, I use an if/else as it's more clear to other programmers.
Of further interest to you may be the ?? operator.
The advantage of the conditional operator is that it is an operator. In other words, it returns a value. Since if is a statement, it cannot return a value.
There is some performance benefit of using the the ? operator in eg. MS Visual C++, but this is a really a compiler specific thing. The compiler can actually optimize out the conditional branch in some cases.
The scenario I most find myself using it is for defaulting values and especially in returns
return someIndex < maxIndex ? someIndex : maxIndex;
Those are really the only places I find it nice, but for them I do.
Though if you're looking for a boolean this might sometimes look like an appropriate thing to do:
bool hey = whatever < whatever_else ? true : false;
Because it's so easy to read and understand, but that idea should always be tossed for the more obvious:
bool hey = (whatever < whatever_else);
If you need multiple branches on the same condition, use an if:
if (A == 6)
f(1, 2, 3);
else
f(4, 5, 6);
If you need multiple branches with different conditions, then if statement count would snowball, you'll want to use the ternary:
f( (A == 6)? 1: 4, (B == 6)? 2: 5, (C == 6)? 3: 6 );
Also, you can use the ternary operator in initialization.
const int i = (A == 6)? 1 : 4;
Doing that with if is very messy:
int i_temp;
if (A == 6)
i_temp = 1;
else
i_temp = 4;
const int i = i_temp;
You can't put the initialization inside the if/else, because it changes the scope. But references and const variables can only be bound at initialization.
The ternary operator can be included within an rvalue, whereas an if-then-else cannot; on the other hand, an if-then-else can execute loops and other statements, whereas the ternary operator can only execute (possibly void) rvalues.
On a related note, the && and || operators allow some execution patterns which are harder to implement with if-then-else. For example, if one has several functions to call and wishes to execute a piece of code if any of them fail, it can be done nicely using the && operator. Doing it without that operator will either require redundant code, a goto, or an extra flag variable.
With C# 7, you can use the new ref locals feature to simplify the conditional assignment of ref-compatible variables. So now, not only can you do:
int i = 0;
T b = default(T), c = default(T);
// initialization of C#7 'ref-local' variable using a conditional r-value⁽¹⁾
ref T a = ref (i == 0 ? ref b : ref c);
...but also the extremely wonderful:
// assignment of l-value⁽²⁾ conditioned by C#7 'ref-locals'
(i == 0 ? ref b : ref c) = a;
That line of code assigns the value of a to either b or c, depending on the value of i.
Notes
1. r-value is the right-hand side of an assignment, the value that gets assigned.
2. l-value is the left-hand side of an assignment, the variable that receives the assigned value.
I am trying to verify my code in Dafny and I encountered a problem:
I have a method that is iterating over a sequence and changes it. The method changes the sequence according to the elements in the sequence. I would like to add a post condition like this: "if the elements in the sequence are X then something should happen". The problem is that the method changes the set (adds element etc.) and I want to check the condition of the original sequence. Is there an elegant way of doing that in Dafny? (The only way I could think of right now is keeping global var of the original condition of the sequence, but I am looking for the right way of doing that).
Code example:
method changeSeq(p: class1, s: seq<class1>)
ensures |s| == 10 ==> p in s
{
if (|s| == 10){
s := s + [p];
}
}
In the code, I want the post condition to check original s stat, and not its stat after we changed it.
you can use old for old value of a variable like s == old(s).
Here is one example: http://rise4fun.com/Dafny/fhQgD
From Dafny Documentation 22.18. Old Expressions
OldExpression_ = "old" "(" Expression(allowLemma: true, allowLambda: true) ")"
An old expression is used in postconditions. old(e) evaluates to the value expression e had on entry to the current method. Note that old only affects heap dereferences, like o.f and a[i]. In particular, old has no effect on the value returned for local variables or out-parameters.
i am reading a c++ source code,
converting infix to postfix
i am using turbo c++
#include <stdio.h>
typedef struct node
{
float data;
struct node *next;
} stack;
void StackInitiate(stack **head)
{
//error
if(*head=(stack *)malloc(sizeof(stack))==NULL)
exit(1);
(*head)->next=NULL;
}
// I'm getting .. cannot convert 'int' to 'node *' ...
can anybody tell me why so . and how to solve it regards.
full source code here
Because of operator precedence the expression
*head=(stack *)malloc(sizeof(stack))==NULL
is actually equivalent to
*head=((stack *)malloc(sizeof(stack))==NULL)
That is, you assign the value of the comparison to *head.
You need to put in your own parentheses to make it correct:
(*head=(stack *)malloc(sizeof(stack)))==NULL
Or better yet use the new operator which is what you rreally should use to allocate object dynamically in C++:
(*head=new stack)==NULL
Because of operator precedence rules, the following:
if(*head=(stack *)malloc(sizeof(stack))==NULL)
is parsed as
if(*head=((stack *)malloc(sizeof(stack))==NULL))
which is assigning the result of the == NULL comparison to *head.
To compare the new value of *head to null instead, use parentheses to change the precedence:
if((*head=(stack *)malloc(sizeof(stack)))==NULL)
Or even better, separate the assignment from the if condition:
*head = (stack *) malloc(sizeof(stack));
if(*head == NULL)
This convention is more readable and would also avoid similar bugs happening in the future.
Consider below struct:
typedef struct _Index {
NSInteger category;
NSInteger item;
} Index;
If I use this struct as a property:
#property (nonatomic, assign) Index aIndex;
When I access it without any initialization right after a view controller alloc init, LLDB print it as:
(lldb) po vc.aIndex
(category = 0, item = 0)
(lldb) po &_aIndex
0x000000014e2bcf70
I am a little confused, the struct already has valid memory address, even before I want to allocate one. Does Objective-C initialize struct automatically? If it is a NSObject, I have to do alloc init to get a valid object, but for C struct, I get a valid struct even before I tried to initialize it.
Could somebody explains, and is it ok like this, not manually initializing it?
To answer the subquestion, why you cannot assign to a structure component returned from a getter:
(As a motivation this is, because I have read this Q several times.)
A. This has nothing to do with Cbjective-C. It is a behavior stated in the C standard. You can check it for simple C code:
NSMakeSize( 1.0, 2.0 ).width = 3.0; // Error
B. No, it is not an improvement of the compiler. If it would be so, a warning would be the result, not an error. A compiler developer does not have the liberty to decide what an error is. (There are some cases, in which they have the liberty, but this are explicitly mentioned.)
C. The reason for this error is quite easy:
An assignment to the expression
NSMakeSize( 1.0, 2.0 ).width
would be legal, if that expression is a l-value. A . operator's result is an l-value, if the structure is an l-value:
A postfix expression followed by the . operator and an identifier designates a member of a structure or union object. The value is that of the named member,82) and is an lvalue if the first expression is an lvalue.
ISO/IEC 9899:TC3, 6.5.2.3
Therefore it would be assignable, if the expression
NSMakeSize( 1.0, 2.0 )
is an l-value. It is not. The reason is a little bit more complex. To understand that you have to know the links between ., -> and &:
In contrast to ., -> always is an l-value.
A postfix expression followed by the -> operator and an identifier designates a member of a structure or union object. The value is that of the named member of the object to which the first expression points, and is an lvalue. 83)
Therefore - that is what footnote 83 explains – ->, &, and . has a link:
If you can calculate the address of a structure S having a component C with the & operator, the expression (&S)->C is equivalent to S.C. This requires that you can calculate the address of S. But you can never do that with a return value, even it is a simple integer …
int f(void)
{
return 1;
}
f()=5; // Error
… or a pointer …
int *f(void)
{
return NULL;
}
f()=NULL; // Error
You always get the same error: It is not assignable. Because it is a r-value. This is obvious, because it is not clear,
a) whether the way the compiler returns a value, esp. whether he does it in address space.
b) when the time the life time of the returned value is over
Going back to the structure that means that the return value is a r-value. Therefore the result of the . operator on that is a r-value. You are not allowed to assign a value to a r-value.
D. The solution
There is a solution to assign to a "returned structure". One might decide, whether it is good or not. Since -> always is an l-value, you can return a pointer to the structure. Dereferencing this pointer with the -> operator has always an l-value as result, so you can assign a value to it:
// obj.aIndex returns a pointer
obj.aIndex->category = 1;
You do not need #public for that. (What really is a bad idea.)
The semantics of the property are to copy the struct, so it doesn't need to be allocated and initialized like an Objective-C object would. It's given its own space like a primitive type is.
You will need to be careful updating it, as this won't work:
obj.aIndex.category = 1;
Instead you will need to do this:
Index index = obj.aIndex;
index.category = 1;
obj.aIndex = index;
This is because the property getter will return a copy of the struct and not a reference to it (the first snippet is like the second snippet, without the last line that assigns the copy back to the object).
So you might be better off making it a first class object, depending on how it will be used.