if(a .feq. 5.0_dp) then **** if(a .fne. 5.2_dp) then ***
I come across some codes like this. What does the .feq. or .fne. mean? Is it "=" or "\ =" ?
In Fortran, operators (unary or binary) can take this form, a string of letters (up to 63) with a . at either end. So .feq. and .fne. are operators.
We'll also see operators such as .not., .eq. and so on.
Some operators, such as the two just mentioned, are standard intrinsic operators, some may be non-standard intrinsic operators, and we can even have user defined operators.
.feq. and .fne. are not (Fortran 2018) standard intrinsic operators. They may be non-standard intrinsic operators, but most likely they are user-defined. As they are not standard operators, we cannot say what they do (although as veryreverie comments, we can make reasonable guesses).
You will need to read the documentation for the project (or compiler, for the case of non-standard intrinsic operators), or you can look at the available source code.
How will you find what a user-defined operator does? For .feq. for example you should find an interface block with the OPERATOR(...) syntax:
interface operator (.feq.)
...
end interface operator (.feq.)
Inside that interface block you'll find mention of one or more specific functions, much as you would with other generic functions. Check these functions until you find one with the right number of arguments (one for a unary operator, two for a binary) of the right type (first argument matching the one after the .feq. if it's unary; or to the left if it's binary, with the second argument the right). You can then see what this function does.
You may also find your IDE or other tools will tell you how the operator is resolved.
Related
The operator '&' can be used in both of following way int a; scanf("%d",&a);
and printf("%d",1&2).
But different behaviour (for first as address operator and second time as bit-wise operator).
I know operator overloading is not there in C. Then how it works ?. Also highlight for c++.
I know operator overloading is not there in C.
This is incorrect. a + b performs integer addition if a and b are integers, floating-point addition of a and b are floating-point numbers, and pointer arithmetic if a or b is a pointer.
C has operator overloading built into the language. It does not support custom operator overloading defined by the program.
In the case of & being an operator for taking an address and for performing a bitwise AND, the distinction is made by the language grammar. The & for taking an address can appear only applied to a cast-expression in the grammar. The & for bitwise AND can appear only after an AND-expression and before an equality-expression. These “tokens” (cast-expression, AND-expression, and equality-expression) may be unfamiliar to you, but they are formally defined in the grammar for the C language, and, as the compiler is parsing source code, it recognizes the structure of expressions and matches the source code to the tokens of the grammar. This is also true for C++ except for a minor technical difference: In C++, one of the tokens is and-expression instead of AND-expression.
The definition of the grammar is such that recognition of these tokens always uniquely distinguishes how the & operator is being used.
In "C" language, operators have different meaning when they are used as prefix to expression, suffix to expression or "infix" (between two expressions).
Consider '*', which performs multiplication as 'infix' operator, and pointer indirection when used as a prefix. Similarily, the '-' operator, which performs subtraction as 'infix' operator, and negation when used as a prefix.
Basically, it's not about overriding, it if the operator appears between two expressions, or as a prefix to a single expression.
In the same way, The "C" compiler knows if the '&' is bit-wise and, or address-of, based on it's position is the expression: If it is between two expressions, it's the AND, if it is before an expression, it is 'address-of'.
See https://en.wikipedia.org/wiki/Infix_notation about infix.
C does not support operator overloading (beyond what it built into the language).
As you can see in this Wikipedia Operators in C
The Address-of ("address of a") "&a" is defined as R* K::operator &();
whereas
The Bitwise AND "a & b" is defined as R K::operator &(S b);
So basically the "&" operator has different meaning when used as a unary operator and as a binary operator operator. The same goes for various other operators like, "*" , "-", etc.
It has simple different meanings when applied to the lvalue (it the unary operator in this case) or when it is used in the math expression with two operands.
"The operator & can be used in both of following way int a; scanf("%d",&a); and printf("%d",1&2)."
Note that in the case of C++ you forgot another important third use. The & operator can also be used to declare references.
int x = 24;
int& r_x = x;
the language divides operators based on its operands first. In one category itself overloading can be in-built. Your example is about the first division. Address-of is a unary operator and bitwise-AND is a binary operator. when you write operator function in c++ you will see the difference of these two categories.
Operator overloading is inbuilt to languages. example simple arithmetic addition operator. it can work with simple one-byte integer data as well as float (significant & exponent). Basically it is there with maths. so while making C language, they just translated those into functionality. In C specification, you cannot find overloading as a keyword for this behavior. According to them, after formula expression, anything has to be expressed as different functions. Each function should be named based on the functionality that it offers. When C++ introduced an opportunity to create new types, operators with its basic n-nary form allowed to operate with new types. In a nutshell, C's philosophy was different.
About halfway down this page there is a "Restrictions" section which states: "New operators such as **, <>, or &| cannot be created".
I know that ** is the exponentiation operator from python, and I know that <> is the "spaceship" (compareTo) operator from PHP. What on earth is &|?
correction <=> is the "spaceship", <> is a common substitute for !=
What the page is trying to say is that you can't create operators in C++ that doesn't already exist as operators. In other words, <> is not an operator in C++ today, so you can't do T operator<>(U x, U y) to define what this operator does. It's an invalid combination. Likewise for the other "symbols" listed. C++ defines a given set of operators, +, -, %, /, *, <<, >> and many more. These are allowed to override (with user-defined types). But the language does not allow the programmer to invent his/her own symbols and making them into operators.
This helps the parsing of the code by having a fixed set of operators that are lexically well-defined. That's aside from the confusions that this would cause when you read
if (x <&^> y) ...
or
x = y +- 11;
[What the heck is <&^> or +- supposed to mean?]
Sure, it's also possible to cause confusion with the existing operators, but it's at least somewhat less confusing than the programmer making up his/her own versions and variants.
The fact that some other language does have meanings for other symbols is up to those languages. Until the symbol is available in C++ (and some new operators are introduced at times, although quite rarely), they are not valid as C++ operators.
The only usage of &| is called: bit wise operators..
It's not used in a programming language, but to express something..like: I want to know if the result and/or (&|) matches something..
Let's have two lines of code:
&car->speed
&(car->speed)
Are these two lines equivalent? Will I get in both cases address to the speed?
If they are equivalents, what is better to choose as coding convention?
Are these two lines equivalent? Will I get in both cases address to the speed?
Yes. -> has higher precedence than that of unary &, therefore &car->speed and &(car->speed) are equivalent.
If they are equivalents, what is better to choose as coding convention?
Go with second as it shows the intended behaviour that you are interested in the address of speed.
This question already asked here several times. Postfix expression operators have higher priority than unary operators. So these two expressions
&car->speed
&(car->speed)
are equivalent.
Or another similar example with other unary operator !
!car->speed
!(car->speed)
As for coding convention I would prefer
&car->speed
and if you want to ampersandify car alone, use this:
( &car )->speed
Defined this way, we can do neither ++x++ nor ++x--. But on the other hand, both (++x)++ and (++x)-- are useful expressions: (++x)++ increments x by two and returns the value "in the middle", while (++x)-- is essentially equivalent to x+1 but completely avoids having to call operator+, which can be quite useful sometimes.
So why is the precedence not defined to have ++x++ automatically expand to (++x)++ rather than ++(x++)? Is there some hidden meaning to the latter which I don't understand, or is it just to keep the precedence a simple list with all prefix operators making up one single level?
EDIT Ok, I didn't explicitly say it, but: of course I meant x to be user-defined type. For built-in types, (x+=2)-1 is of course better than (++x)++, and x+1 is a lot better than (++x)--. The situation that I have in mind is an iterator to a rather complicated type of semi-associative container, where operators += and + (being designed for random access) have to rebuild a cache in order to work efficiently for general requests, and are therefore an order of magnitude slower than ++. But of course I can modify them to always check first if the argument is a very small integer, and in that case just call operator++ repeatedly instead of doing the random-access procedure. That should work fine here, though I could imagine I might at some point have a situation in which I want operator+= to always go the random-access way, regardless of how small numbers I present it.
So... for me, I would conclude the answer to be:
the advantage of having a simple and well-memorizeable precedence list in which all postfix operators come before any of the prefix operators is sufficient to tolerate the minor drawback of always having to use parentheses to compose pre- and postfix operators ++/--, as this composition is used very seldom.
The simpler "C does it this way", while it seems likely to be the real reason, is far less satisfying in to me, because since ++x++ was not allowed at all in C it would have been possible to redefine this very composition without damaging any existing code.
Anyway, I will go on using (++x)--, as the parentheses really do not hurt so much.
(++x)++ increments x by two and returns the value "in the middle"
Why not (x += 2) - 1 or (++x, x++)? Both seem to be clearer. For scalars, both are well-defined also in C++03, as opposed to your proposed expression.
(++x)-- is essentially equivalent to x+1 but completely avoids having to call operator+, which can be quite useful sometimes.
That's an arbitrary statement without any explanation. So I'm going to throw into the pool:
x+1 is essentially equivalent to (++x)-- but completely avoids having to call operator++ and operator-- which can be useful sometimes.
So why is the precedence not defined to have ++x++ automatically expand to (++x)++ rather than ++(x++)
Just to make such arcane corner cases not error out? No way. Can you please recite man operator for me? If you cannot do that, better not try and write ++x++ in your code.
C++ standard just kept the C rules and obviously those weren't fixed considering operator overloading and idioms yet be invented in a yet to be invented language.
Looking at what is available in D.M. Ritchie Home Page, on see that this precedence is already present in B (Unary operators are bound right to left. Thus -!x++ is bound -(!(x++)) in Users' Reference to B) and I didn't see increment operators in BCPL.
Both (++x)++ and (++x)-- invoke undefined behaviour [assuming x to be a primitive type]. For user defined type the scenario would be different. However it is not generally recommended to use such confusing expressions in your code.
As far as the precendence is concerned this answer explains why post increment has higher precedence than pre increment.
after this question, I don't know what to think.
In OCaml, if you do something like -1.0**2.0 (because of the typing you need to have float), you obtain 1.00. According to the standard order of operations, the result should be -1 (as in python).
I wasn't able to find the reason or a clear definition of the operator precedence in OCaml...
Is this because of the type system ? or the fact that there's a binding underneath with pow ?
As the very page you quote says, "The order in which the unary operator − (usually read "minus") acts is often problematical." -- it quotes Excel and bc as having the same priority for it as O'CAML, but also says "In written or printed mathematics" it works as in Python. So, essentially, there's no universal consensus on this specific issue.
Operator precedence is syntax-directed in OCaml, which means that the first character of the function identifier (and whether it's unary or binary) determines the operator precedence according to a fixed sequence. Contrast this with languages like Haskell, where the operator precedence can be specified at function definition regardless of which characters are used to form the function identifier.