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What is the difference between the dot (.) operator and -> in C++? [duplicate]
(14 answers)
Closed 3 years ago.
I am learning C++ in a advanced programming class from my work since I have only worked in Web and .NET languages so far.
In a midway test the instructor has marked all of my uses of (*a).b as wrong and deducted points for it, which could negatively affect my final score and I need a near perfect score to transision in work from web stack to application stack, so could some of you help me resolve this dispute?
If a is a pointer, there's no difference in functionality at all, and in fact one is expressed in terms of the other [expr.ref§2]:
The expression E1->E2 is converted to the equivalent form (*(E1)).E2; the remainder of [expr.ref] will address only the first option (dot).
If a is an instance of a class with overloaded operators * and ->, there could be a difference. But such a discrepancy would be surprising, and I'd consider the class to have a bug because of this.
In the end, it's all about convention and readability then. The -> operator exists as a shorthand for the */. pair, as it is shorter and has better precedence rules (no need for parentheses). Thus, one would indeed use it rather than */..
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Double Negation in C++
(14 answers)
What is the usage of "!!" (negating twice)? [duplicate]
(3 answers)
Confused by use of double logical not (!!) operator [duplicate]
(5 answers)
Closed 2 years ago.
I often see experienced programmers write !!x, even though the expected expression is a Boolean (i.e., zero or not zero) and not an integer.
For example, a line from boost:
BOOST_ASSERT(!!p); // where `p` is a pointer
What's the point of !!p when just p will do?
What I understand by a Boolean parameter is an expression converted to a value of integral type, and the value is compared against zero, explicitly or implicitly (with if or its ternary operator equivalent).
Thus, anything that takes a Boolean and expects only 0 or 1 is wrongly implemented, if my understanding of Boolean is correct.
For clarification: it's obvious that ! converts to bool; the question is explicitly asking for why.
By default, BOOST_ASSERT(expr) expands to assert(expr). That C++ assert function takes a scalar argument, not bool. Thus, your statement, "the expected expression is a boolean," is not correct.
Classes in C++ can implement operator bool(); the basic_stream classes are examples of such: assert(std::cin) will not work. The !! operator forces the use of std::cin::operator bool(), and bool after !! will be evaluated to int 0 or 1. This is shorter than assert(bool(std::cin)).
If I were a boost author, I would expand BOOST_ASSERT(expr) to assert(!!(expr)) - as is done for ( BOOST_LIKELY(!!(expr)) ? ((void)0) ).
Closely related to the question
Warnings.
Some compilers issue warnings.
Overflow of the integer storing the Boolean.
I'll just quote the comment mentioning the behaviour
The compiler would treat the values the same either way if it's only used as an operand to an if or &&, but using !! may help on some compilers if if (!!(number & mask)) gets replaced with bit triggered = !!(number & mask); if (triggered); on some embedded compilers with bit types, assigning e.g. 256 to a bit type will yield zero. Without the !!, the apparently-safe transformation (copying the if condition to a variable and then branching) won't be safe.
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Some consider defining operator bool() to be "unsafe". And to have an idiomatical way to convert to bool, they define bool operator !().
So writing !!obj is to support such users. And it doesn't even hurt to do so!
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Multiple preincrement operations on a variable in C++(C ?)
(2 answers)
Closed 6 years ago.
I have some code with lines which increment a counter.
++ count;
Sometimes I have an if condition which means I should increment count by 2.
count += 2;
Does "double increment'ing" work in the same way?
++ ++ count;
It would be helpful to know if both C and C++ compilers interpret this the same way.
As this is clearly syntactically correct, the question that remains is: "Is this UB because of unsequenced writes?"
It is not (in C++11 and later) because
5) The side effect of the built-in pre-increment and pre-decrement operators is sequenced before its value computation (implicit rule due to definition as compound assignment)
(From here)
So the code is fine as of C++11.
However, the sequencing rules were different before that, and pre-C++11, the code actually has UB.
In C, that code does not even compile.
The fact that the behavior is different between C and C++ and even between different C++ standards and that this question arises in the first place is a hint that the simple count += 2; is the safer and more readable version. You should prefer it over the "cute and clever" ++ ++count;.
There are two methods. One that obviously works, one where you have to ask a question on StackOverflow. There are comments saying "in C++ 11 or later"... How sure are you that your C++ code runs with C++ 11 and not something older? Even if you use extensions that were not part of an earlier language, your language could be C++0x with some extensions.
Clearly you shouldn't care whether the second method works or not, but use the one that obviously works.
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Evaluate mathematical expression from a string using VB
(3 answers)
Closed 4 years ago.
I am developping a calculator function in Visual Basic. I think there is no basic one in the default .NET libraries.
I use System.Data.DataTable.Compute() to calculate the normal Math expression. But I want my function to solve functions like Sinus() or Round() as well.
Currently I am using Regex for this. For example for functions with one argument I use
Dim expr As String = Regex.Replace(Term, "(?<func>[A-Za-z]*?)\((?<arg1>[0-9,.]*?)\)", AddressOf FunctionLibrary.Funcs1)
the Sinus() function would look like this: sin(Number).
But this only allows to write Doubles or Integers into the argument. I can not write inner functions or even another math expression between the parentheses.
If I would write more functions as an argument, Regex would detect the first ")" inside the function, which is the closing parenthese of the inner function, as the end of the outer function.
Is there any way to make Regex recognize that theres an inner function as well?
If anyone knows an Evaluate() function for Visual Basic which is in the default .NET libraries, this might help me as well
Is there any way to make Regex recognize that theres an inner function as well?
No, there is not. Regular expressions, as the name implies, solve Regular grammars and simple mathematical expressions are Context-Free. Regular expressions do not have a stack to match arbitrary expressions. For example, distinguishing between (()) and ()() require at least one character of lookahead (or backtracking). Yes, PCRE-style regular expressions can let you create a fixed number of lookahead characters, but so far as I know you have to specify the number of characters, and anyway this is not going to solve your problem.
Evaluating arithmetic expressions require handling precedence, subexpressions, possibly variables and types. Regular expressions cannot do this, attempting to do it with regular expressions will lead you into a pit of failure.
Nor are they even necessary. Evaluating mathematical expressions is a solved problem and there are dozens of parsers and evaluators written, tested, and ready for you to drop into your application. You have not given us enough information to decide which one would be best for you, and anyway Stack Overflow is not a tool advocacy site. You could start by going through the list at Gary Beene's Equation Parser review.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What does the word “literal” mean?
Often when reading literature about C++, I encounter the word "literal". It is a bit unclear to me what exactly this term means in C++.
A literal is some data that's presented directly in the code, rather than indirectly through a variable or function call.
Here are some examples, one per line:
42
128
3.1415
'a'
"hello world"
The data constituting a literal cannot be modified by a program, but it may be copied into a variable for further use:
int a = 42; // creates variable `a` with the same value as the literal `42`
This concept is by no means unique to C++.
The term "literal" comes from the fact that you've written data literally into your program, i.e. exactly as written, not "hidden" behind a variable name.
Wikipedia gives you quickly this about literals.
In your C or C++ source code, Things like 1234, nullptr (in recent C++), "abcd" are literals.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
C++ Comma Operator
that probably a trivial question, but I don't know the answer. And this has been troubling me this afternoon.
I was just writing a function to convert RVB to YUV. Nothing really special, but mistakenly used the comma (,) instead of a dot in my numbers.
It compiles but the result was not what I expected, for example "-3713796" instead of a 0-255 range number.
(0,615*(double) 61) - (0,51498*(double) 61) - (0,10001*(double) 61)
So what does it mean ?
If it's not a compilation error, it's probably usefull for something but what?
Ps: I was using C++ with Qt.
You've accidentally used the comma operator. It evaluates its first operand and discards the result, and then evaluates the second operand and returns its value and type.
In my experience, it's most commonly used in loop statements. For other uses, see Uses of C comma operator
It's the comma operator which evaluates each operand and returns the rightmost. In this case it evaluated 0, then 615*(double) 61 and resulted in a value of that product.