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XOR using mathematical operators

By : hiren
Date : October 15 2020, 11:12 AM
like below fixes the issue (a − b)²
code :

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C++: overloading mathematical operators

By : user2077833
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further If you declare oparator+ as an instance function then the first argument is being passed as this object and thus you need only one more argument. Read this for more info, in particular try to understand the const concept:
code :
class Person
    Person &operator+=(const Person &i);
    const Person operator+(const Person &i) const;

Person &Person::operator+=(const Person &i) {
    ...   // Do the compound assignment work.
    return *this;

const Person Person::operator+(const Person &i) const {    
    Person result = *this;     // Make a copy of myself.  Same as MyClass result(*this);
    result += i;            // Use += to add other to the copy.
    return result;              // All done!

split a mathematical expression on operators and include the operators in the output array

By : neha garg
Date : March 29 2020, 07:55 AM
Does that help I'm trying to split the mathematical strings on maths operators. for example , jsfiddle
code :
var expression = "7.2*6+3/2-5*6+(7-2)*5";
var copy = expression;

expression = expression.replace(/[0-9]+/g, "#").replace(/[\(|\|\.)]/g, "");
var numbers = copy.split(/[^0-9\.]+/);
var operators = expression.split("#").filter(function(n){return n});
var result = [];

for(i = 0; i < numbers.length; i++){
     if (i < operators.length) result.push(operators[i]);


how do I format mathematical operators to str

By : melihkorkmaz
Date : March 29 2020, 07:55 AM
this one helps. You don't. * and = are not objects, they are operators, part of the language.
Just use strings for those.
code :
print '{} * {} = 1'.format(x, y)

Mathematical operators to use to give following result

By : user2646404
Date : March 29 2020, 07:55 AM
I hope this helps . In languages such as C, C++, Objective-C, Java, etc, with a right shift operator (>>) you could just do this:
code :
y = x >> 1;
y = x / 2;

token_get_all and mathematical operators

By : user3710766
Date : March 29 2020, 07:55 AM
it should still fix some issue Since they are single chars, they are already terminal symbols. No need to make a token out of it. You can find the list of available parser tokens here: http://php.net/manual/en/tokens.php
Have a look at a (pseudo) grammar:
code :
# Using a token

# Using the plain char
product := T_NUMBER '*' T_NUMBER
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