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: http://www.cs.caltech.edu/courses/cs11/material/cpp/donnie/cppops.html 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+3/25*6+(72)*5";
var copy = expression;
expression = expression.replace(/[09]+/g, "#").replace(/[\(\\.)]/g, "");
var numbers = copy.split(/[^09\.]+/);
var operators = expression.split("#").filter(function(n){return n});
var result = [];
for(i = 0; i < numbers.length; i++){
result.push(numbers[i]);
if (i < operators.length) result.push(operators[i]);
}
console.log(result);

how do I format mathematical operators to str
By : melihkorkmaz
Date : March 29 2020, 07:55 AM

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++, ObjectiveC, 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.phpHave a look at a (pseudo) grammar: code :
# Using a token
product := T_NUMBER T_MULT_OPERATOR T_NUMBER
# Using the plain char
product := T_NUMBER '*' T_NUMBER

