operator-precedence

Background I am learning the sicp according to an online course and got confused by its lecture notes. In the lecture notes, the applicative order seems to equal cbv and normal order to cbn. Confusion But the wiki points out that, beside evaluation orders(left to right, right to left, or simultaneous), there is a difference between the applicative order and cbv: Unlike call-by-value, applicative order evaluation reduces terms within a function body as much as possible before the function is applied. I don't understand what does it mean by reduced. Aren't applicative order and cbv both going to

Closed. This question is off-topic . It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 4 years ago . A quick search for C++ precedence yields many attempts. The disconcerting part is that they are all different. Most are assuredly wrong, albeit in minor details. I will include three. The first, from cppreference.com claims there are 16 levels of precedence. Learn.cpp has 18. A simpler table at university of Purdue is much simpler. http://en.cppreference.com/w/cpp/language/operator_precedence http://www.learncpp

Say I have an expression as follows (where ⨁ and ⨂ are binary operators which have the same precedence level but not the same associativity): x ⨁ y ⨂ z Would y belong to ⨁ or ⨂ , and based on what criteria? According to the Edsgar Dijkstra's Shunting-yard algorithm if neighboring two operators in an expressions have the same precedence level then the expression is disambiguated based on the associativity of the second operator. If the second operator is left associative then the operand belongs to the first operator. If the second operator is right associative then the operand belongs to the

Consider a class MyClass with: a member function myClass& myFunction1(int) that modifies the object and returns *this a member function int myFunction2() const that does not modify the object Does the C++11/14 standard guarantee that: myclass.myFunction1(myclass.myFunction2()).myFunction1(myclass.myFunction2()); is equivalent to: myclass.myFunction1(myclass.myFunction2()); myclass.myFunction1(myclass.myFunction2()); No. The compiler can first call myclass.myFunction2() twice and then do the two myFunction1 calls in the first example code. But not in the second example code. There is nothing

The code: int r=1; System.out.println(r + (r=2)); The output is: 3. But I expected 4 because I thought the code inside the parentheses is executed first? Official Docs on Operators says All binary operators except for the assignment operators are evaluated from left to right; assignment operators are evaluated right to left. So + is evaluated left-to-right ,where as assignment operators are evaluated right to left. Use this if you want 4 int r=1; System.out.println((r=2) + r); // execute + is left-to-right Associativity of + is left-to-right , and the value of the expression (r=2) is 2 . Refer

After porting some obfuscated C code into C++ (namely Fairy-Max chess engine by Harm Geert Muller), I get lots of warnings similar to these: suggest parentheses around comparison in operand of '&' [-Werror=parentheses] suggest parentheses around '+' in operand of '&' While turning off the warnings is not an option, the solution is to add parenthesis in expressions according to the operator precedence . For example: if(z&S&&!ab&K==INF&d>2&v>V&v<Beta){ needs to be transformed into this: if((z&S) && ((!ab)&(K==INF)&(d>2)&(v>V)&(v<Beta))) { But doing this manually is quite time-consuming. I tried