operator-precedence

问题 This question already has answers here : Short circuit behavior of logical expressions in C in this example (1 answer) Increment and logical operators precedence [duplicate] (3 answers) Closed 4 years ago . #include <stdio.h> int main(void) { int i = -3, j = 2, k = 0, m; m = ++i && ++j || ++k; printf("%d %d %d %d\n",i,j,k,m); return 0; } I am trying to learn about associativity and precedence of operators in C. Here, The output comes out to be -2 3 0 1 , but I think the output should be -2 3

问题 I want to handle geographic names i.e /new_york or /new-york etc and since new-york is django-slugify for New york then maybe I should use the slugifed names even if names with underscores look better since I may want to automate the URL creation via an algorithm such as django slugify. A guess is that ([A-Za-z]+) or simply ([\w-]+) can work but to be safe I ask you which regex is best choice in this case. I've already got a regex that handles number connecting numbers to a class: ('/([0-9]*)

问题 I am totally lost why I get these results: int i = 1; int[] a = new int[6]; a[0] = 0; a[1] = 1; a[2] = 2; a[3] = 3; a[4] = 4; a[5] = 5; i += i + a[i++] + a[i++]; //i is 5 i = 1; i += i + a[i++] + a[i++] + a[i++]; // i is 8 I (wrongly) thought that there are these options: i = i(=1) + a[i++] + etc - meaning that i = 1 is cached and not changed then. Expression evaluation order is exactly from left to right, I guess (?!). i is increased, which results (for first example) in i = i(=3) + a[1] + a

问题 I'm learning about operator precedence in Haskell. Several places across the web mention that function application has higher precedence than operators, but I couldn't find a definitive source for that. Here is one such mention from A Gentle Introduction To Haskell: Function application has higher precedence than any infix operator There is a section in the Haskell 98 Report that alludes to it: normal constructor application has higher precedence than infix constructor application Where is a

问题 As stated in this question: LLVM and GCC, different output same code, LLVM and GCC result in different output for the same code. #include <stdio.h> #define MAX(a,b) ( (a) > (b) ? (a) : (b) ) int increment() { static int i = 42; i += 5; printf("increment returns %d\n",i); return i; } int main( int argc, char ** argv ) { int x = 50; printf("max of %d and %d is %d\n", x,increment(),MAX(x, increment())); printf("max of %d and %d is %d\n", x,increment(),MAX(x, increment())); return 0; } The LLVM

问题 This question already has answers here : SQL Logic Operator Precedence: And and Or (4 answers) Closed last year . I'm debugging some code and came across this. Could someone help me by putting parens around this statement according to SQL Server ordering. Is it just me, or is this bad coding? WHERE T1.C1 = @VAR1 AND T1.C2 = @VAR2 AND T1.C3 NOT IN(@VAR3) OR T2.C1 = @VAR2 OR T3.C1 = @VAR2 回答1: As is, it will be parsed like WHERE (T1.C1 = @VAR1 AND T1.C2 = @VAR2 AND T1.C3 NOT IN(@VAR3)) OR (T2

问题 This question already has answers here : SQL Logic Operator Precedence: And and Or (4 answers) Closed last year . I'm debugging some code and came across this. Could someone help me by putting parens around this statement according to SQL Server ordering. Is it just me, or is this bad coding? WHERE T1.C1 = @VAR1 AND T1.C2 = @VAR2 AND T1.C3 NOT IN(@VAR3) OR T2.C1 = @VAR2 OR T3.C1 = @VAR2 回答1: As is, it will be parsed like WHERE (T1.C1 = @VAR1 AND T1.C2 = @VAR2 AND T1.C3 NOT IN(@VAR3)) OR (T2

问题 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()); 回答1: No. The compiler can first call myclass.myFunction2() twice and then do

问题 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)

问题 In any programming language textbooks, we are always told how each operator in that language has either left or right associativity. It seems that associativity is a fundamental property of any operator regardless of the number of operands it takes. It also seems to me that we can assign any associativity to any operator regardless of how we assign associativity to other operators. But why is it the case? Perhaps an example is better. Suppose I want to design a hypothetical programming