operator-precedence

How does cout's << operator work with regard to operator precedence? [duplicate]

余生长醉 提交于 2019-12-23 16:06:55
问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Unexpected order of evaluation (compiler bug?) I couldn't predict the output for this program : #include<iostream> using namespace std; int *p(int *a) { (*a)++; return a; } int main() { int i=0; cout<<i++<<" "<<(*p(&i))++<<" "<<i++<<" "<<i<<endl; return 0; } When compiled in vs2008, it outputs 3 2 0 4 . Can anybody explain why it's not 0 2 3 4 ? Note: It works great if there is no function call to p . Thanks in

How does cout's << operator work with regard to operator precedence? [duplicate]

你。 提交于 2019-12-23 16:04:00
问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Unexpected order of evaluation (compiler bug?) I couldn't predict the output for this program : #include<iostream> using namespace std; int *p(int *a) { (*a)++; return a; } int main() { int i=0; cout<<i++<<" "<<(*p(&i))++<<" "<<i++<<" "<<i<<endl; return 0; } When compiled in vs2008, it outputs 3 2 0 4 . Can anybody explain why it's not 0 2 3 4 ? Note: It works great if there is no function call to p . Thanks in

Where 'foo' OR 'bar' AND 'lol' OR 'rofl' MySQL

折月煮酒 提交于 2019-12-23 12:51:01
问题 In what order would this be evaluated. My intension is that if it finds either foo or bar, it would also search for lol and rofl. Is this totally in the woods? And if so, how would one evaluate an expression like that. 回答1: The AND operator has higher precedence than OR in MySql, so your current expression evaluates as: WHERE 'foo' OR ('bar' AND 'lol') OR 'rofl' Add parentheses to the expression if you want to force the evaluation order: WHERE ('foo' OR 'bar') AND ('lol' OR 'rofl') 回答2: AND

Why is the boolean expression “1 in (1, 2, 3) == True” False? [duplicate]

心已入冬 提交于 2019-12-23 12:18:59
问题 This question already has answers here : Why does (1 in [1,0] == True) evaluate to False? [duplicate] (1 answer) Why does the expression 0 < 0 == 0 return False in Python? (9 answers) Closed 3 years ago . Why does the statement 1 in (1, 2, 3) == True return False in Python? Is the operator priority in Python ambiguous? 回答1: Because, per the documentation on operator precedence: Note that comparisons, membership tests, and identity tests, all have the same precedence and have a left-to-right

Why do you have to add parentheses to + - operations when concatenating?

試著忘記壹切 提交于 2019-12-23 10:48:09
问题 I was writing a small program when I encountered something strange. If I wanted PHP to present an arithmetic operations of addition or subtraction with an echo statement and the outcome of the operation I had to add parentheses or the html page wouldn't present the operation but just the outcome. Below is a reduced example. first case (without parentheses): $a = 10; $b = 5; echo "$a + $b = ".$a + $b."<br>"; // 15 echo "$a - $b = ".$a - $b."<br>"; // 5 echo "$a * $b = ".$a * $b."<br>"; // 10 *

List Operator Precedence in Perl

安稳与你 提交于 2019-12-23 08:38:42
问题 I'm reading the "Beginning Perl" book, and it gives these two statements: print "Test one: ", 6 > 3 && 3 > 4, "\n"; print "Test two: ", 6 > 3 and 3 > 4, "\n"; The first line prints nothing with a new line, the second line prints a 1 with no new line. I'm confused about the output. According to the author, the second statement gives weird output because it's just like saying: print ("Test two: ", 6 > 3) and 3 > 4, "\n"; However, why is the first statement not the same? I thought it had

Fixity of backtick operators?

主宰稳场 提交于 2019-12-23 06:48:14
问题 What is the fixity of backtick operators? For instance in this code from Real World Haskell: ghci> (1+) `fmap` [1,2,3] ++ [4,5,6] [2,3,4,4,5,6] It's evident the backtick operator `fmap` has a higher fixity than ++ , but none is given by GHCi. 回答1: §4.4.2 of the Haskell Report states that Any operator lacking a fixity declaration is assumed to be infixl 9 "Any operator" includes normal function names in backticks. Your example shows that `fmap` does have higher fixity than ++ , because ++ acts

Operator Precedence.. () and ++

白昼怎懂夜的黑 提交于 2019-12-23 04:06:23
问题 Salute.. I have an unusual problem. Here in this table in MSDN library we can see that precedence of () is higher than ++ (Pre-increment) . but when I run this code, it seems that precedence of ++(prefex) is higher: int main() { int a=3,b=2,x; x=++a + (a-b); cout<<"x= "<<x; return 0; } and the answer is : x=6 This happens with prefex ++ only and works as I expect with post-increment . Is there any reason? Regards.. int main() { int a=3,b=2,x; x=a++ + (a-b); cout<<"x= "<<x; return 0; } x=4 (I

Java - parentheses and assignment

余生颓废 提交于 2019-12-22 20:45:33
问题 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? 回答1: 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. 回答2: Use this if you want 4 int r=1; System.out.println((r=2) + r); //

Precedence between member access from a pointer and cast

陌路散爱 提交于 2019-12-22 11:24:25
问题 If I have typedef struct{ int i; } typeB; typeA *p; then: What is the precedence between member access from a pointer and cast? (typeB *)p->i Is it actually ((typeB *)p)->i or (typeB *)(p->i) ? 回答1: An operator precedence table would show that a -> binds more tightly than the cast. typedef void *typeA; typeB b; typeA a = &b; (typeB *)a->i; /* wrong: attempt to dereference a void pointer */ ((typeB *)a)->i; /* ok */ A complete operator precedence table for your future reference is provided