operator-precedence | 易学教程

How is “std::cin>>value” evaluated in a while loop?

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问题 Currently I'm self-learning C++ Primer 5th. Here comes something I'm not sure. (I couldn't find the exact relevant question on F.A.Q). Consider this while loop: while(std::cin>>value){...} \\value here was defined as int. The text book says: That expression reads the next number from the standard input and stores that number in value. The input operator (§ 1.2, p. 8) returns its left operand, which in this case is std::cin. This condition, therefore, tests std::cin.When we use an istream as a

Does the C/C++ ternary operator actually have the same precedence as assignment operators?

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问题 Almost all C/C++ operator precedence tables I have consulted list the ternary conditional operator as having higher precedence than the assignment operators. There are a few tables, however, such as the one on wikipedia, and the one at operator-precedence.com, that place them on the same precedence level. Which is it, higher or same? 回答1: In the C++ grammar, assignment-expression: conditional-expression logical-or-expression assignment-operator initializer-clause throw-expression conditional

a += a++ * a++ * a++ in Java. How does it get evaluated?

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问题 I came across this problem in this website, and tried it in Eclipse but couldn't understand how exactly they are evaluated. int x = 3, y = 7, z = 4; x += x++ * x++ * x++; // gives x = 63 System.out.println(x); y = y * y++; System.out.println(y); // gives y = 49 z = z++ + z; System.out.println(z); // gives z = 9 According to a comment in the website, x += x++ * x++ * x++ resolves to x = x+((x+2)*(x+1)*x) which turns out to be true. I think I am missing something about this operator precedence.

Why does 1+++2 = 3?

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How does Python evaluate the expression 1+++2 ? How many ever + I put in between, it is printing 3 as the answer. Please can anyone explain this behavior And for 1--2 it is printing 3 and for 1---2 it is printing -1 Greg Hewgill Your expression is the same as: 1+(+(+2)) Any numeric expression can be preceded by - to make it negative, or + to do nothing (the option is present for symmetry). With negative signs: 1-(-(2)) = 1-(-2) = 1+2 = 3 and 1-(-(-2)) = 1-(2) = -1 I see you clarified your question to say that you come from a C background. In Python, there are no increment operators like ++ and

How are java increment statements evaluated in complex expressions

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问题 What is the output of the following code: int x = 2; x += x++ * x++ * x++; System.out.println(x); I understand that ++variableName is pre-increment operator and the value of variableName is incremented before it is used in the expression whereas variableName++ increments its value after the expression is executed. What I want to know is - how does this logic apply here? 回答1: Its easier to see the what is going on with x = 1 instead of 2. The output for x=1 is 7. The key to the understanding

php string number concatenation messed up

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问题 I got some php code here: <?php echo 'hello ' . 1 + 2 . '34'; ?> which outputs 234, but when I add a number 11 before "hello": <?php echo '11hello ' . 1 + 2 . '34'; ?> It outputs 1334 rather than 245(which I expected it to), why is that? 回答1: That's strange... But <?php echo '11hello ' . (1 + 2) . '34'; ?> OR <?php echo '11hello ', 1 + 2, '34'; ?> fixing issue. UPDv1: Finally managed to get proper answer: 'hello' = 0 (contains no leading digits, so PHP assumes it is zero). So 'hello' . 1 + 2

How do we explain the result of the expression (++x)+(++x)+(++x)?

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x = 1; std::cout << ((++x)+(++x)+(++x)); I expect the output to be 11 , but it's actually 12 . Why? CB Bailey We explain it by expecting undefined behaviour rather than any particular result. As the expression attempts to modify x multiple times without an intervening sequence point its behaviour is undefined . Artelius As others have said, the C and C++ standards do not define the behaviour that this will produce. But for those people who don't see why the standards would do such a thing, let's go through a "real world" example: 1 * 2 + 3 + 4 * 5 There's nothing wrong with calculating 1 * 2 +

“IF” argument evaluation order?

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if(a && b) { do something; } is there any possibility to evaluate arguments from right to left(b -> a)? if "yes", what influences the evaluation order? (i'm using VS2008) The evaluation order is specified by the standard and is left-to-right . The left-most expression will always be evaluated first with the && clause. If you want b to be evaluated first: if(b && a) { //do something } If both arguments are methods and you want both of them to be evaluated regardless of their result: bool rb = b(); bool ra = a(); if ( ra && rb ) { //do something } With C++ there are only a few operators that

++i + ++i + ++i in Java vs C

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问题 int i=2; i = ++i + ++i + ++i; Which is more correct? Java's result of 12 or C = 13. Or if not a matter of correctness, please elaborate. 回答1: There is nothing like more correct. It is actually undefined and its called Sequence Point Error. http://en.wikipedia.org/wiki/Sequence_point 回答2: Java guarantees (§15.7.1) that it will be evaluated left-to-right, giving 12. Specifically, ++ has higher precedence that + . So it first binds those, then it associates the addition operations left to right

Reordering factor gives different results, depending on which packages are loaded

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I wanted to create a barplot in which the bars were ordered by height rather than alphabetically by category. This worked fine when the only package I loaded was ggplot2. However, when I loaded a few more packages and ran the same code that created, sorted, and plotted my data frame, the bars had reverted to being sorted alphabetically again. I checked the data frame each time using str() and it turned out that the attributes of the data frame were now different, even though I'd run the same code each time. My code and output are listed below. Can anyone explain the differing behavior? Why