conditional-operator

How to overload the conditional operator? [duplicate]

…衆ロ難τιáo~ 提交于 2019-11-26 17:23:44
问题 This question already has answers here : Closed 7 years ago . Possible Duplicate: Operator overloading I was wonder how can I over load the Conditional operator in cpp? int a,b,c; a=10; b=11; c = (a>b) ? a : b; Is it possible? 回答1: Several operators cannot be overloaded. These operators take a name, rather than an object, as their right operand: Direct member access (.) Deference pointer to class member (.*) Scope resolution (::) Size of (sizeof) The conditional operator (?:) also cannot be

What is ?: in PHP 5.3? [duplicate]

99封情书 提交于 2019-11-26 17:23:11
Possible Duplicate: What are the PHP operators “?” and “:” called and what do they do? From http://twitto.org/ <?PHP require __DIR__.'/c.php'; if (!is_callable($c = @$_GET['c'] ?: function() { echo 'Woah!'; })) throw new Exception('Error'); $c(); ?> Twitto uses several new features available as of PHP 5.3: The DIR constant The ?: operator Anonymous functions What does number 2 do with the ?: in PHP 5.3? Also, what do they mean by anonymous functions? Wasn't that something that has existed for a while? Ben James ?: is a form of the conditional operator which was previously available only as:

Using return in ternary operator

你。 提交于 2019-11-26 16:59:23
问题 I'm trying to use return in a ternary operator, but receive an error: Parse error: syntax error, unexpected T_RETURN Here's the code: $e = $this->return_errors(); (!$e) ? '' : return array('false', $e); Is this possible? Thanks! 回答1: This is the correct syntax: return !$e ? '' : array('false', $e); 回答2: Close. You'd want return condition?a:b 回答3: It doesn't work in most languages because return is a statement (like if , while , etc.), rather than an operator that can be nested in an

Ternary operator in JSTL/EL

半城伤御伤魂 提交于 2019-11-26 16:20:40
问题 The following tag of JSTL can be used to set a value to a variable in a request scope. <c:set var="value" scope="request" value="someValue"/> I want to check conditionally, if the variable value being set is empty or not and display the result accordingly something like the following, using <c:when>...</c:when> . <c:choose> <c:when test="${not empty value}"> <c:out default="None" value="${value}"/> </c:when> <c:otherwise> <c:out default="None" value="None"/> </c:otherwise> </c:choose> I want

Ternary conditional and assignment operator precedence

爱⌒轻易说出口 提交于 2019-11-26 16:12:54
问题 I'm confused about direct assignment and ternary conditional operators precedence: #include<stdio.h> int main(void) { int j, k; j = k = 0; (1 ? j : k) = 1; // first printf("%d %d\n", j, k); j = k = 0; 1 ? j : k = 1; // second printf("%d %d\n", j, k); return 0; } I would expect the output to be: 1 0 1 0 But it happens to be: 1 0 0 0 Plus I get this warning: main.cpp:20: warning: statement has no effect which is about the line I commented as second. Since the direct assignment operator has less

?: operator (the &#39;Elvis operator&#39;) in PHP

|▌冷眼眸甩不掉的悲伤 提交于 2019-11-26 15:41:42
I saw this today in some PHP code: $items = $items ?: $this->_handle->result('next', $this->_result, $this); I'm not familiar with the ?: operator being used here. It looks like a ternary operator, but the expression to evaluate to if the predicate is true has been omitted. What does it mean? BalusC It evaluates to the left operand if the left operand is truthy , and the right operand otherwise. In pseudocode, foo = bar ?: baz; roughly resolves to foo = bar ? bar : baz; or if (bar) { foo = bar; } else { foo = baz; } with the difference that bar will only be evaluated once. You can also use

Error: lvalue required in this simple C code? (Ternary with assignment?)

和自甴很熟 提交于 2019-11-26 14:40:51
I have : #include<stdio.h> int main() { int a=5,b=6; (a>b)?b=a:b=b; // Here is the error return 0; } But if I replace : (a>b)?b=a:b=b; // Error with (a>b)?(b=a):(b=b); // No-Error I understand the lvalue is a value to which something can be assigned and how is it different from rvalue , but why is the extra parenthesis making the difference. Assignment has a lower precedence than the ternary operator so the line evaluates like: ((a>b)?b=a:b)=b; use: b=(a>b)?a:b; Nawaz Actually, in C, this code (a>b)?b=a:b=b; is parsed by many compilers as ((a>b)?b=a:b)=b; which is an error, as the expression (

should std::common_type use std::decay?

老子叫甜甜 提交于 2019-11-26 14:24:16
问题 Given types A,B , I am concerned with the exact definition of std::common_type<A,B> , disregarding the variadic case std::common_type<A...> for arbitrary types A... . So let using T = decltype(true ? std::declval<A>() : std::declval<B>()); using C = std::common_type<A,B>; Now, according to a number of sources, I have found the following relations (skipping typename for brevity): cppreference.com: C::type = std::decay<T>::type cplusplus.com: C::type = T GCC 4.8.1 <type_traits> implementation:

Compare multiple values in PHP

房东的猫 提交于 2019-11-26 13:45:45
I'd like to go from this: if($var == 3 || $var == 4 || $var == 5 || $var =='string' || $var == '2010-05-16') { // execute code here } to this: if($var == (3, 4, 5, 'string', '2010-05-16')) { // execute code here } Seems very redundant to keep typing $var , and I find that it makes it a bit cumbersome to read. Is there a way in PHP to do simplify it in this way? I read on a post here that when using XQuery you can use the = operator as in $var = (1,2,3,4,5) etc. Place the values in an array, then use the function in_array() to check if they exist. $checkVars = array(3, 4, 5, "string", "2010-05

?: ternary conditional operator behaviour when leaving one expression empty

我与影子孤独终老i 提交于 2019-11-26 13:29:52
I was writing a console application that would try to "guess" a number by trial and error, it worked fine and all but it left me wondering about a certain part that I wrote absentmindedly, The code is: #include <stdio.h> #include <stdlib.h> int main() { int x,i,a,cc; for(;;){ scanf("%d",&x); a=50; i=100/a; for(cc=0;;cc++) { if(x<a) { printf("%d was too big\n",a); a=a-((100/(i<<=1))?:1); } else if (x>a) { printf("%d was too small\n",a); a=a+((100/(i<<=1))?:1); } else { printf("%d was the right number\n-----------------%d---------------------\n",a,cc); break; } } } return 0; } More specifically