operator-precedence | 易学教程

Operator precedence (bitwise '&' lower than '==')

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In the C programing language, why do the bitwise operators (& and |) have lower precedence than the equality operator (==)? It does not make sense to me. Caladain You need to ask Brian Kernighan or Dennis Ritchie. From this forum: http://bytes.com/topic/c/answers/167377-operator-precedence The && and || operators were added later for their "short-circuiting" behavior. Dennis Ritchie admits in retrospect that the precedence of the bitwise operators should have been changed when the logical operators were added. But with several hundred kilobytes of C source code in existence at that point and

Why is the order of evaluation for function parameters unspecified in c++?

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问题 The standard doesn't specify the order of evaluation of arguments with this line: The order of evaluation of arguments is unspecified. What does Better code can be generated in the absence of restrictions on expression evaluation order imply? What is the drawback in asking all the compilers to evaluate the function arguments Left to Right for example? What kinds of optimizations do compilers perform because of this unspecified spec? 回答1: Allowing the compiler to re-order the evaluation of the

Is this code well-defined?

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问题 This code is taken from a discussion going on here. someInstance.Fun(++k).Gun(10).Sun(k).Tun(); Is this code well-defined? Is ++k in Fun() evaluated before k in Sun()? What if k is user-defined type, not built-in type? And in what ways the above function calls order is different from this: eat(++k);drink(10);sleep(k); As far as I know, in both situations, there exists a sequence point after each function call . If so, then why can't the first case is also well-defined like the second one?

Order of execution of parameters guarantees in Java?

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问题 Given the following function call in C : fooFunc( barFunc(), bazFunc() ); The order of execution of barFunc and BazFunc is not specified, so barFunc() may be called before bazFunc() or bazFunc() before barFunc() in C . Does Java specify an order of execution of function argument expressions or like C is that unspecified? 回答1: From the Java Language Specification (on Expressions): 15.7.4 Argument Lists are Evaluated Left-to-Right In a method or constructor invocation or class instance creation

Ternary conditional and assignment operator precedence

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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

Where is it legal to use ruby splat operator?

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问题 Splats are cool. They're not just for exploding arrays, although that is fun. They can also cast to Array and flatten arrays (See http://github.com/mischa/splat/tree/master for an exhaustive list of what they do.) It looks like one cannot perform additional operations on the splat, but in 1.8.6/1.9 the following code throws "unexpected tSTAR": foo = bar || *zap #=> unexpected tSTAR Whereas this works: foo = *zap || bar #=> works, but of limited value Where can the splat appear in an

Does this code from “The C++ Programming Language” 4th edition section 36.3.6 have well-defined behavior?

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In Bjarne Stroustrup's The C++ Programming Language 4th edition section 36.3.6 STL-like Operations the following code is used as an example of chaining : void f2() { std::string s = "but I have heard it works even if you don't believe in it" ; s.replace(0, 4, "" ).replace( s.find( "even" ), 4, "only" ) .replace( s.find( " don't" ), 6, "" ); assert( s == "I have heard it works only if you believe in it" ) ; } The assert fails in gcc ( see it live ) and Visual Studio ( see it live ), but it does not fail when using Clang ( see it live ). Why am I getting different results? Are any of these

Haskell operator vs function precedence

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问题 I am trying to verify something for myself about operator and function precedence in Haskell. For instance, the following code list = map foo $ xs can be rewritten as list = (map foo) $ (xs) and will eventually be list = map foo xs My question used to be, why the first formulation would not be rewritten as list = (map foo $) xs since function precedence is always higher than operator precedence, but I think that I have found the answer: operators are simply not allowed to be arguments of

Post-increment on a dereferenced pointer?

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Trying to understand the behaviour of pointers in C, I was a little surprised by the following (example code below): #include <stdio.h> void add_one_v1(int *our_var_ptr) { *our_var_ptr = *our_var_ptr +1; } void add_one_v2(int *our_var_ptr) { *our_var_ptr++; } int main() { int testvar; testvar = 63; add_one_v1(&(testvar)); /* Try first version of the function */ printf("%d\n", testvar); /* Prints out 64 */ printf("@ %p\n\n", &(testvar)); testvar = 63; add_one_v2(&(testvar)); /* Try first version of the function */ printf("%d\n", testvar); /* Prints 63 ? */ printf("@ %p\n", &(testvar)); /*

Operator precedence with Javascript Ternary operator

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I cant seem to wrap my head around the first part of this code ( += ) in combination with the ternary operator. h.className += h.className ? ' error' : 'error' The way i think this code works is as following: h.className = h.className + h.className ? ' error' : 'error' But that isn't correct because that gives a error in my console. So my question is how should i interpet this code correctly? Kobi h.className = h.className + (h.className ? ' error' : 'error') You want the operator to work for h.className , better be specific about it. Of course, no harm should come from h.className += ' error'