unary-operator

Some information source on operator precedence like this says that unary operators like ! , ~ , + , - have higher precedence than assignment = . However, the following expressions are possible: !a = true # => false (with warning) a # => true ~a = 1 # => -2 a # => 1 +a = 1 # => 1 a # => 1 -a = 1 # => -1 a # => 1 Considering these results, the only possible explanation I can think of is that these unary operator have lower precedence than the assignment. If that is the case, then it would mean that the information I mentioned above is wrong. Which is correct? Is there a different explanation? My

Confusion rose because of this post . The author updated his post, and the result became clear. Conclusion: Java evaluates expressions from left to right Closed! As evaluation of expression is done from right to left the following code should store 5 in j : int i=2; int j=++i+i++; System.out.println(j); But I get 6 as the output, which forces me to re-think the right to left evaluation idea. Kindly explain the theory here. int i = 2; int j = ++i + i++; is the same as int i = 2; // This part is from ++i i = i + 1; int left = i; // 3 // This part is from i++ int right = i; // 3 i = i + 1; int j

I am currently converting some OpenCV code from C++ to Java. I can't use JavaCV, as we need the conversion in native Java, not a JNA. At one point in the code, I get the following assignment: dst[x] = (uchar)(-(kHit >= kForeground)); Where dst is uchar* , kHit and kForeground are int s. I've been unable to find anything about how this works, and Java will not recognize it as an operation. There is an operation on these two variables at another point in the code, and it stores one of two values: 255 or 0. The code in question comes from opencv/video/src/bgfg_gaussmix.cpp . In C++ a boolean

There's the following bit of Python code in a project I have to maintain: # If the `factor` decimal is given, compute new price and a delta factor = +factor.quantize(TWOPLACES) new_price = +Decimal(old_price * factor).quantize(TWOPLACES) delta = new_price - old_price The question here is the purpose of + in front of a variable. Python docs call it unary plus operator , which “yields its numeric argument unchanged”. Can it be safely removed then? (Incidentally, the code was written by me some time ago, hopefully I've learned the lesson—it wouldn't be a question if tests existed, or if the use

What is the difference between these unary operators in C# ? . Can you provide me with example? Please provide the name of each. :) += vs =+ ++x vs x++ This has no doubt been answered before, but anyway... They differ in how they change the value and how they return the result. The first two += and =+ behave in the way that the first increments a variable, the other sets a variable. They are not related. Observe the following code: // += x = 1; printf( x += 1 ); // outputs 2, the same as x = x+1 printf( x ); // outputs 2 // =+ x = 1; printf( x =+ 1 ); // outputs 1, the same as x = 1; printf( x

Till now I was believing that there is no use of unary + operator. But then I came across with following example: char ch; short sh; int i; printf("%d %d %d",sizeof(ch),sizeof(sh),sizeof(i)); // output: 1 2 4 printf("%d %d %d",sizeof(+ch),sizeof(+sh),sizeof(i)); // output: 4 4 4 Does it mean + is doing type conversion here? Because it is behaving same as following printf("%d %d %d",sizeof((int)ch),sizeof((int)sh),sizeof(i)); // output: 4 4 4 This forces me to think + is doing type conversion. But then I try it on double double f; printf("%d %d",sizeof(+f),sizeof((int)f),sizeof(f)); // output:

Possible Duplicate: Undefined Behavior and Sequence Points Pleae explain the behaviour of following statements int b=3; cout<<b++*++b<<endl; How will it be calculated? Greg Howell The behavior here is undefined. See this question Relevant standard quote: §5/4.1 Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression. The most common sequence point is the end of a statement. Also worth noting from the standard: §5.2.2/8 The order of evaluation of arguments is unspecified. The standard says this is