dereference

I am using PHP 7.2. I come across the following note from the arrays chapter of PHP Manual Array dereferencing a scalar value which is not a string silently yields NULL , i.e. without issuing an error message. I understand how to dereference an array literal but I'm not able to understand how the "array dereferencing" works on a scalar value of type boolean/integer/float/string? If you look at the code example from the PHP manual itself, you can notice the contradiction as it's not the value of integer type is not silently yielding NULL according to the manual. <?php function getArray() {

In the hope of gaining a better understanding of the answers given in this post, can someone please explain to me if the following circular buffer implementation is possible, and if not, why not. #define CB_TYPE_CHAR 0 #define CB_TYPE_FLOAT 1 ... typedef struct CBUFF { uint16 total; /* Total number of array elements */ uint16 size; /* Size of each array element */ uint16 type; /* Array element type */ uint16 used; /* Number of array elements in use */ uint16 start; /* Array index of first unread element */ void *elements; /* Pointer to array of elements */ } CBUFF; ... void cbRead(CBUFF

To avoid keep having to use -> and instead work directly with the object, is it acceptable practice to do: obj x = *(new obj(...)); ... delete &obj; This is not just poor practice, but: Leaking memory (most likely, unless you are using some pattern that is not visible from the code you provided), since obj will store a copy of the original object created by the new expression, and the pointer to that object returned by new is lost; Most importantly, undefined behavior , since you are passing to delete a pointer to an object that was not allocated with new . Per paragraph 5.3.5/2 of the C++11

Despite thoroughly reading the documentation, I'm rather confused about the meaning of the & and * symbol in Rust, and more generally about what is a Rust reference exactly. In this example, it seems to be similar to a C++ reference (that is, an address that is automatically dereferenced when used): fn main() { let c: i32 = 5; let rc = &c; let next = rc + 1; println!("{}", next); // 6 } However, the following code works exactly the same: fn main() { let c: i32 = 5; let rc = &c; let next = *rc + 1; println!("{}", next); // 6 } Using * to dereference a reference wouldn't be correct in C++. So I

I've seen code like this in a couple of old projects: class Class { static void Method() {} }; ((Class*)0)->Method(); This code contains undefined behavior because it includes dereferencing a null pointer (no matter what happens afterwards). It really makes no sense - the cast is there to feed the type name to the compiler and whoever wrote the code above could have written this instead: Class::Method(); and the latter would be okay. Why would anyone write the former code? Is it a known idiom from some good old days or what? Static member functions were added into C++ in 1989, in Release 2.0

I am trying following hql query to execute SELECT count(*) FROM BillDetails as bd WHERE bd.billProductSet.product.id = 1002 AND bd.client.id = 1 But it is showing org.hibernate.QueryException: illegal attempt to dereference collection [billdetail0_.bill_no.billProductSet] with element property reference [product] [select count(*) from iland.hbm.BillDetails as bd where bd.billProductSet.product.id=1001 and bd.client.id=1] at org.hibernate.hql.ast.tree.DotNode$1.buildIllegalCollectionDereferenceException(DotNode.java:68) at org.hibernate.hql.ast.tree.DotNode.checkLhsIsNotCollection(DotNode.java

I'm struggling with the pointer sign *, I find it very confusing in how it's used in both declarations and expressions. For example: int *i; // i is a pointer to an int But what is the logic behind the syntax? What does the * just before the i mean? Let's take the following example. Please correct me where I'm wrong: char **s; char *(*s); // added parentheses to highlight precedence And this is where I lose track. The *s between the parantheses means: s is a pointer? But a pointer to what? And what does the * outside the parentheses mean: a pointer to what s is pointing? So the meaning of this