Use the format specifier %p: The standard requires that the argument is of type void* for %p specifier. Since, printf is a variadic function, there’s no implicit conversion to void * from T * which would happen implicitly for any non-variadic functions in C. Hence, the cast is required. To quote the standard: 7.21.6 Formatted … Read more
I get that compilation error because of this line which intended to increase the pointer by 0x200 (to point to the next segment) I’v seen this but I didn’t use any illegal symbol! P.S. The initialization of the pointer:
Method 1 (using new) Allocates memory for the object on the free store (This is frequently the same thing as the heap) Requires you to explicitly delete your object later. (If you don’t delete it, you could create a memory leak) Memory stays allocated until you delete it. (i.e. you could return an object that you created using new) The example in the question will leak memory unless … Read more
The variables with the * are pointers. A ‘normal’ variable, for example a char or an int, contains the value of that datatype itself – the variable can hold a character, or an integer. A pointer is a special kind of variable; it doesn’t hold the value itself, it contains the address of a value in memory. … Read more
I’ll interpret your question as two questions: 1) why -> even exists, and 2) why . does not automatically dereference the pointer. Answers to both questions have historical roots. Why does -> even exist? In one of the very first versions of C language (which I will refer as CRM for “C Reference Manual“, which came with 6th Edition Unix in … Read more
Because you’re passing the value of the pointer to the method and then dereferencing it to get the integer that is pointed to.
To start, note that is a special pointer ( == memory address) to the class its in. First, an object is instantiated: Next, a pointer is instantiated: Next, the memory address of a is assigned to the pointer b: Next, the method CDummy::isitme(CDummy ¶m) is called: A test is evaluated inside this method: Here’s the tricky part. param is an … Read more
and Are just the same thing. so does: And once you declared the reference, you can think of it just like the actual variable itself. so you don’t need particular operator to dereference it. when passing variables through function arguments, the raw object is always shallow copied in memory to the new variable in the … Read more
foo->bar is equivalent to (*foo).bar, i.e. it gets the member called bar from the struct that foo points to.
When you have an assignment operator in a statement, the LHS of the operator must be something the language calls an lvalue. If the LHS of the operator does not evaluate to an lvalue, the value from the RHS cannot be assigned to the LHS. You cannot use: since 10 does not evaluate to an lvalue. You can use: since i does … Read more