It is a pointer to a char. When declaring a pointer, the asterisk goes after the type and before the identifier, with whitespace being insignificant. These all declare char pointers: To make things even more confusing, when declaring multiple variables at once, the asterisk only applies to a single identifier (on its right). E.g.: It is primarily for … Read more
Using pointer2 or pointer3 produce the same binary except manipulations as ++pointer2 as pointed out by WhozCraig. I recommend using typedef (producing same binary code as above pointer3) Note: Since C++11, you can also use keyword using instead of typedef in your example: Note: If the array tab1 is used within a function body => this array will be placed within the call stack memory. But the stack size is limited. Using … Read more
As written, NULL isn’t defined in your program. Usually, that’s defined in a standard header file — specifically <cstddef> or <stddef.h>. Since you’re restricted to iostream, if yours doesn’t get NULL implicitly from that header, you can use 0 or, in C++11, nullptr, which is a keyword and doesn’t require a header. (It is not recommended to define NULL yourself. It might work sometimes, but it is technically … Read more
Pointer to an integer value Pointer to a pointer to an integer value (Ie, in the second case you will require two dereferrences to access the integer’s value)
At worst, your lecturer is wrong. At best, he was simplifying terminology, and confusing you in the process. This is reasonably commonplace in software education, unfortunately. The truth is, many books get this wrong as well; the array is not “passed” at all, either “by pointer” or “by reference”. In fact, because arrays cannot be passed by … Read more
this means pointer to the object, so *this is an object. So you are returning an object ie, *this returns a reference to the object.
f2 is taking it’s arguments by reference, which is essentially an alias for the arguments you pass. The difference between pointer and reference is that a reference cannot be NULL. With the f you need to pass the address (using & operator) of the parameters you’re passing to the pointer, where when you pass by reference you just pass the parameters and … Read more
Yes, you can initialize pointers to a value on declaration, however you can’t do: & is the address of operator and you can’t apply that to a constant (although if you could, that would be interesting). Try using another variable: or type-casting:
The most straightforward answer is: The difference here is that will place Hello world in the read-only parts of the memory and making s a pointer to that, making any writing operation on this memory illegal. While doing: puts the literal string in read-only memory and copies the string to newly allocated memory on the … Read more
printf(“\nlVptr[60 ] is %d \n”, *(int*)lVptr); This will cast the void pointer to a pointer to an int and then dereference it correctly. If you want to treat it as an array (of one), you could do a slightly ugly ((int *)lVptr)[0]. Using [1] is out of bounds, and therefore not a good idea (as … Read more