Ptr->Hello(); /* Versus */ (*Ptr).Hello();

Pointer expressions: *ptr++, *++ptr and ++*ptr

Here's a detailed explanation which I hope will be helpful. Let's begin with your program, as it's the simplest to explain.

int main()
{
    char *p = "Hello";
    while(*p++)
        printf("%c",*p);
    return 0;
}

The first statement:

char* p = "Hello";

declares p as a pointer to char. When we say "pointer to a char", what does that mean? It means that the value of p is the address of a char; p tells us where in memory there is some space set aside to hold a char.

The statement also initializes p to point to the first character in the string literal "Hello". For the sake of this exercise, it's important to understand p as pointing not to the entire string, but only to the first character, 'H'. After all, p is a pointer to one char, not to the entire string. The value of p is the address of the 'H' in "Hello".

Then you set up a loop:

while (*p++)

What does the loop condition *p++ mean? Three things are at work here that make this puzzling (at least until familiarity sets in):

1. The precedence of the two operators, postfix ++ and indirection *
2. The value of a postfix increment expression
3. The side effect of a postfix increment expression

1. Precedence. A quick glance at the precedence table for operators will tell you that postfix increment has a higher precedence (16) than dereference / indirection (15). This means that the complex expression *p++ is going to be grouped as: *(p++). That is to say, the * part will be applied to the value of the p++ part. So let's take the p++ part first.

2. Postfix expression value. The value of p++ is the value of p before the increment. If you have:

int i = 7;
printf ("%d\n", i++);
printf ("%d\n", i);

the output will be:

7
8

because i++ evaluates to i before the increment. Similarly p++ is going to evaluate to the current value of p. As we know, the current value of p is the address of 'H'.

So now the p++ part of *p++ has been evaluated; it's the current value of p. Then the * part happens. *(current value of p) means: access the value at the address held by p. We know that the value at that address is 'H'. So the expression *p++ evaluates to 'H'.

Now hold on a minute, you're saying. If *p++ evaluates to 'H', why doesn't that 'H' print in the above code? That's where side effects come in.

3. Postfix expression side effects. The postfix ++ has the value of the current operand, but it has the side effect of incrementing that operand. Huh? Take a look at that int code again:

int i = 7;
printf ("%d\n", i++);
printf ("%d\n", i);

As noted earlier, the output will be:

7
8

When i++ is evaluated in the first printf(), it evaluates to 7. But the C standard guarantees that at some point before the second printf() begins executing, the side effect of the ++ operator will have taken place. That is to say, before the second printf() happens, i will have been incremented as a result of the ++ operator in the first printf(). This, by the way, is one of the few guarantees the standard gives about the timing of side effects.

In your code, then, when the expression *p++is evaluated, it evaluates to 'H'. But by the time you get to this:

printf ("%c", *p)

that pesky side-effect has occurred. p has been incremented. Whoa! It no longer points to 'H', but to one character past 'H': to the 'e', in other words. That explains your cockneyfied output:

ello

Hence the chorus of helpful (and accurate) suggestions in the other answers: to print the Received Pronunciation "Hello" and not its cockney counterpart, you need something like

while (*p)
    printf ("%c", *p++);

So much for that. What about the rest? You ask about the meanings of these:

*ptr++
*++ptr
++*ptr

We just talked about the first, so let's look at the second: *++ptr.

We saw in our earlier explanation that postfix increment p++ has a certain precedence, a value, and a side effect. The prefix increment ++p has the same side effect as its postfix counterpart: it increments its operand by 1. However, it has a different precedence and a different value.

The prefix increment has lower precedence than the postfix; it has precedence 15. In other words, it has the same precedence as the dereference / indirection operator *. In an expression like

*++ptr

what matters is not precedence: the two operators are identical in precedence. So associativity kicks in. The prefix increment and the indirection operator have right-left associativity. Because of that associativity, the operand ptr is going to be grouped with the rightmost operator ++ before the operator more to the left, *. In other words, the expression is going to be grouped *(++ptr). So, as with *ptr++ but for a different reason, here too the * part is going to be applied to the value of the ++ptr part.

So what is that value? The value of the prefix increment expression is the value of the operand after the increment. This makes it a very different beast from the postfix increment operator. Let's say you have:

int i = 7;
printf ("%d\n", ++i);
printf ("%d\n", i);

The output will be:

8
8

... different from what we saw with the postfix operator. Similarly, if you have:

char* p = "Hello";
printf ("%c ", *p);    // note space in format string
printf ("%c ", *++p);  // value of ++p is p after the increment
printf ("%c ", *p++);  // value of p++ is p before the increment
printf ("%c ", *p);    // value of p has been incremented as a side effect of p++

the output will be:

H e e l                // good dog

Do you see why?

Now we get to the third expression you asked about, ++*ptr. That's the trickiest of the lot, actually. Both operators have the same precedence, and right-left associativity. This means the expression will be grouped ++(*ptr). The ++ part will be applied to the value of the *ptr part.

So if we have:

char q[] = "Hello";
char* p = q;
printf ("%c", ++*p);

the surprisingly egotistical output is going to be:

I

What?! Okay, so the *p part is going to evaluate to 'H'. Then the ++ comes into play, at which point, it's going to be applied to the 'H', not to the pointer at all! What happens when you add 1 to 'H'? You get 1 plus the ASCII value of 'H', 72; you get 73. Represent that as a char, and you get the char with the ASCII value of 73: 'I'.

That takes care of the three expressions you asked about in your question. Here is another, mentioned in the first comment to your question:

(*ptr)++ 

That one is interesting too. If you have:

char q[] = "Hello";
char* p = q;
printf ("%c", (*p)++);
printf ("%c\n", *p);

it will give you this enthusiastic output:

HI
    

What's going on? Again, it's a matter of precedence, expression value, and side effects. Because of the parentheses, the *p part is treated as a primary expression. Primary expressions trump everything else; they get evaluated first. And *p, as you know, evaluates to 'H'. The rest of the expression, the ++ part, is applied to that value. So, in this case, (*p)++ becomes 'H'++.

What is the value of 'H'++? If you said 'I', you've forgotten (already!) our discussion of value vs. side effect with postfix increment. Remember, 'H'++ evaluates to the current value of 'H'. So that first printf() is going to print 'H'. Then, as a side effect, that 'H' is going to be incremented to 'I'. The second printf() prints that 'I'. And you have your cheery greeting.

All right, but in those last two cases, why do I need

char q[] = "Hello";
char* p = q;

Why can't I just have something like

char* p = "Hello";
printf ("%c", ++*p);   // attempting to change string literal!

Because "Hello" is a string literal. If you try ++*p, you're trying to change the 'H' in the string to 'I', making the whole string "Iello". In C, string literals are read-only; attempting to modify them invokes undefined behavior. "Iello" is undefined in English as well, but that's just coincidence.

Conversely, you can't have

char p[] = "Hello";
printf ("%c", *++p);  // attempting to modify value of array identifier!

Why not? Because in this instance, p is an array. An array is not a modifiable l-value; you can't change where p points by pre- or post- increment or decrement, because the name of the array works as though it's a constant pointer. (That's not what it actually is; that's just a convenient way to look at it.)

To sum up, here are the three things you asked about:

*ptr++   // effectively dereferences the pointer, then increments the pointer
*++ptr   // effectively increments the pointer, then dereferences the pointer
++*ptr   // effectively dereferences the pointer, then increments dereferenced value

And here's a fourth, every bit as much fun as the other three:

(*ptr)++ // effectively forces a dereference, then increments dereferenced value

The first and second will crash if ptr is actually an array identifier. The third and fourth will crash if ptr points to a string literal.

There you have it. I hope it's all crystal now. You've been a great audience, and I'll be here all week.

What is the rationale for difference between - and . in c/c++?

When you have a language that is, at its core, intended to be a "portable assembler" you don't try to hide implementation details from the user. Of course the compiler can figure out that in the expression a.b the a in question is a pointer. Can you? Reliably? All the time? In hairy, complicated code? And can you not envision a circumstance where not being able to quickly note that a is a pointer could be a problem?

Stop thinking in terms of "hard to write" and start thinking in terms of "easy to read". (Yeah, I know. This goes against the whole C programmer mythos!) You'll be reading code a couple of orders of magnitude more often than you'll be writing it.

What is the difference between char *ptr = hi; int *ptr = 100?

Unless you're really know what you're doing and have some special plan, what you're doing does not make sense.

int *ptr = "100" is basically just a shortcut for this:

char *tmp = "100";
int  *ptr = tmp;

To avoid the warning, you need to do this:

int  *ptr = (int*) tmp;

But don't do this.

C++ member access/indirection operator equivalence

The conversion from E1 -> E2 to (*(E1)).E2 only applies to raw pointer types. For class types, E1 -> E2 evaluates to (E1).operator->().E2, which potentially might recursively expand out even more copies of operator-> if the return type of operator-> is not itself a pointer type. You can see this by creating a type that supports operator* but not operator-> and trying to use the arrow operator on it; you'll get an error that operator-> is undefined.

As a follow-up, it's common to implement operator -> in terms of operator * in a way that makes the semantics of -> match the semantics for pointers. You often see things like this:

PointerType ClassType::operator-> () const {
    return &**this;
}

This expression is interpreted as

&(*(*this)),

meaning "take this object (*this), dereference it (*(*this)), and get the address of what you find (&(*(*this)).)." Now, if you use the rule that E1 -> E2 should be equivalent to (*(E1)).E2, you can see that you end up getting something equivalent.

Correct usage of free() function in C

You should call free only on pointers which have been assigned memory returned by malloc, calloc, or realloc.

char* ptr = malloc(10); 

// use memory pointed by ptr 
// e.g., strcpy(ptr,"hello");

free(ptr); // free memory pointed by ptr when you don't need it anymore

Things to keep in mind:

• Never free memory twice. This can happen for example if you call free on ptr twice and value of ptr wasn't changed since first call to free. Or you have two (or more) different pointers pointing to same memory: if you call free on one, you are not allowed to call free on other pointers now too.

• When you free a pointer you are not even allowed to read its value; e.g., if (ptr) not allowed after freeing unless you initialize ptr to a new value

• You should not dereference freed pointer

• Passing null pointer to free is fine, no operation is performed.

How do function pointers in C work?

Function pointers in C

Let's start with a basic function which we will be pointing to:

int addInt(int n, int m) {
    return n+m;
}

First thing, let's define a pointer to a function which receives 2 ints and returns an int:

int (*functionPtr)(int,int);

Now we can safely point to our function:

functionPtr = &addInt;

Now that we have a pointer to the function, let's use it:

int sum = (*functionPtr)(2, 3); // sum == 5

Passing the pointer to another function is basically the same:

int add2to3(int (*functionPtr)(int, int)) {
    return (*functionPtr)(2, 3);
}

We can use function pointers in return values as well (try to keep up, it gets messy):

// this is a function called functionFactory which receives parameter n
// and returns a pointer to another function which receives two ints
// and it returns another int
int (*functionFactory(int n))(int, int) {
    printf("Got parameter %d", n);
    int (*functionPtr)(int,int) = &addInt;
    return functionPtr;
}

But it's much nicer to use a typedef:

typedef int (*myFuncDef)(int, int);
// note that the typedef name is indeed myFuncDef

myFuncDef functionFactory(int n) {
    printf("Got parameter %d", n);
    myFuncDef functionPtr = &addInt;
    return functionPtr;
}

Reading a String as an int pointer

Your problem is that you're casting the pointer to an int* pointer.. which is 32 bits.. not 16 like the char*.

Therefore, each increment is 32 bits. Here's a picture (praise my artwork if you must):

^{Sorry about the dodgy arrows.. I think my mouse batteries are dying}

When you're reading via a char pointer.. you're reading character by character at 16 bits.

When you cast it to an int pointer.. you're reading at 32-bit increments. That means, ptr[0] is both H and e (but points at the base of the H). ptr[1] is both l's..

That is why you are essentially skipping a character in your output.

When you cast it back to a char here:

Console.WriteLine((char)*ptr);

..only the first 16 bits will result from that conversion, which is the first character in each pair.

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