Trouble With Pointer Arithmetic

I’ve attached two short c programs which I’ll be referencing
in the text below, if you have a moment please ctex.zip find in this
LINK>> WeTransfer

Attachment also here:
Attachment 15458

It contains the files:


listing1.6.2Indirection.c and listing1.6.1.c




Regarding listing1.6.2Indirection.c


If ptr is the name given to a pointer
which is pointing to the beginning of the buffer array
i.e. buffer[0].


And I’ve set buffer[0] to 435.


why does:


printf(“ %f \n”, ptr);


print 0.00 to the console? Im expecting it to print 435?
In fact I was initially expecting some long random value
because ptr is just a variable name?


whats adds to my confusion is
also because…


In listing1.6.1.c


ptr is pointing to the last element in buffer[1024];


but when I say:


buffer[1003] = 500;
printf(“%f \n”, ptr);


the console prints 500?


This is a very confusing result when ptr is
pointing to the last element of buffer here.
When I’ve changed a different element althogether?


What am I not getting ?


Many Many thanks

Quote:

---

Originally Posted by Fauveboy

I’ve attached two short c programs which I’ll be referencing

---

Don't do that. Come up with the smallest and simplest (and compilable) program that demonstrates the problem, then post the code of that here in [code][/code] bbcode tags. If you really cannot do that, it would be better to post the code to a pastebin and link to the pastebin entry.

Quote:

---

Originally Posted by Fauveboy

If ptr is the name given to a pointer
which is pointing to the beginning of the buffer array
i.e. buffer[0].


And I’ve set buffer[0] to 435.


why does:


printf(“ %f \n”, ptr);


print 0.00 to the console? Im expecting it to print 435?
In fact I was initially expecting some long random value
because ptr is just a variable name?

---

The correct format specifier to print a pointer is %p, and you would cast the pointer to void*:

Code:

---

printf(" %p \n", (void*)ptr);

---

Having done that, you would expect the address of buffer[0] to be printed (i.e., the "some long random value"), not its value (i.e., 435).

sorry for the previous attachments, ive now posted the code below.
I understand that there is an appropriate formatter for pointers and
by using %p you get a number which is the address of what the pointer
is referencing.


But I’m curious about the following behaviour thats occuring.


where the code reads:


buffer[100] = 765;
…
printf(“%f \n”, ptr);


how does ptr manage to print 765?


if it was to print a value surely
it would be a random one as its set to point to the first
element in buffer?


With my understanding so far I was initially expecting ptr to print
a random long value because I’m not using the dereference syntax “*”
which is used to get any value from it?


What seems to happen when I use *ptr with prinf and %f (occuring as
the last instruction at the end of the code) is that a long random
value prints to the console (opposite to what i expected)




Whys this behaviour happening?

Code:

---



 int main()
  {
      double* ptr;
      double buffer[1024];
 
      buffer[0] = 435;
     
      /* ptr prints 765 why? */
      printf("%f \n", ptr);
 
    /*set ptr to the first elements address of buffer */
      ptr = buffer;
     
      buffer[100] = 765;
 
      /* ptr prints 765 why? */
      printf("%f \n", ptr);
     
      /* this prints the address that ptr points to */
      printf("%p \n", ptr);
 
      /* why is void used here? */
      printf("%p \n", (void*)ptr);
 
      /* ppoint ptr to the addres of the last element in buffer */
      ptr = &buffer[1024];
 
      buffer[1003] = 500;
 
      printf("%f \n", ptr);
 
      printf("%f \n", *ptr);
  }

---

Quote:

---

Originally Posted by Fauveboy

where the code reads:


buffer[100] = 765;
…
printf(“%f \n”, ptr);


how does ptr manage to print 765?

---

That is a mix of two problems:

• At that point ptr had not been given an initial value, so its value is indeterminate, hence any result is to be expected.
• Trying to print a pointer using %f results in undefined behaviour. In practice, what you're looking at is probably some attempt to interpret the bytes of the pointer value (i.e., the address) as a double, and the result ended up as 765. Or it could be something else, but then you would have to say, examine the assembly output, and maybe even examine the source code of the compiler.

Quote:

---

Originally Posted by Fauveboy

With my understanding so far I was initially expecting ptr to print
a random long value because I’m not using the dereference syntax “*”
which is used to get any value from it?

---

Yes, you are right for the second printf call, and that's why you should use %p to print pointers, and %f to print doubles. Mixing up the two merely causes you unnecessary confusion.

Quote:

---

Originally Posted by Fauveboy

What seems to happen when I use *ptr with prinf and %f (occuring as
the last instruction at the end of the code) is that a long random
value prints to the console (opposite to what i expected)

---

You set ptr to &buffer[1024]. That means it points one past the end of the buffer. This is a technique used when traversing a range with pointers (i.e., you use another pointer to point to the first element, then iterate by incrementing this second pointer until its value is equal to the one past the end pointer), but one must take care never to deference the pointer. You then dereferenced the pointer, so of course you ended up with a problem, since your pointer doesn't point to a valid object in the buffer.