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i am just started learning pointers in c. I have following few doubts. If i find the answers for the below questions. It Will be really useful for me to understand the concept of pointers in c. Thanks in advance.

i)

char *cptr;
int value = 2345;
cptr = (char *)value;

whats the use of (char *) and what it mean in the above code snippet.

ii) 

char *cptr;
int value = 2345;
cptr = value;

This also compiles without any error .then whats the difference between i & ii code snippet

iii) &value is returning address of the variable. Is it a virtual memory address in RAM? Suppose another c program running in parallel, will that program can have same memory address as &value. Will each process can have duplicate memory address same as in other process and it is independent of each other?

iv)

#define MY_REGISTER (*(volatile unsigned char*)0x1234)

void main()
{
    MY_REGISTER=12;
    printf("value in the address tamil  is %d",(MY_REGISTER));
}

The above snippet compiled successfully. But it outputs segmentation fault error. I don't know what's the mistake I am doing. I want to know how to access the value of random address, using pointers. Is there any way? Will program have the address 0x1234 for real?

v) printf("value at the address %d",*(236632));//consider the address 236632 available in

//stack

why does the above printf statement showing error?

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  • For the fifth point, what do you mean by "statement fails"? Commented Nov 20, 2013 at 7:47
  • The first question has a typo in it; cptr is declared, but ptr is used. Commented Nov 20, 2013 at 7:50
  • Removed 3 tags that are unrelated to the question. Commented Nov 20, 2013 at 9:25

4 Answers 4

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  1. That's a type cast, it tells the compiler to treat one type as some other (possibly unrelated) type. As for the result, see point 2 below.

  2. That makes cptr point to the address 2345.

  3. Modern operating systems isolate the processes. The address of one variable in one process is not valid in another process, even if started with the same program. In fact, the second process may have a completely different memory map due to Address Space Layout Randomisation (ASLR).

  4. It's because you try to write to address 0x1234 which might be a valid address on some systems, but not on most, and almost never on a PC running e.g. Windows or Linux.

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3 Comments

as per my knowledge,each running process consider that it has whole access to the entire memory(RAM).let us say two process Aprocess, Bprocess ,consider a address location 0x0000, will both Aprocess & Bprocess will have separate 0x0000 virtual address ?.correct me if am wrong.thanks
@tamil_innov That's the point of virtual addresses, it doesn't have to be the same as a physical address. On system with swap-space, an address doesn't even have to be in physical memory. Also, even if two processes of the same program happen to have the same memory map, it's still two different processes with two separate memory maps, with two separate allocations. Memory is not shared between processes.
@JoachimPileborg: That's a bit strong. In particular, the memory used for code is often shared. It's read-only anyway.
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i)

(char *) means, that you cast the data stored in value to a pointer ptr, which points to a char. Which means, that ptr points to the memory location 2345. In your code snipet ptr is undefined though. I guess there is more in that program.

ii)

The difference is, that you now write to cptr, which is (as you defined) a pointer pointing to a char. There is not much of a difference as in i) except, that you write to a different variable, and that you use a implicit cast, which gets resolved by the compiler. Again, cptr points now to the location 2345 and expects there to be a char

iii)

Yes you can say it is a virtual address. Also segmentation plays some parts in this game, but at your stage you don't need to worry about it at all. The OS will resolve that for you and makes sure, that you only overwrite variables in the memory space dedicated to your program. So if you run a program twice at the same time, and you print a pointer, it is most likely the same value, but they won't point at the same value in memory.

iv)

Didn't see the write instruction at first. You can't just write anywhere into memory, as you could overwrite another program's value.

v)

Similar issue as above. You cannot just dereference any number you want to, you first need to cast it to a pointer, otherwise neither the compiler, your OS nor your CPU will have a clue, to what exactely it is pointing to

Hope I could help you, but I recommend, that you dive again in some books about pointers in C.

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i.) Type cast, you cast the integer to a char

ii.) You point to the address of 2345.

iii.) Refer to answer from Joachim Pileborg. ^ ASLR

iv.) You can't directly write into an address without knowing if there's already something in / if it even exists.

v.) Because you're actually using a pointer to print a normal integer out, which should throw the error C2100: illegal indirection.

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You may think pointers like numbers on mailboxes. When you set a value to a pointer, e.g cptr = 2345 is like you move in front of mailbox 2345. That's ok, no actual interaction with the memory, hence no crash. When you state something like *cptr, this refers to the actual "content of the mailbox". Setting a value for *cptr is like trying to put something in the mailbox in front of you (memory location). If you don't know who it belongs to (how the application uses that memory), it's probably a bad idea. You could use "malloc" to initialize a pointer / allocate memory, and "free" to cleanup after you finish the job.

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