What is the difference between *x=i and x=&i?
Code:
int i=2;
int *x;
*x=i; //what is the difference between this...
x=&i; //...and this??
//Also, what happens when I do these? Not really important but curious.
x=i;
*x=*i;
*x=i; //what is the difference between this...
This assigns the value of i to an integer stored at the address pointed to by x
x=&i; //...and this??
This assigns the address of i to x.
Note that in your example x is unassigned, so the behavior of *x=i is undefined.
Here is a better illustration:
int i = 2, j = 5;
printf("%d %d\n", i, j); // Prints 2 5
int *x = &j;
*x = i;
printf("%d %d\n", i, j); // Prints 2 2
int i=2;
int *x;
*x=i; // x is not initialized, this is undefined behavior
x=&i; // this assigns the address of i to x
and
*x=&i; // is invalid C. You cannot assign a pointer to an
// integer without an explicit conversion.
*x=i changes the value stored in the memory location pointed to by x to be equal to the value stored in i.
x=&i changes the memory location x is pointing to to i's memory location.
x=i is wrong. You will most likely get a segfault.
*x=*i is wrong. You cannot dereference i, because i is not a pointer.
*x=&i (actually, more properly, *x=(int)&i) will store the memory location of i as an integer in the memory location pointed to by x.
*x=i; dereferences the pointer x and assigns i.
x=&i makes the pointer x point at the variable i.
*x = i changes the value stored at the address which is stored in x. In your case, unedited, that will crash, because the address x is likely junk, possibly NULL. You would need a malloc() or new or something.
x = &i changes the address stored in x, so that is is the address of the variable i. This is perfectly safe in your example.
Well, when you say
*x = i
You're saying: make the variable x points to the value i.
When you say
x = &i
You're saying make the address x points to the address of i.
And I guess you should be able to figure the other ones out by yourself!
*x = i
This actually assigns the value of i to the memory location pointed to by x.
x = &i
This assigns the address of variable i to a pointer variable x. x should be a pointer.
When you do x = i, this will give you a runtime error, as you are trying to erroneously assign an address(which in this case is 2) which does not belong to the address space of your process. To successfully do this, i should also be a pointer and must point to an address which is in your address space.
When you do *x = *i, in your case will again give an error. If i is a pointer, then the address pointed to by x will get the value present at the address pointed to by i.
*x = i assigns 2 to the memory address pointed to by x. Note that this would probably crash because x hasn't been initialized to a memory address via malloc or an assignment to a buffer or address of a stack variable.
x = &i assigns the pointer x to the address of variable i.
x = i would assign the pointer x to the value of 2, which would most likely point to an invalid memory address and would require a cast.
*x = *i would depend on the current value of x. And since i is not a pointer you cannot dereference it.
*x = &i would write the address of i to the memory address pointed to by x, which would depend on the code preceding it. It would likely crash if you didn't assign x to a valid address.
Some of these calls would require a cast to be syntactically correct.
Related
Im fairly new to C programming and I am confused as to how pointers work. How do you use ONLY pointers to copy values for example ... use only pointers to copy the value in x into y.
#include <stdio.h>
int main (void)
{
int x,y;
int *ptr1;
ptr1 = &x;
printf("Input a number: \n");
scanf("%d",&x);
y = ptr1;
printf("Y : %d \n",y);
return 0;
}
It is quite simple. & returns the address of a variable. So when you do:
ptr1 = &x;
ptr1 is pointing to x, or holding variable x's address.
Now lets say you want to copy the value from the variable ptr1 is pointing to. You need to use *. When you write
y = ptr1;
the value of ptr1 is in y, not the value ptr1 was pointing to. To put the value of the variable, ptr1 is pointing to, use *:
y = *ptr1;
This will put the value of the variable ptr1 was pointing to in y, or in simple terms, put the value of x in y. This is because ptr1 is pointing to x.
To solve simple issues like this next time, enable all warnings and errors of your compiler, during compilation.
If you're using gcc, use -Wall and -Wextra. -Wall will enable all warnings and -Wextra will turn all warnings into errors, confirming that you do not ignore the warnings.
What's a pointer??
A pointer is a special primitive-type in C. As well as the int type stored decimals, a pointer stored memory address.
How to create pointers
For all types and user-types (i.e. structures, unions) you must do:
Type * pointer_name;
int * pointer_to_int;
MyStruct * pointer_to_myStruct;
How to assing pointers
As I said, i pointer stored memory address, so the & operator returns the memory address of a variable.
int a = 26;
int *pointer1 = &a, *pointer2, *pointer3; // pointer1 points to a
pointer2 = &a; // pointer2 points to a
pointer3 = pointer2; // pointer3 points to the memory address that pointer2 too points, so pointer3 points to a :)
How to use a pointer value
If you want to access to the value of a pointer you must to use the * operator:
int y = *pointer1; // Ok, y = a. So y = 25 ;)
int y = pointer1; // Error, y can't store memory address.
Editing value of a variable points by a pointer
To change the value of a variable through a pointer, first, you must to access to the value and then change it.
*pointer1++; // Ok, a = 27;
*pointer1 = 12; // Ok, a = 12;
pointer1 = 12; // Noo, pointer1 points to the memory address 12. It's a problem and maybe it does crush your program.
pointer1++; // Only when you use pointer and arrays ;).
Long Winded Explanation of Pointers
When explaining what pointers are to people who already know how to program, I find that it's really easy to introduce them using array terminology.
Below all abstraction, your computer's memory is really just a big array, which we will call mem. mem[0] is the first byte in memory, mem[1] is the second, and so forth.
When your program is running, almost all variables are stored in memory somewhere. The way variables are seen in code is pretty simple. Your CPU knows a number which is an index in mem (which I'll call base) where your program's data is, and the actual code just refers to variables using base and an offset.
For a hypothetical bit of code, let's look at this:
byte foo(byte a, byte b){
byte c = a + b;
return c;
}
A naive but good example of what this actually ends up looking like after compiling is something along the lines of:
Move base to make room for three new bytes
Set mem[base+0] (variable a) to the value of a
Set mem[base+1] (variable b) to the value of b
Set mem[base+2] (variable c) to the sum mem[base+0] + mem[base+1]
Set the return value to mem[base+2]
Move base back to where it was before calling the function
The exact details of what happens is platform and convention specific, but will generally look like that without any optimizations.
As the example illustrates, the notion of a b and c being special entities kind of goes out the window. The compiler calculates what offset to give the variables when generating relevant code, but the end result just deals with base and hard-coded offsets.
What is a pointer?
A pointer is just a fancy way to refer to an index within the mem array. In fact, a pointer is really just a number. That's all it is; C just gives you some syntax to make it a little more obvious that it's supposed to be an index in the mem array rather than some arbitrary number.
What a does referencing and dereferencing mean?
When you reference a variable (like &var) the compiler retrieves the offset it calculated for the variable, and then emits some code that roughly means "Return the sum of base and the variable's offset"
Here's another bit of code:
void foo(byte a){
byte bar = a;
byte *ptr = &bar;
}
(Yes, it doesn't do anything, but it's for illustration of basic concepts)
This roughly translates to:
Move base to make room for two bytes and a pointer
Set mem[base+0] (variable a) to the value of a
Set mem[base+1] (variable bar) to the value of mem[base+0]
Set mem[base+2] (variable ptr) to the value of base+1 (since 1 was the offset used for bar)
Move base back to where it had been earlier
In this example you can see that when you reference a variable, the compiler just uses the memory index as the value, rather than the value found in mem at that index.
Now, when you dereference a pointer (like *ptr) the compiler uses the value stored in the pointer as the index in mem. Example:
void foo(byte* a){
byte value = *a;
}
Explanation:
Move base to make room for a pointer and a byte
Set mem[base+0] (variable a) to the value of a
Set mem[base+1] (variable value) to mem[mem[base+0]]
Move base back to where it started
In this example, the compiler uses the value in memory where the index of that value is specified by another value in memory. This can go as deep as you want, but usually only ever goes one or two levels deep.
A few notes
Since referenced variables are really just numbers, you can't reference a reference or assign a value to a reference, since base+offset is the value we get from the first reference, which is not stored in memory, and thus we cannot get the location where that is stored in memory. (&var = value; and &&var are illegal statements). However, you can dereference a reference, but that just puts you back where you started (*&var is legal).
On the flipside, since a dereferenced variable is a value in memory, you can reference a dereferenced value, dereference a dereferenced value, and assign data to a dereferenced variable. (*var = value;, &*var, and **var are all legal statements.)
Also, not all types are one byte large, but I simplified the examples to make it a bit more easy to grasp. In reality, a pointer would occupy several bytes in memory on most machines, but I kept it at one byte to avoid confusing the issue. The general principle is the same.
Summed up
Memory is just a big array I'm calling mem.
Each variable is stored in memory at a location I'm calling varlocation which is specified by the compiler for every variable.
When the computer refers to a variable normally, it ends up looking like mem[varlocation] in the end code.
When you reference the variable, you just get the numerical value of varlocation in the end code.
When you dereference the variable, you get the value of mem[mem[varlocation]] in the code.
tl;dr - To actually answer the question...
//Your variables x and y and ptr
int x, y;
int *ptr;
//Store the location of x (x_location) in the ptr variable
ptr = &x; //Roughly: mem[ptr_location] = x_location;
//Initialize your x value with scanf
//Notice scanf takes the location of (a.k.a. pointer to) x to know where
//to put the value in memory
scanf("%d", &x);
y = *ptr; //Roughly: mem[y_location] = mem[mem[ptr_location]]
//Since 'mem[ptr_location]' was set to the value 'x_location',
//then that line turns into 'mem[y_location] = mem[x_location]'
//which is the same thing as 'y = x;'
Overall, you just missed the star to dereference the variable, as others have already pointed out.
Simply change y = ptr1; to y = *ptr1;.
This is because ptr1 is a pointer to x, and to get the value of x, you have to dereference ptr1 by adding a leading *.
For the code below
int main()
{
int x = 2;
int *ip = &x;
printf("%d",*ip); // Printing the value of *ip gives 2
//Now if the value of ip is incremented
ip++;
printf("%d",*ip); //Printing the value of *ip gives the incremented memory
}
Can someone please explain as how the value of *ip is getting printed as the incremented memory location. *ip being the deferencing operator should return the value at the address right?
I see two things wrong with the code. One is the line:
int *ip = x;
This attempts to assign the value of an int to a pointer. This can be forced with compiler flags, but should give an error. Nevertheless, I suspect this may be a typo in your question, since dereferencing an address of 2 should almost certainly crash, but you've claimed it ran and produced an output of 2. I'm going to assume you actually have:
int *ip = &x;
The second problem is:
ip++;
Here you are incrementing the pointer, rather than the int it points to. I think what you intended (to increment 2 to 3) would be:
(*ip)++;
That is, increment the value the pointer points to.
If you increment the pointer, it will now point to somewhere else on your stack, which could be another local variable (like the pointer itself), or the return address, or the stack frame pointer, or some other hidden value. There is a good chance the new value will be some sort of pointer which a value similar to the address of your pointer, or will otherwise look like the output of printing a pointer. Naturally, you should not be incrementing a pointer that points to a single element. It's never a good idea.
This code may behave unpredictably, because
int *ip = x;
assigns the value of x, to a pointer.
Basically it hard codes 2 into the pointer ip.
It may compile with warnings.
It may later lead to memory errors at run time.
I'm very new to C. I come from a Java background, and I'm having a hard time understanding pointers. My understanding of what *x = 1 is take the memory address of x and assign it to 1, where as x = 1 means assign the variable x to the value 1.
Am I correct?
Well, as written you have it completely backwards, since you're saying assign x to the value 1 etc.
x=1 means store the value 1 to the variable x.
*x=1 means store the value 1 at the memory address x points to.
*x = 1;
means that x contains a memory address, assign 1 to that memory address.
x = 1;
means assign 1 to the variable x.
x = VARIABLE
*x = POINTER TO ADDRESS
By saying x=1 means you are directly assigning the value 1 into the x. Where as *x=1 has a bit different approach.
Say, int y = 10 and x is a pointer which is pointing to the address of y by defining int *x = &y. After declaration of *x, through out the program, *x will be treated as the value at the address of y. So, when *x=1 is used, the the value at the address of y, which was 10 earlier, will be changed to 1 now. Thus, y=1 and *x=1 are internally doing same thing.
You can follow this link for detailed understanding in simpler way about C pointers.
In the question Find size of array without using sizeof in C the asker treats an int array like an array of int arrays by taking the address and then specifying an array index of 1:
int arr[100];
printf ("%d\n", (&arr)[1] - arr);
The value ends up being the address of the first element in the "next" array of 100 elements after arr. When I try this similar code it doesn't seem to do the same thing:
int *y = NULL;
printf("y = %d\n", y);
printf("(&y)[0] = %d\n", (&y)[0]);
printf("(&y)[1] = %d\n", (&y)[1]);
I end up getting:
y = 1552652636
(&y)[0] = 1552652636
(&y)[1] = 0
Why isn't (&y)[1] the address of the "next" pointer to an int after y?
Here:
printf("(&y)[1] = %d\n", (&y)[1]);
You say first: take address of y. Then afterwards you say: add 1 times so many bytes as the size of the thing which is pointed to - which is pointer to int, and hence probably 4 bytes are added - and dereference whatever is that on that address. But you don't know what is on that new memory address and you can't/shouldn't access that.
Arrays are not pointers, and pointers are not arrays.
The "array size" code calculates the distance between two arrays, which will be the size of an array.
Your code attempts to calculate the distance between two pointers, which should be the size of a pointer.
I believe the source of confusion is that (&y)[1] is the value of the "next" pointer to an int after y, not its address.
Its address is &y + 1.
In the same way, the address of y is &y, and (&y)[0] - or, equivalently *(&y) - is y's value.
(In the "array size" code, (&arr)[1] is also the "next" value, but since this value is an array, it gets implicitly converted to a pointer to the array's first element — &((&array)[1])[0].)
If you run this:
int *y = NULL;
printf("y = %p\n", y);
printf("&y = %p\n", &y + 0);
printf("&y + 1 = %p\n", &y + 1);
the output looks somewhat like this:
y = (nil)
&y = 0xbf86718c
&y + 1 = 0xbf867190
and 0xbf867190 - 0xbf86718c = 4, which makes sense with 32-bit pointers.
Accessing (&y)[1] (i.e. *(&y + 1)) is undefined and probably results in some random garbage.
Thanks for the answers, I think the simplest way to answer the question is to understand what the value of each expression is. First we must know the type, then we can determine the value
I used the c compiler to generate a warning by assigning the values to the wrong type (a char) so I could see exactly what it thinks the types are.
Given the declaration int arr[100], the type of (&arr)[1] is int [100].
Given the declaration int *ptr, the type of (&ptr)[1] is int *.
The value of a int[100] is the constant memory address of where the array starts. I don't know all the history of why that is exactly.
The value of a int * on the other hand is whatever memory address that pointer happens to be holding at the time.
So they are very different things. To get the constant memory address of where a pointer starts you must dereference it. So &(&ptr)[1] is int ** which the constant memory address of where the int pointer starts.
I'm experiencing a mind-blowing doubt concerning the use of pointers in C. So, I've searched a lot about this, but no satisfatory answer was presented to me. Here is the thing:
I declare a pointer of type INT, and a variable of type INT (e.g int x, *pointer). So, let's suppose that both of them occupy sequential addresses in RAM, like 0x102 and 0x106, respectively. No surprises so far. Then, I declare x = 5. My memory map should be like this, shouldn't?
int x, *pointer;
x = 5;
Ok. In the college I learned to assign a pointer this way:
pointer = &x;
And my memory map should be like this:
So far so good. But the question is: if, instead of above, I assign a pointer like this:
*pointer = x;
The memory address of 'x' shouldn't be stored in the pointer's memory address? I always wondered something like the "Memory Map 2", but the result is the same of the "Memory Map 1", that is, the 0x106 address holds a garbage number. So how the program KNOWS where I'd like to point to, if the memory address of 'x' isn't stored at pointer's memory address? Where this information is stored?
It looks like a simple question, but I can't understand. :(
Thanks in advance! :)
Getting how pointers work is tricky. Here's something that might help.
You wrote
int x, *pointer;
which is not just idiomatic; this is telling you something important. It is telling you that the expression x is a variable that can hold an integer; that is hopefully clear. It is also telling you that the expression *pointer is also a variable that can hold an integer.
When you say
x = 123;
that means "store the value 123 in the variable x".
And so when you say
*pointer = 456;
that means "store the value 456 in the variable *pointer".
When you say
pointer = &x;
that means "the expression *pointer -- which remember is a variable that can hold an integer -- is the same variable as x". They are aliases -- two names for the same variable.
So your question is:
How does the program know where I'd like to point to, if the memory address of x isn't stored at pointer's memory address?
Let me rephrase that question using the terminology I've established:
How does the program know which variable *pointer refers to if I do not initialize pointer?
It does not know. If you say:
int x;
printf("%d", x);
then you can get any integer printed; this is undefined behavior. You haven't said what value you want the variable x to have, so it can have any value. When you say:
int *pointer;
*pointer = 123;
Then you are saying "store 123 in variable *pointer", but you haven't said what variable *pointer is. So, just as x can have any value, *pointer can be any variable. Again, we have undefined behavior.
Is that now clear?
int x, *pointer;
*pointer = x;
This is undefined behaviour, because pointer does not point to a valid memory location.
int x, *pointer;
x = 5;
pointer = &x;
*pointer = x;
The last line of this is basically the same as x = x, because pointer points to x;
Since * is the dereference operator, then you will try to:
store the value of x (since there's no addressof, & operator before its name)
to the memory pointed to by pointer (and not into the pointer itself, which wouldn't make sense anyway), which is indeterminate, since your pointer hasn't yet been initialized.
This is not storing the address of x into pointer; what you want is achieved solely by writing pointer = &x;, nothing else will do that. By the way, because of the assignment to the memory pointed to by an uninitialized pointer, *pointer = x invokes undefined behavior.
Doing
*somePtr = someVar
assigns the value of somevar in the place pointed by somePtr ..
To be more comprehensive,
int x = 2, y =3;
int * pointer;
pointer = &x;
*pointer = y;
asiigns the value of y in x . That's it!