I was wondering what the difference in is between calling a method like :
int x;
mymethod(x);
and
mymethod(&x);
Because C always does call-by-value, if you want the function to be able to change the x inside the function itself, you have to pass the address of x.
mymethod(x);
will pass x, for example if x is 2, you could as well have written mymethod(2)
mymethod(&x)
will pass the address to x. Now the method can change the value stored at that address, so after the function is completed, the actual value of x might have been changed.
Now you can also declare a pointer:
int* y; //y is now a pointer to a memory address
y = &x; //y now points to the memory address of x;
*y = 5; will set the "value found at the address y" to 5, thus will set x to 5;
mymethod(x) passes in the integer x as a parameter. mymethod(&x) passes in the address in memory of x. If you required a pointer-to-int as an argument, then you would use the second one.
In few words, when you declare a pointer, you precede it by an asterisk:
int *ptr;
When you pass &x instead of x, you are passing the memory address.
Please, read this useful introduction to the pointers.
Regards.
In the line:
int x;
you allocate a piece of memory in the size of "int".
for this explanation we will assume the size of int is 4 bytes (it might not be 4 bytes).
so after the line "int x;" there are 4 bytes in the memory assigned to "x".
the value of "x" is inside these 4 bytes:
for example, if x=4 then it will look in the memory like this:
[0, 0, 0, 4] or in binary [0000000, 00000000, 00000000, 00000010].
(In real life it could also be [4, 0, 0, 0], but I won't get into that).
so the VALUE of x is 4.
But lets say I want the address of "x", where it is placed in the memory.
This is where the operator "&" comes into play, using this operator I request the address of x.
so if [0, 0, 0, 4] starts at the location "0x12341234" in the memory, &x will return that (0x12341234).
Now, if I want to store the address in a variable, this variable's type is not "int", but it is something that points to the address of int which is being marked as "int*".
So:
int x = 4; // <-- allocates memory of 4 bytes and fills its value with the number 4.
int* pointer_to_x = &x; // <-- pointer_to_x points to the address where x is located in the memory.
if there is a method declared like so:
void mymethod(int x) than we pass THE VALUE of x, so the method is being called mymethod(x).
if there is a method declared like so:
void mymethod(int* x) than we pass a POINTER to the address of x, so the method is being called mymethod(&x).
It is really the tip of the iceberg, and I really tried to keep it simple as I could, so if you have further questions, just ask!
There are also terms called "by value" and "by reference" but you still need to understand better the difference between int and int* and than "by value" and "by reference" will be quite natural.
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 *.
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.
Given the following Code
int main()
{
int z;
int **x;
int * y;
x = (int**)malloc(sizeof(int*));
y = (int*)malloc(sizeof(int));
*y = 6;
*x = y; // Point 1
z = 3;
*x = &z; // Point 2
}
I am to draw box-and-circle diagrams of variables at point 1 and point 2.
Following is what i have got point 1.
Following is what i have got for point 2.
can anyone confirm if my approach is correct and my solution? sorry i am new to pointers and c.
Let's take this step by step. First, we reserve a few locations on the stack for the variables.
Next, allocate a small block the size of int pointer.
The newly allocated block should eventually be assigned an address of an int since X is a pointer to a pointer to an int. Next, allocate another small block.
Now put the address of y into the location pointed to by x
Lastly, assign 3 to z and change the value that x is pointing to, which will now be the address of z.
Hope this helps.
At point 1:
a. y contains the address of a dynamically allocated block of memory (let's call this "Block A") that contains the value 6
b. x contains the address of a different dynamically allocated block of memory (let's call this "Block B"), and "Block B" contains the address of "Block A".
c. z is an uninitialized int
At point 2:
a. y is unchanged from point 1
b. z now contains the value 3
c. x still contains the address "Block B", but "Block B" now contains the address of z rather than the address of "Block A".
Diagrammatically, where circles are variables with automatic storage duration (i.e. x, y and z) and rectangles are dynamically allocated blocks of memory:
NO!! This should clear your doubt. Consider the address(y)=300, address(z)=400. content(300)=6, content(400)=3
*x = y; // Point 1
The above statement says that the content of pointer x will be the address of y. So the pointer x is pointing to pointer y. (i.e. content(x)=300 )
*x = &z; // Point 2
The above statement says that the content of pointer x will be the address of z. So the pointer x is pointing to pointer z. (i.e. content(x)=400 )
NOTE: In order to access the values you have to do **x which will finally access the values.
We know that (*) is dereference operator.
Now lets try to access the values using above terminology.
Point 1: **x => *(*x) => *(address(y)) => *(300) => content(300) => 6.
Point 2: **x => *(*x) => *(address(z)) => *(400) => content(400) => 3.
I'm not wholly familiar with this, but x is of type pointer-to-pointer-to-int - two layers of indirection. Y is of type pointer-to-int - one layer. However, the diagram shows them both having the same indirection from the underlying storage. You do this for the second diagram where you take the address of z, but taking the address of z is the same level of indirection as assigning y to *x, as in both cases, you're telling x to point to a memory address that points directly to memory.
So in short, I think your "point 1" diagram should be similar to your "point 2" diagram, with x's "pointer box" pointing to y's "pointer box".
Edit: I'm also not sure about the "point 2" diagram in that it seems to imply that x is pointer-to-int, as its pointer points directly to z's memory - but it's been a long time since I've seen box-and-pointer diagrams (which this essentially seems to be) to know whether there's an alternative way to add that layer of indirection in. I actually think "Z" and "3" should be in the same box as in box-and-pointer diagrams, since z directly refers to the underlying storage.
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.