Code
short **p = (short **)malloc(sizeof(short *));
*p = malloc(sizeof(short));
**p = 10;
printf("**p = %d", **p);
Output
**p = 10
In this code, a multiple pointer **p is declared and *p is used without any declaration(maybe it's by **p).
What does *p mean in my case? Sorry for very simple question.
I saw C standard and stack overflow, but I couldn't find out something.
For any array or pointer p and index i, the expression p[i] is exactly equal to *(p + i) (where * is the unary dereference operator, the result of it on a pointer is the value that the pointer is pointing to).
So if we have p[0] that's then exactly equal to *(p + 0), which is equal to *(p) which is equal to *p. Going backwards from that, *p is equal to p[0].
So
*p = malloc(sizeof(short));
is equal to
p[0] = malloc(sizeof(short));
And
**p = 10;
is equal to
p[0][0] = 10;
(**p is equal to *(*(p + 0) + 0) which is equal to *(p[0] + 0) which is then equal to p[0][0])
It's important to note that the asterisk * can mean different things in different contexts.
It can be used when declaring a variable, and then it means "declare as pointer":
int *p; // Declare p as a pointer to an int value
It can be used to dereference a pointer, to get the value the pointer is pointing to:
*p = 0; // Equal to p[0] = 0
And it can be used as the multiplication operator:
r = a * b; // Multiply the values in a and b, store the resulting value in r
short **p = (short **)malloc(sizeof(short *));
This line declares a pointer to a pointer p. Additionally the value of p is set to the return value from malloc. It is equivalent to
short **p;
p = (short **)malloc(sizeof(short *));
The second line
*p = malloc(sizeof(short));
Here *p is the value of p. *p is of type pointer. *p is set to the return value of malloc. It is equivalent to
p[0] = malloc(sizeof(short));
The third line
**p = 10;
**p is the value of the value of p. It is of type short. It is equivalent to
p[0][0] = 10
In effect what the code above does is to allocate a 2D array of short, then allocate memory for the first row, and then set the element p[0][0] to 10.
As a general comment on your code, you should not use typecast in malloc. See Do I cast the result of malloc?
What does *p mean when **p is already declared?
short **p = (short **)malloc(sizeof(short *));
(better written as)
short **p = malloc (sizeof *p);
Declares the pointer-to-pointer-to short p and allocates storage for a signle pointer with malloc and assigns the beginning address for that block of memory to p. See: In C, there is no need to cast the return of malloc, it is unnecessary. See: Do I cast the result of malloc?
*p = malloc(sizeof(short));
(equivalent to)
p[0] = malloc (sizeof *p[0]);
Allocates storage for a single short and assigns the starting address for that block of memory to p[0].
**p = 10;
(equivalent to)
*p[0] = 10;
(or)
p[0][0] = 10;
Assigns the value 10 to the dereference pointer *p[0] (or **p or p[0][0]) updating the value at that memory address to 10.
printf("**p = %d", **p);
Prints the value stored in the block of memory pointed to by p[0] (the value accessed by dereferencing the pointer as *p[0] or **p)
The way to keep this straight in your head, is p is a single pointer of type pointer-to-pointer-to short. There are 2-level of indirection (e.g. pointer-to-pointer). To remove one level of indirection, you use the unary * operator, e.g.
*p /* has type pointer-to short */
or the [..] also acts as a dereference such that:
p[0] /* also has type pointer-to short */
You still have a pointer-to so you must remove one more level of indirection to refernce the value stored at the memory location pointed to by the pointer. (e.g. the pointer holds the address where the short is stored as its value). So you need:
**p /* has type short */
and
*p[0] /* also has type short */
as would
p[0][0] /* also has type short */
The other piece to keep straight is the type controls pointer-arithmetic. So p++ adds 8-bytes to the pointer-to-ponter address so it now points to the next pointer. If you do short *q = (*p)++; (or short *q = p[0]++, adds 2-bytes to the address for the pointer-to-short, soqnow points to the nextshortin the block of memory beginning at*p(orp[0]`). (there is no 2nd short because you only allocated 1 -- but you get the point)
Let me know if you have further questions.
Let me put it in different way,
consider an example,
int x;
int *y = &x;
int **z = &y;
x = 10;
Which simplifies to this,
Note: Only for illustration purpose I have chosen address of x,y,z as 0x1000,0x2000,0x3000 respectively.
What does *p mean in my case?
In short the snippetshort **p = (short **)malloc(sizeof(short *)); is dynamically allocating a pointer to a pointer of type short i.e same asy in my example.
Related
Refer the following C program and while incrementing the pointer (i.e. p), it is correctly incrementing by 4 bytes. While if I try to increment the pointer to pointer (i.e. pp), then same is incrementing by 8 bytes. And I am not understanding why is it happening in this way and may be i have misunderstanding in the concept.
#include <stdio.h>
int main()
{
float a = 5, *p, **pp;
p = &a;
pp = &p;
printf("a=%f, p=%p, pp=%p\n", a, p, pp);
a = a + 1;
p = p + 1;
pp = pp + 1;
printf("a=%f, p=%p, pp=%p\n", a, p, pp);
return 0;
}
output:
a=5.000000, p=0x7ffc93c93374, pp=0x7ffc93c93368
a=6.000000, p=0x7ffc93c93378, pp=0x7ffc93c93370
Pointer arithmetic is done in units of the size of the type that the pointer points to. On your system, sizeof(float) is 4, so incrementing p adds 4 bytes to it. But sizeof(float*) is 8 because it's a 64-bit system, so incrementing pp adds 8 bytes to it.
To append the answer of #Barmar I would like to point out that if you have an array
T a[N];
where T is some type and N is some value then after such a declaration of a pointer like
T *p = a;
The pointer p will point to the first element of the array a. This declaration is equivalent to
T *p = &a[0];
If to increment the pointer p it is naturally to assume that it will point to the second element of the array a that is its value will be the value of the expression &a[1]. S0 you need to add to the original value of the pointer p the value that is equal to the value of the size of an element of the array a that is the value equal to sizeof( T ).
Such a calculation is named the pointer arithmetic.
Thus the expression
p + 1
or
++p
means to add the value sizeof( T ) to the value stored in the pointer p. As a result the pointer expression will point to the next element of the array.
Say I have the following problem:
main(void) {
int * p;
int nums [3] = {1,5,9};
char c [3] = {'s','t','u'};
p = nums [2];
*p = (int) *c;
}
What does the last line mean?
Let's break it down: *p = (int) *c;
c is a char array.
*c is the first element of the char array, because c[0] = *(c+0) = *(c) = *c
(int) *c casts the first element of the char array c to an integer. This is required, because with...
*p = (int) *c you assign the to an integer casted char to the content of pointer p.
This code will not work, or will cause problems if it does.
the line;
p = nums[2];
sets the value of the pointer p to the value 9. This is not likely a legal value for your pointer. If it were, then the memory location 9 would be set to 115 which is the integer value of 's'.
*c → Decay c to pointer-to-first-element, and access the pointed-to value. Same as c[0].
(int) *c → cast that value to int.
*p = (int) *c → assign that to what p points to.
There are many issues in this code, let's address them first.
Firstly, main(void) is not conforming code. You need to change that to int main(void).
Secondly, p = nums [2]; is wrong. p is of type int *, and nums[2] is of type int. You may not assign an int to a int * and expect something fruitful to happen. Maybe what you meant to write is p = &nums[2];. Without this modification, going further will invoke undefined behavior as you will try to access a memory location that is invalid to your program.
Then, coming to your question,
*p = (int) *c;
it basically dereference cNOTE to get the value, then cast it to an int type and store into the memory location pointed by p. However, in C, this casting is not required. The above statement is equivalent to
*p = *c;
anyway.
NOTE: Array name decays to the pointer to the first element of the array, i.e., in this code, using c is the same as &c[0], roughly.
I always see two kinds of functions like the following:
void Function_1(
int** buff
)
{
int* retNb = null;
retNb = (int*) malloc(42 * sizeof(int));
*buff = retNb;
}
void Function_2(
int* retNb
)
{
retNb = (int*) malloc(42 * sizeof(int));
}
What is the difference between function_1 and function_2 ? What are their use case ?
A really big difference, they're really different - by ANY means.
Let's keep it simple:
void Function_1(int** buff)
Parameter in function 1 is a pointer to a pointer to an int named buff and what this function does is the following:
int* retNb = null;
Declaring an int pointer called retNb, the assignment to NULL here is really not necessary, though, because next line is the following:
retNb = (int*) malloc(42 * sizeof(int));
retNb is called to get the value returned from malloc. malloc allocated 42*sizeof(int) spot in memory and returned it to retNb and now it can be treated as a simple array with 42 integers and can be accessed with square brackets [] as retNb[i].
*buff = retNb;
in pointers * is the way to get to the content of the address in the variable (variable == pointer in this case) so when using *buff when buff is **buff (a pointer to a pointer) you're actually asking for the pointer buff points to. Let's say we have the code:
int** myPointerToPointer = NULL;
int* myPointer = malloc(sizeof(int));
*myPointer = 4; //or myPointer[0] = 4;
printf("%d\n", *myPointer);
myPointerToPointer = &myPointer;
printf("%d\n", *(*myPointerToPointer));
printf("%d\n", &myPointer);
printf("%d\n", *myPointerToPointer);
then the output would be:
4
4
SOME_ADDRESS
SAME_ADDRESS
because the first printing is the value of myPointer (4) and the second is the value of the value of myPointerToPointer which value is myPointer which value is 4 :P
Third and fourth outputs are the same two because the address of myPointer is actually where myPointerToPointer is pointing to.
About the second function:
void Function_2(int* retNb)
It gets an int pointer - not a pointer to a pointer - just a pointer - means it contains an address of an integer variable that can be accessed with the opeartor *.
retNb = (int*) malloc(42 * sizeof(int));
this line is assigning dynamic memory (again 42 times sizeof(int)) and now can be treated as a regular int array with 42 spots - BUT that's the only thing it does, it won't have anything point to it, just assign memory and that's it.
Hope you understand :P
In function two u are have a pointer to an array of 42 ints.
int function 1 you set buff to point to this array.
I am a total beginner to C so please, work with my ignorance. Why does a normal pointer
int* ptr = &a; has two spaces in memory (one for the pointer variable and one for the value it points to) and an array pointer int a[] = {5}; only has one memory space (if I print out
printf("\n%p\n", a) I get the same address as if I printed out: printf("\n%p\n", &a).
The question is, shouldn't there be a memory space for the pointer variable a and one for its value which points to the first array element? It does it with the regular pointer int* ptr = &a;
It's a little unclear from your question (and assuming no compiler optimization), but if you first declare a variable and then a pointer to that variable,
int a = 4;
int *p = &a;
then you have two different variables, it makes sense that there are two memory slots. You might change p to point to something else, and still want to refer to a later
int a = 4;
int b = 5;
int *p = &a; // p points to a
// ...
p = &b; // now p points to b
a = 6; // but you can still use a
The array declaration just allocates memory on the stack. If you wanted to do the same with a pointer, on the heap, you would use something like malloc or calloc (or new in c++)
int *p = (int*)malloc(1 * sizeof(int));
*p = 4;
but of course remember to free it later (delete in c++)
free(p);
p = 0;
The main misunderstanding here is that &a return not pointer to pointer as it expected that's because in C language there some difference between [] and * (Explanation here: Difference between [] and *)
If you try to &a if a was an pointer (e.g. int *a) then you obtain a new memory place but when your use a static array (i.e. int a[]) then it return address of the first array element. I'll also try to clarify this by mean of the next code block.
#include <stdio.h>
int main(int argc, char *argv[])
{
// for cycles
int k;
printf("That is a pointer case:\n");
// Allocate memory for 4 bytes (one int is four bytes on x86 platform,
// can be differ for microcontroller e.g.)
int c = 0xDEADBEEF;
unsigned char *b = (unsigned char*) &c;
printf("Value c: %p\n", c);
printf("Pointer to c: %p\n", &c);
printf("Pointer b (eq. to c): %p\n", b);
// Reverse order (little-endian in case of x86)
for (k = 0; k < 4; k++)
printf("b[%d] = 0x%02X\n", k, b[k]);
// MAIN DIFFERENCE HERE: (see below)
unsigned char **p_b = &b;
// And now if we use & one more we obtain pointer to the pointer
// 0xDEADBEEF <-- b <-- &p_b
// This pointer different then b itself
printf("Pointer to the pointer b: %p\n", p_b);
printf("\nOther case, now we use array that defined by []:\n");
int a[] = {5,1};
int *ptr = &a;
// 'a' is array but physically it also pointer to location
// logically it's treat differ other then real pointer
printf("'a' is array: %x\n", a);
// MAIN DIFFERENCE HERE: we obtain not a pointer to pointer
printf("Pointer to 'a' result also 'a'%x\n", &a);
printf("Same as 'a': %x\n", ptr);
printf("Access to memory that 'a' pointes to: \n%x\n", *a);
return 0;
}
This is very simple. In first case,
int* ptr = &a;
you have one variable a already declared and hence present in memory. Now you declare another variable ptr (to hold the address, in C variables which hold address of another variable are called pointers), which again requires memory in the same way as a required.
In second case,
int a[] = {5};
You just declare one variable (which will hold a collection of ints), hence memory is allocated accordingly for a[].
In this expression, int* p = &a; p has only one memory location, of the WORD size of your CPU, most probably, and it is to store the address (memory location) of another variable.
When you do *p you are dereferencing p, which means you are getting the value of what p points to. In this particular case that would be the value of a. a has its own location in memory, and p only points to it, but does not itself store as content.
When you have an array, like int a[] = {5};, you have a series (or one) of memory locations, and they are filled with values. These are actual locations.
Arrays in C can decay to a pointer, so when you printf like you did with your array, you get the same address, whether you do a or &a. This is because of array to pointer decay.
a is still the same location, and is only that location. &a actually returns a pointer to a, but that pointer sits else where in memory. If you did int* b = &a; then b here would not have the same location as a, however, it would point to a.
ptr is a variable containing a memory address. You can assign various memory addresses to ptr. a is a constant representing a fixed memory address of the first element of the array. As such you can do:
ptr = a;
but not
a = ptr;
Pointers point to an area in memory. Pointers to int point to an area large enough to hold a value of int type.
If you have an array of int and make a pointer point to the array first element
int array[42];
int *p = array;
the pointer still points to a space wide enough for an int.
On the other hand, if you make a different pointer point to the whole array, this new pointer points to a larger area that starts at the same address
int (*q)[42]; // q is a pointer to an array of 42 ints
q = &array;
the address of both p and q is the same, but they point to differently sized areas.
I am working with C and I'm a bit rusty. I am aware that * has three uses:
Declaring a pointer.
Dereferencing a pointer.
Multiplication
However, what does it mean when there are two asterisks (**) before a variable declaration:
char **aPointer = ...
Thanks,
Scott
It declares a pointer to a char pointer.
The usage of such a pointer would be to do such things like:
void setCharPointerToX(char ** character) {
*character = "x"; //using the dereference operator (*) to get the value that character points to (in this case a char pointer
}
char *y;
setCharPointerToX(&y); //using the address-of (&) operator here
printf("%s", y); //x
Here's another example:
char *original = "awesomeness";
char **pointer_to_original = &original;
(*pointer_to_original) = "is awesome";
printf("%s", original); //is awesome
Use of ** with arrays:
char** array = malloc(sizeof(*array) * 2); //2 elements
(*array) = "Hey"; //equivalent to array[0]
*(array + 1) = "There"; //array[1]
printf("%s", array[1]); //outputs There
The [] operator on arrays does essentially pointer arithmetic on the front pointer, so, the way array[1] would be evaluated is as follows:
array[1] == *(array + 1);
This is one of the reasons why array indices start from 0, because:
array[0] == *(array + 0) == *(array);
C and C++ allows the use of pointers that point to pointers (say that five times fast). Take a look at the following code:
char a;
char *b;
char **c;
a = 'Z';
b = &a; // read as "address of a"
c = &b; // read as "address of b"
The variable a holds a character. The variable b points to a location in memory that contains a character. The variable c points to a location in memory that contains a pointer that points to a location in memory that contains a character.
Suppose that the variable a stores its data at address 1000 (BEWARE: example memory locations are totally made up). Suppose that the variable b stores its data at address 2000, and that the variable c stores its data at address 3000. Given all of this, we have the following memory layout:
MEMORY LOCATION 1000 (variable a): 'Z'
MEMORY LOCATION 2000 (variable b): 1000 <--- points to memory location 1000
MEMORY LOCATION 3000 (variable c): 2000 <--- points to memory location 2000
It declares aPointer as a pointer to a pointer to char.
Declarations in C are centered around the types of expressions; the common name for it is "declaration mimics use". As a simple example, suppose we have a pointer to int named p and we want to access the integer value it's currently pointing to. We would dereference the pointer with the unary * operator, like so:
x = *p;
The type of the expression *p is int, so the declaration of the pointer variable p is
int *p;
In this case, aPointer is a pointer to a pointer to char; if we want to get to the character value it's currently pointing to, we would have to dereference it twice:
c = **aPointer;
So, going by the logic above, the declaration of the pointer variable aPointer is
char **aPointer;
because the type of the expression **aPointer is char.
Why would you ever have a pointer to a pointer? It shows up in several contexts:
You want a function to modify a pointer value; one example is the strtol library function, whose prototype (as of C99) is
long strtol(const char * restrict str, char ** restrict ptr, int base);
The second argument is a pointer to a pointer to char; when you call strtol, you pass the address of a pointer to char as the second argument, and after the call it will point to the first character in the string that wasn't converted.
Remember that in most contexts, an expression of type "N-element array of T" is implicitly converted to type "pointer to T", and its value is the address of the first element of the array. If "T" is "pointer to char", then an expression of type "N-element array of pointer to char" will be converted to "pointer to pointer to char". For example:
void foo(char **arr)
{
size_t i = 0;
for (i = 0; arr[i] != NULL; i++)
printf("%s\n", arr[i]);
}
void bar(void)
{
char *ptrs[N] = {"foo", "bar", "bletch", NULL};
foo(ptrs); // ptrs decays from char *[N] to char **
}
You want to dynamically allocate a multi-dimensional array:
#define ROWS ...
#define COLS ...
...
char **arr = malloc(sizeof *arr * ROWS);
if (arr)
{
size_t i;
for (i = 0; i < ROWS; i++)
{
arr[i] = malloc(sizeof *arr[i] * COLS);
if (arr[i])
{
size_t j;
for (j = 0; j < COLS; j++)
{
arr[i][j] = ...;
}
}
}
}
It means that aPointer points to a char pointer.
So
aPointer: pointer to char pointer
*aPointer :pointer to char
**aPointer: char
An example of its usage is creating a dynamic array of c strings
char **aPointer = (char**) malloc(num_strings);
aPointer gives you a char, which can be used to represent a zero-terminated string.
*aPointer = (char*)malloc( string_len + 1); //aPointer[0]
*(aPointer + 1) = (char*)malloc( string_len + 1); //aPointer[1]
This is a pointer to a pointer to char.