I have a problem, I have a function and I do not understand a specific thing. The function is:
int F( int* x , int n ){
int i , m=0
for (i=0;i<n; i++){
m=x[ i ] + m;
}
return m * m ;
}
I call the function with a pointer and with an integer. Later I do a "for", but I do not understand the line:
m=x[ i ] + m;
Because x is a pointer not an array.
Could you please help me.
Then x points to the memory position then to +1. For example if i call the function with
n=10
x=&n
F(x,n)
the function returns somenthing strange.
X points to the position memory to n, later to the position memory to n+1??
Since x is a pointer, when you pass the array to the function, x points to the first element of the array. Since array is a contigous allocation of memory, The pointer can be made to point to consecutive elements of the array. Thats why
m=x[i]+m
x[i] implies to the ith index from the first element of the array
main()
{
int x[10]={1,2,3,4,5,6,7,8,9,10},sum;
sum=function(x,10);
return 0;
}
This function sends the array to the function, with 10, the size of the array
Arrays are represented as contiguous memory and the array variable gets interpreted as a pointer to the base of that memory (e.g. &(x[0])). Array offset syntax gets translated into pointer arithmetic.
See this post, which clarifies the difference between pointer and arrays:
[] - indexed dereference
a[b] is equivalent to *(a + b). This means a and b must be a pointer to an array element and an integer; not necessarily respectively, because a[b] == *(a + b) == *(b + a) == b[a]. Another important equivalence is p[0] == 0[p] == *p.
The function might be equivalently declared (with more clarity perhaps):
int F(int x[], int n);
and you would call it like so:
int data[3] = {1, 2, 3};
int value = F(data, 3);
Related
I'm new in programming and learning about pointers in array. I'm a bit confused right now. Have a look at the program below:
#include <stdio.h>
int fun();
int main()
{
int num[3][3]={23,32,478,55,0,56,25,13, 80};
printf("%d\n",*(*(num+0)+1));
fun(num);
printf("%d\n", *(*(num+0)+1));
*(*(num+0)+0)=23;
printf("%d\n",*(*(num+0)));
return 0;
}
int fun(*p) // Compilation error
{
*(p+0)=0;
return 0;
}
This was the program written in my teacher's notes. Here in the main() function, in the printf() function dereference operator is being used two times because num is pointer to array so first time dereference operator will give pointer to int and then second one will give the value at which the pointer is pointing to.
My question is that when I'm passing the array name as argument to the function fun() then why *p is used; why not **p as num is a pointer to array?
Second thing why *(p+0) is used to change the value of zeroth element of the array; why not *(*(p+0)+0)=0 as in the main() function *(*(num+0)+0) is used to change the value of zeroth element?
The whole thing is very confusing for me but I have to understand it anyway. I have searched about this and found that there is a difference between pointer to array and pointer to pointer but I couldn't understand much.
The trick is the array-pointer-decay: When you mention the name of an array, it will decay into a pointer to its first element in almost all contexts. That is num is simply an array of three arrays of three integers (type = int [3][3]).
Lets analyse the expression *(*(num + 1) + 2).
When you mention num in the expression *(num + 1), it decays into a pointer to its first element which is an array of three integers (type = int (*)[3]). On this pointer pointer arithmetic is performed, and the size of whatever the pointer points to is added to the value of the pointer. In this case it is the size of an array of three integers (that's 12 bytes on many machines). After dereferencing the pointer, you are left with a type of int [3].
However, this dereferencing only concerns the type, because right after the dereferencing operation, we see expression *(/*expression of type int[3]*/ + 2), so the inner expression decays back into a pointer to the first array element. This pointer contains the same address as the pointer that results from num + 1, but it has a different type: int*. Consequently, the pointer arithmetic on this pointer advances the pointer by two integers (8 bytes). So the expression *(*(num + 1) + 2) yields the integer element at an offset of 12 + 8 = 20 bytes, which is the sixth integer in the array.
Regarding your question about the call of fun(), that call is actually broken, and only works because your teacher did not include the arguments in the forward declaration of fun(). The code
int fun(int* arg);
int main() {
int num[3][3] = ...;
...
fun(num);
}
would have generated a compile time error due to the wrong pointer type. The code of your teacher "works", because the pointer to the first array in num is the same as the pointer to the first element of the first array in num, i. e. his code is equivalent to
int fun(int* arg);
int main() {
int num[3][3] = ...;
...
//both calls are equivalent
fun(num[0]);
fun(&num[0][0]);
}
which would compile without error.
This example shows a matrix, pointers to the first integers of arrays, and pointer to pointer
#include<stdio.h>
int fun(int (*p)[3]); /* p is pointer to array of 3 ints */
int main()
{
/* matrix */
int num[3][3]={{23,32,478},{55,0,56},{25,13, 80}};
/* three pointers to first integer of array */
int *pnum[3] = {num[0], num[1], num[2]};
/* pointer to pointer */
int **ppnum = pnum;
printf("%d\n", *(*(num+1)+2));
fun(num);
printf("%d\n", *(*(num+1)+2));
pnum[1][2] = 2;
printf("%d\n", *(*(num+1)+2));
ppnum[1][2] = 3;
printf("%d\n", *(*(num+1)+2));
return 0;
}
int fun(int (*p)[3])
{
p[1][2]=1;
return 0;
}
You do not actually need any pointers to print anything here.
Your int num[3][3] is actually an array of three elements, each of which is an array of three integers. Thus num[0][0] = 23, num[1][1] = 0, and so on. Thus you can say printf("%d", num[0][0]) to print the first element of the array.
Pointer to variable:
Pointer is variable which stores the address( of a variable). Every one know that.
Pointer to Array:
An array is a variable which has the starting point(address) of group of same objects.
And the pointer is a variable which stores the starting point(address) of an Array.
For example:
int iArray[3];
iArray is a variable which has an address value of three integers and the memory is allocated statically. And the below syntax is provided in a typical programming languages.
// iArray[0] = *(iArray+0);
// iArray[1] = *(iArray+1);
// iArray[2] = *(iArray+2);
In the above the iArray is a variable through which we can access the three integer variables, using any of the syntax mentioned above.
*(iArray+0); // Here iArray+0 is the address of the first object. and * is to dereference
*(iArray+1); // Here iArray+1 is the address of the second object. and * is to dereference
So simple, what is there to confuse.
The below lines are for your understanding
int iArray1[3];
int iArray2[3][3];
int *ipArray = 0;
ipArray = iArray1; // correct
ipArray = iArray2[0]; // correct
ipArray = iArray2[2]; // correct
int **ippArray = iArray2; // wrong
As per the above last line, compiler will not take it as a valid assignment. So **p is not used.
Pointer arthmatic cannot be applied on double arrays because of the way memory is allocated.
I am trying to get value from "second row" in multidimensional array. But I have some problems with that.
I thought that numbers are stored sequentialy in memory so tab[2][2] is stored same as tab[4]. But it seems that I was wrong.
This is what I tried:
int b[2][2] = {{111,222},{333,444}};
int i = 0;
for(;i < 100; i++)
printf("%d \n", **(b+i));
The problem is that I get only 111 and 333 as the result. There is no 222 or 444 in other 98 results. Why?
The problem is that **(b+i) doesn't do what you think it does. It evaluates to:
b[i][0]
As Matt McNabb noted,
**(b+i)
is equivalent to:
*(*(b+i)+0)
and since *(b+i) is equivalent to b[i], the expression as a whole can be seen as:
*(b[i]+0)
and hence:
b[i][0]
Since your array has only 2 rows, only the values of i for 0 and 1 are in bounds of the array (that's the 111 and 333). The rest was wildly out of bounds.
What you could do is:
#include <stdio.h>
int main(void)
{
int b[2][2] = { { 111, 222 }, { 333, 444 } };
int *base = &b[0][0];
for (int i = 0; i < 4; i++)
printf("%d: %d\n", i, base[i]);
return 0;
}
Output:
0: 111
1: 222
2: 333
3: 444
You can think of a two-dimensional array as a special form of a one-dimensional array. Your array holds 2 elements (!). (Each element happens to be an array of two elements, but let's ignore that for a sec.) Let's make the dimensions different so that we can distinguish them:
int arr2d[2][3] holds 2 elements (each being an array of 3 elements). The "primary index" is written first, that is if you have a one-dimensional array of 3 like int arr1d[3]and want to have an array of those with three elements like arr2d, you have to write arr2d[2][3]. You could arrange that with a typedef which makes it clear that all arrays in C are essentially 1-dimensional:
typedef int arr3IntT[3];
arr3IntT arr2d[2] = { {0,1,2}, {3,4,5} };
Now what does arr2d+i mean? arr2d, like any array, decays to a pointer to its first element (which is an array of 3 ints). arr2d+1 adds the offset of 1 of those elements to the address, so that like always the expression yields the address of the second element (which is the second array of 3 ints). Dereferencing it once like *(arr2d+1) yields that element, i.e. the one-dimensional sub-array of 3 ints. It decays to a pointer to its first element, i.e. a pointer to the first int in that second sub-array. Dereferencing that in the expression **(arr2d+1) yields that int, like always. To summarize: In your original code you iterate from sub-array to sub-array, always referencing the first of their elements, incidentally going out of bounds for i>1.
But principally you are right, elements of n-dimensional arrays in C are lying contiguously in memory, so you can access them one by one if you like to. You just have to index a pointer to int, not one to int[3]. Here is how: The expression arr2d decays to a pointer to its first element which is an array of 3 ints. Dereferencing that gives that first element, a one-dimensional array of 3 ints. Like all arrays, it decays to a pointer to its first element, an int, which is the very first element in the data:
#include<stdio.h>
int main()
{
int arr2d[2][3] = { {0,1,2}, {3,4,5} };
int *p_el1 = *arr2d;
int i, j;
// Sanity check by indexing 2-dimensionally
for(i=0; i<2; i++) for(j=0; j<3; j++) printf("%d\n", arr2d[i][j]);
// iterate the elements 1 by 1
for(i=0; i<3*2; i++) printf("%d\n", p_el1[i]);
}
A multidimensional array is not a fundamentally new type. It's an array type where the elements are themselves arrays. To quote the C99 standard §6.2.5 ¶20 (Types)
An array type describes a contiguously allocated nonempty set of
objects with a particular member object type, called the element type.
Array types are characterized by their element type and by the number
of elements in the array.
int b[2][2] = {{111, 222}, {333, 444}};
The above statement defines b to be an array of 2 elements where each element is the type int[2] - an array of 2 integers. It also initializes the array with the array initializer list. An array is implicitly converted to a pointer to its first element in some cases.
In the printf call, b decays to a pointer to its first element. Therefore, it's equivalent to &b[0] and has type int (*)[2] - a pointer to an array of 2 integers. Please note that it's undefined behaviour to access elements out of the bound of an array. Therefore, the for loop condition i < 100 is wrong. It should be i < 2. Now, let's try to demystify the expression **(b+i).
b -------------> pointer to a[0]
b + i ---------> pointer to a[i]
*(b + i) -----> a[i]
*(*(b + i)) ---> *(a[i]) ----> *(&(a[i])[0]) ----> a[i][0]
As noted, the elements of the array are themselves arrays. Therefore, a[i] is an array. Again, the array decays to a pointer to its first element, i.e., to &(a[i])[0]. Applying indirection operator * on this pointer gets us the value at that address which is a[i][0].
You can access the elements of the array through a pointer but the pointer type must be a pointer to element type of the array.
#include <stdio.h>
int main(void) {
int b[2][2] = {{111, 222}, {333, 444}};
int (*p)[2] = b;
// (sizeof b / sizeof b[0]) evaluates to
// the size of the array b
for(int i = 0; i < (sizeof b / sizeof b[0]); i++)
// (sizeof *p / sizeof (*p)[0]) evaluates to
// the size of element of the array which is
// itself an array.
for(int j = 0; j < (sizeof *p / sizeof (*p)[0]); j++)
printf("%d\n", *(*(p + i) + j));
return 0;
}
Here, the expression *(*(p + i) + j) can be decoded as
p ---------------> pointer to the first element of b, i.e., &b[0]
(p + i) ----------> pointer to b[i], i.e., &b[i]
*(p + i) ---------> the array element b[i] ---> decays to &(b[i])[0]
*(p + i) + j -----> &(b[i])[j]
*(*(p + i) + j) --> the element b[i][j]
Therefore, the expression *(*(p + i) + j) is equivalent to b[i][j]. In fact, the C99 standard §6.5.2.1 ¶2 says -
The definition of the subscript operator [] is that E1[E2] is
identical to (*((E1)+(E2)))
This means we have the following are equivalent with context to the above program -
*(*(p + i) + j)
// equivalent to
p[i][j]
// equivalent to
b[i][j]
// equivalent to
*(*(b + i) + j)
I have this piece of program and there are some parts that I don't understand...
#include <stdio.h>
void increment( int *ptr) {++*ptr; }
int main(){
int a[] = { 5, 10 }, i = 0; // We have: vector a=[5,10] | i=0
increment(a); //This is the same as increment(&a) isn't it?
//So we have: a=[6,10] | i=0
increment(&i); // We increment i so: a=[6,10] | i=1
increment(&a[i]); //We increment the element at the position a[1]: a=[6,11] | i=1
increment(a+i); //OMG! What does it mean? a is an array isn't it? Then how can we
//make "a+1"? and how is this an address?
printf("\nResult: i= %d\n", i);
printf( "a[0] = %d\n" "a[1] = %d\n", a[0], a[1]);
return 0;
}
The printfs return:
i=1
a[0]=6
a[1]=12
How can this be explained?
When you pass an array to a function, the array decays to a pointer to the first element. So in the function, doing e.g. *ptr is dereferencing the first element in the array.
Also, because arrays decays to pointers, doing e.g. a[i] is equal to *(a + i).
Unrelated side-note: Because of the commutative of addition, the expression *(a + i) can be written as *(i + a) which leads to i[a] actually being a valid expression.
Arrays and pointer arithmetic are the same in C. If you've not come across pointer arithmetic here's a tutorial on it:
C Pointer Arithemtic
In essence:
a[3]
is the same as
*(a + 3)
Where a is a pointer.
When we give the name of array in the function call, the address of the arrya's first element is passed
ie; increment(a); is same as increment(&a[0]);
So while giving
incrementa(a+i);
since i and array a[] are both integers effectively that becomes
incrementa ( base address of array a (ie; &a[0]) + value of i )
i just acts as the offset to that array element
That it becomes the address of the second element which is &a[1]
so the 11 is modified to 12 while incrementing the value by the function
Im relatively knew to C, i am used to program in Java so i find C a little bit difficult in what concerns arrays. I still cofuse myself with this cases:
int a [];
int* a;
int *a;
In java, i would do something like this to return an array in a function:
int [] returnArr(int [] a){
... modify a ...
return a;
}
int [] a = {...};
int [] b = returnArr(a); ##
How can i do the same in C, specially the parts with ##.
EDITED:
I have this function:
float *normalizeValues(float *v, float maxY){
int size = sizeof(v) / sizeof(float);
float max = findMax(v);
float ratio = maxY / max;
int i;
for(i = 0; i < size ; ++i){
v[i] = v[i] * ratio;
}
return v;
}
And im doing the following:
float vert [] = {306, 319, 360, 357, 375, 374, 387, 391, 391, 70, 82, 94, 91, 108, 114, 125, 127, 131};
int i = 0;
float *vert2;
vert2 = normalizeValues(vert, 0.7);
for(i = 0; i < sizeof(vert2) / sizeof(float); ++i){
fprintf(stdout,": %f\n",vert2[i]);
}
And the output is only 1 element.
EDIT: To directly answer your updated question: you have to pass in the size of the array. C has no mechanism to store the size of arrays like Java does. If the compiler knows about the size of the array because the array is a global or local variable, not dynamically allocated, then you can use the sizeof() operator. Otherwise, you have to know the size separately, or use sentinel values in your array (such as a 0.0 at the end, or a NULL).
As for arrays, pointers and arguments in general, see below:
You will be returning a pointer to the array, which is indicated with the '*' syntax:
int *returnArr(int[] a) {
// modify a...
return a;
}
int a[] = { ... };
int *b;
b = returnArr(a);
A few things to note:
You can't do assignments in variable declarations that involve non-constant expressions (e.g., function calls). This might have changed in C99, though.
The brackets go after the variable name, unlike in Java where they are part of the type. Even though Java's syntax is more consistent, it doesn't quite make sense in C where you often give the array size in the brackets in the variable declaration:
int a[3] = { ... };
There's no way to specify that a function returns an array as opposed to a plain pointer. In C, array references decay to pointers (though pointers and arrays are NOT the same thing, as is commonly claimed). That means that whenever you pass an array around, C only provides a means to a pass a pointer to the array. The whole array isn't actually copied. As it happens, the name of the array is also a pointer to the first element of the array.
Please also take note of what user268396 says in their answer. If you are planning to create a new array and return it, you'll need to either allocate the array dynamically, or have a pointer to an already allocated array be passed in (which is what it seems like you are kind of doing anyway).
You can't. When the function returns the stack frame will be wiped out (typically) and your generated array will be clobbered by that. You can however edit the function prototype to accept a pointer to the array to modify. That kind of function argument is known as an "output parameter". Example:
void function func(int a, int b, int[2] to_modify)
{
to_modify[0] = a;
to_modify[1] = b;
}
int main()
{
int foo[2];
func(1, 2, foo);
printf("Result: foo[0] = %d, foo[1] = %d\n", foo[0], foo[1]);
return 0;
}
This will print "Result: foo[0] = 1, foo[1] = 2".
Hope this helps
#include<stdio.h>
void change(int *c)/*Pointer c now has the first location of the array a[]*/
{
*(c+0) = 0;/*assign values to the array by adding step-size to the first array position*/
*(c+1) = 1;
*(c+2) = 2;
*(c+3) = 3;
*(c+4) = 4;
}
main()
{
int a[5]={10,20,30,40,50}; /* Declare and Assign an array a[] of size 5.*/
int *b = a; /*Declare and assign a Pointer to the location of the array.*/
change(b); /*pass the pointer(which is now pointing to first position of array) to the change() function.*/
printf("%d,%d,%d,%d,%d,",a[0],a[1],a[2],a[3],a[4]);/*Print the changed value.*/
}
Output: 0,1,2,3,4,
From Java point of view, Pointers are simply like(not exactly) Object references.
Object O;
O = New SomeClassName();
Like Object Reference O is pointing to some Actual Object of type SomeClassName, so does pointers in C:
int *b;
b = &a;
Variable b is simply pointing to the address location to a.
Taking a deep dive into array concepts:
int a[5];
int *b = a;
Here we are just saying like Mr.*b point to the first location of group a i.e. a[0].
Now the power pointer in C is that from now on, here after:
*b means a[0]
*(b+1) means a[1]
*(b+2) means a[2]
*(b+3) means a[3]
*(b+4) means a[4]
This means you change in *(b+4), you're changing a[4].
int* returnArr(int a[]){
//modify a
return a;
}
One need mention is when you use an array in the parameter list of a function, it will be converted into a pointer. So in main(...)'s declaration, char *argv[] and char **argv are actually same. In this case, int a[] and int* a are same. But array and pointer is not the same thing.
Take the following code as an example:
int a[10];
int main(){
int *p = a;
p[5] = 4;
return p[5];
}
p is a pointer, when we access p[i], note that the address of p is not the address of a, the content of p is the address of a. Then the code will:
access the memory to get the content of p, i.e. the address of a.
compute the offset based on i and type of the pointer(int).
access the memory to get the result.
a is an array of int, if we access a[i], the address of a is just the address of a[0], the code will:
Compute the offset based on i and the type int.
Access the memory.
Pointer and array are different types. So if you declare int *p in one file and use it in that file, but define the p as an array in another file, that will cause problem.
You may also wonder about int *p = a, in ANSI, if you use an array(its name) as an expression, the compiler will convert it into a pointer, pointing to the very first element of the array.
Update based on Jim Balter's comments:
If you use an array(its name) as an expression, the compiler will not always convert it into a pointer, pointing to the very first element of the array. For instance, in sizeof(p->q->a), p->q->a is an expression but if a is an array it isn't converted into a pointer.
"Except when it is the operand of the sizeof operator or the unary &
operator, or is a string literal used to initialize an array, an
expression that has type ‘‘array of type’’ is converted to an
expression with type ‘‘pointer to type’’ that points to the initial
element of the array object.
In C, you can only return a pointer of an array in a function.
For example, if you want to return a string(array of char) in a function, you can return a pointer to a null-ended string. If you want to return an array of some other type(int, user-defined struct, etc), you can alloc some memory to store the array, and return the pointer of the array, return the size of the array in the parameter.
example:
int *function(int *size)
{
*size = 10;
int *intP = (int *)malloc((*size)*sizeof(int));
return intP;
}
So I have the following code snippet:
#include <stdio.h>
void pointer_shift(int *a, int n);
int main(void) {
int a[] = {100, 101, 102};
pointer_shift(a1, 3);
}
void pointer_shift(int *a, int n) {
int i;
for (i = 0; i != n - 1; i++) {
*(a + i) = *(a + i + 1);
}
}
I just want to clarify how the pointers work in this snippet. So pointer_shift takes in 'a', a pointer to an int, correct? a1 is passed in to this parameter, and since arrays decay to a pointer to their first element, it works.
First of all, hopefully what I said in the above paragraph is correct. Secondly, what does *(a + i) = *(a + i + 1); actually do? Say we're on the first iteration of the for loop, and i = 0. Then the left side, *a, accesses what, exactly? Does it represent a pointer? I thought * was the dereferencing operator, and accesses the object that a pointer points to... And so then it sets *a = *(a + 1). (a + 1) is the next element in the array, but what exactly does this assignment do, and why?
Thanks!
It is actually not pointer shift, but value shift, *(a+i) is of same effect as a[i], so what it does is a[i] = a[i+1]
*(a + i) = *(a + i + 1); is copying array elements within the array using a bit of pointer arithmetic.
*(a + i) is equivalent to a[i], so the statement is equivalent to a[i] = a[i + 1];. The loop is moving the array values "to the left" in the array: a[0] = a[1]; a[1] = a[2]; and so on.
Your understanding of the function call is correct.
I just want to clarify how the pointers work in this snippet. So pointer_shift takes in 'a', a pointer to an int, correct? a1 is passed in to this parameter, and since arrays decay to a pointer to their first element, it works.
Yes, when you pass an array to a function it degrades to a pointer. An array is not a pointer in an object sense, but it is a pointer in a value sense. When you pass it to a function its value is passed, i.e., a pointer to the first element.
array indexing is the same as pointer arithmetic, so the last two lines in this snippet are equivalent:
int arr[] = {1, 2, 3};
arr[0] = 10;
*arr = 10;
as are these:
arr[1] = 20;
*(arr + 1) = 20;
The expression a + i is pointer arithmetic, incrementing the memory address stored in a by i units of the pointer size of a. So if a pointer to an int takes four bytes on your system, and if the current memory address is, say, 0x1234 the value of a + 1 would be 0x1238.
What the asterisk does is dereference that address and access the actual value at that address. If you have 100 stored at a or a[0], and 101 stored at a + 1 or a[1], then *(a + i) = *(a + i + 1) replaces 100 with 101 at a[0], for i = 0.
Basically, you want to read this C tutorial on pointers and arrays.