#include<stdio.h>
void main()
{
int a[2][2][2]={1,2,3,4,5,6,7,8};
printf("%d\t%d",*(*(*(a+1)+1)+1),a[1][1][1]);
}
The output of the code is 8 8 but the expression *(*(*(a+i)+j)+k); & a[i][j][k]; are equal I know how to evolve the expression a[i][j][k]; but I don't have an idea how to evolve the expression *(*(*(a+i)+j)+k); with proper priority operations of * operator & () operator. Please explain it with taking some values of i,j,k.
In pointer arithmetic,
array[i] = *(array + i).
array[i][j] = *(*(array + i) + j).
array[i][j][k] = *(*(*(array + i) + j) + k)
Initially, a is pointing towards 1. When you do a+1 it moves to the 5, and when you dereference (*) you cause the next +1 to make it move only 2 positions, to the 7. The second * causes the third +1 to move it to the last element, 8, and the final * actually gets the value from that address (which is 8).
Related
I've been facing difficulties to understand the fourth line of code after the first curly brace,
#include<stdio.h>
int main()
{
int arr[] = {10,20,36,72,45,36};
int *j,*k;
j = &arr[4];
k = (arr+4);
if(j==k)
printf("The two pointers are pointing at the same location");
else
printf("The two pointers are not pointing at the same location");
}
I just wanted to know what the fourth line of code after the first curly brace i.e. k = (arr+4); does?
Since k was a pointer it was supposed to point at something that had an "address of operator" ? I can still understand that if it doesn't have the "address of operator" then whatever does the part of the code k = (arr+4) do?
For any array or pointer arr and index i, the expression arr[i] is exactly equal to *(arr + i).
Now considering that arrays naturally can decay to pointers to their first element, arr + i is a pointer to element i.
Without & or the sizeof operator an array converts to a pointer to the first element of the array.In your case array will convert to a pointer to int and will point to the first element.arr + 1 will point to the second element,arr + 2 will point to the third element etc..arr+1 means increment arr with sizeof(int).
This is simplified diagram of the first 2 elements of the array.Let's assume int is 4 bytes long.
| first element | second element |
-------------------------------------------------
| | | | | | | | |
| | | | | | | | |
-------------------------------------------------
0x00 0x01 0x02 0x03 0x04 0x05 0x06 0x07
arr will contain 0x00
arr + 1 will contain 0x04
*arr will mean take the value from adress 0x00.It's equivalent to *(arr+0), *(0+arr), arr[0] and 0[arr].Since arr is of type int* it will take a four bytes long value.
With int* k = array k will contain the same address with array.
k = (arr + 4) will contain the address of the 5th element.
j = &arr[4]; will also store the address of the 5th element
As k is defined as pointer, it can be use to store address as value and can point to a location.
Here (arr+4) will return address of arr and plus 4.
If int takes 4 bytes then it will point to 2nd element in arr, so it depends on system(32bit/64 bit), that how much it takes to store int.
Something that you should aware of (C Standards#6.3.2.1p3):
Except when it is the operand of the sizeof operator, the _Alignof 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 and is not an lvalue. If the array object has register storage class, the behavior is undefined.
The statement:
int arr[] = {10,20,36,72,45,36};
arr is an array of int.
The expression arr[i], can also be written as:
*(arr+i)
So, &a[i] can be written as:
&(*(arr+i))
The operator & is used to get the address and the operator * is used for dereferencing. These operators cancel the effect of each other when used one after another. Hence, &(*(arr+i)) is equivalent to arr+i.
I just wanted to know what the fourth line of code after the first curly brace i.e. k = (arr+4); does?
In the statement:
k = (arr+4);
none of the operators - sizeof, _Alignof and unary & is used. So,
arr will convert to a pointer to type int. That means, arr+4 will give the address of the four element past the object (i.e. int) pointed to by arr, which is nothing but &a[4].
&arr[4] --> &(*(arr+4)) --> (arr+4)
You need to understand pointer arithmetic here.
arr[i] is interpreted as *(arr + i),
where * is 'dereferencing' or 'value at' operator and arr always represents the address of first element of the array i.e. the address of array itself, which literally means,
valueat(starting address of arr + i)
Now suppose address of arr is 100 and you are adding i elements of type arr in address of array and not the value of i, that is,
valueat(100 + i)
Now according to your code, here you are assigning address of 4th element to pointer j ,
j = &arr[4];
j = &(valueat(100+ 4 elements of type int));
j = &(valueat(100+ 16)); ->
j = &(valueat(116)); ->
j = &(45) that is j = 116
Now when you do
k = (arr+4);
k = (starting address of arr + 4 elements of type int);
k = (100 + 16); that is k = 116, and that is why the output,
The two pointers are pointing at the same location
Hope this helps.
This question already has answers here:
With arrays, why is it the case that a[5] == 5[a]?
(20 answers)
Closed 5 years ago.
Array declaration:
int arr [ ]={34, 65, 23, 75, 76, 33};
Four notations: (consider i=0)
arr[i]
and
*(arr+i)
and
*(i+arr)
and
i[arr]
Lets take a look at how your array is laid out in memory:
low address high address
| |
v v
+----+----+----+----+----+----+
| 34 | 65 | 23 | 75 | 76 | 33 |
+----+----+----+----+----+----+
^ ^ ^ ^
| | | ...etc
| | |
| | arr[2]
| |
| arr[1]
|
arr[0]
That the first elements is arr[0], the second arr[1] is pretty clear, that's what everybody learns. What is less clear is that the compiler actually translates an expression such as arr[i] to *(arr + i).
What *(arr + i) does is first get a pointer to the first element, then do pointer arithmetic to get a pointer to the wanted element at index i, and then dereference the pointer to get its value.
Due to the commutative property of addition, the expression *(arr + i) is equal to *(i + arr) which due to the above mentioned translation is equal to i[arr].
The equivalence of arr[i] and *(arr + i) is also what's behind the decay of an array to a pointer to its first element.
The pointer to the arrays first element would be &arr[0]. Now we know that arr[0] should be equal to *(arr + 0) which means &arr[0] has to be equal to &*(arr + 0). Adding zero to anything is a no-op, so leading to the expression &*(arr). Parentheses with only one term and no operator can also be removed, leaving &*arr. And lastly the address-of and dereference operator are each other opposites and cancel out each other, leaving us with simply arr. So &arr[0] is equal to arr.
Each element in the array, have a position in memory. The positions in the arrays are sequential. The arrays in C are pointers and always point the first direction on memory for the collection (first element of the array).
arr[i] => Gets value of "i-position" in the array. It is the same that arr[i] = *(arr + i)
*(arr+i) => Gets value that is in memory by adding the position in memory that point arr and i value.
*(i+arr) => Is the same that *(arr+i). The sum is commutative.
i[arr] => Is the same that *(i+arr). It's another way of representing.
They are the same because the C language specification says so. Read n1570
The notation a[i] is syntactic sugar for *(a+i).
The first one is mathematical syntax (symbolics closer of what human brain is educated with) while the second one corresponds directly to one assembler instruction.
On the other hand *(a+i)=*(i+a)=i[a] because the arithmetic of pointers is commutative.
These are the same because of how the array subscript operator [] is defined.
From sectino 6.5.2.1 of the C standard:
2 A postfix expression followed by an expression in square brackets []
is a subscripted designation of an element of an array object. The
definition of the subscript operator [] is that E1[E2] is
identical to (*((E1)+(E2))). Because of the conversion rules that
apply to the binary + operator, if E1 is an array object
(equivalently, a pointer to the initial element of an array object)
and E2 is an integer, E1[E2] designates the E2-th element of
E1 (counting from zero).
The expression arr[i] in your example is of the form E1[E2]. Because the standard states that this is the same as *(E1+E2) that means that arr[i] is the same as *(arr + i).
Because of the commutative property of addition, *(arr + i) is the same as *(i + arr). Applying the equivalence rule above to this expression gives i[arr].
So in short, those 4 expressions are equivalent because of how the standard defines array subscripting and because of the commutative property of addition.
It works because an array variable in C (i.e. arr in your example) is just a pointer to the beginning of an array of memory locations. A pointer is number which represents the address of a specific memory location. When you put and '*' in front of a pointer, it means "give me the data in that memory location".
So, if arr is a pointer to the beginning of the array, *(arr) or *(arr + 0) is the data in the 0th index of the array, and *(arr + 1) is the data in the 1st index, and so on.
An expression which looks like A[B] essentially gets translated into something like *(A+B). So, arr[0] = *(arr + 0) and arr[i] = *(arr+i), etc.
And because A+B = B+A, the two are interchangeable. Meaning *(arr+i) = *(i+arr).
And because arr[i] = *(arr+i) and *(arr+i) = *(i+arr), it should make sense that arr[i] = i[arr].
This question already has answers here:
With arrays, why is it the case that a[5] == 5[a]?
(20 answers)
Closed 9 years ago.
consider following C code
int a[]={1,2,3,4};
printf("%d",2[a]);
this prints "3".How is it possible? I know in a[2] a is the base address of array.But in 2[a]
what is 2? and how it accessses array a?I am totally confused with this representation of array.
There are two things to remember here:
The first is that array access is basically just a fancy way of using pointer arithmetic. For example, if you have the array
int a[10];
then
a[3] = 5;
is equal to
*(a + 3) = 5;
The second thing to remember is that addition (like in a + 3 above) is commutative, so a + 3 is the same as 3 + a. This leads to e.g.
*(3 + a) = 5;
which can be interpreted as
3[a] = 5;
int a[]={1,2,3,4}; is an integer array containing 4 elements and a is the Base Address, Let the Base Address be denoted by X . Now a[1] means element at address X + sizeOf(int) * 2 = Y (suppose) i.e. element at address Y, likewise 2[a] means element at adsress sizeOf(int) * 2 * X = Y.
Thus even if you write a[2] or 2[a] eventually complier recognizes it as Y and refres to the element at address Y which is 3 in our case.
Hope it addresses the problem right.
*(expr1+expr2) is equivalent to expr1[expr2] or expr2[expr1].
*(expr2+expr1) is equivalent to expr2[expr1] or expr1[expr2].
It is just another way to write an element of an array.
In int a[]={1,2,3,4};, element at index 3 can be referenced by many methods:
As an array element: a[3]
Using pointer: *(a + 3) or *(3 + a) [Addition is commutative in arithmetics and in C]
Now you can write the second way of representation using pointer as 3[a], i.e. a[3] is equal to *(a + 3) is equal to *(3 + a) is equal to 3[a].
If I have the following code in a function:
int A[5][5];
int i; int j;
for(i=0;i<5;i++){
for(j=0;j<5;j++){
A[i][j]=i+j;
printf("%d\n", A[i][j]);
}
}
This simply prints out the sum of each index. What I want to know is if it's possible to access each index in the static array in a similar fashion to dynamic array. So for example, if I wanted to access A[2][2], can I say:
*(A+(2*5+2)*sizeof(int))?
I want to perform some matrix operations on statically allocated matrices and I feel like the method used to dereference dynamic matrices would work the best for my purposes. Any ideas? Thank you.
That's the way to do it: A[i][j].
It prints out the sum of the indexes because, well, you set the element A[i][j] to the sum of the indexes: A[i][j] = i+j.
You can use:
*(*(A + 2) + 2)
for A[2][2]. Pointer arithmetics is done in unit of the pointed type not in unit of char.
Of course, the preferred way is to use A[2][2] in your program.
The subscript operation a[i] is defined as *(a + i) - you compute an offset of i elements (not bytes) from a and then dereference the result. For a 2D array, you just apply that definition recursively:
a[i][j] == *(a[i] + j) == *(*(a + i) + j)
If the array is allocated contiguously, you could also just write *(a + i * rows + j).
When doing pointer arithmetic, the size of the base type is taken into account. Given a pointer
T *p;
the expression p + 1 will evaluate to the address of the next object of type T, which is sizeof T bytes after p.
Note that using pointer arithmetic may not be any faster than using the subscript operator (code up both versions and run them through a profiler to be sure). It will definitely be less readable.
Pointer arithmetic can be tricky.
You are on the right track, however there are some differences between pointer and normal arithmetic.
For example consider this code
int I = 0;
float F = 0;
double D = 0;
int* PI = 0;
float* PF = 0;
double* PD = 0;
cout<<I<<" "<<F<<" "<<D<<" "<<PI<<" "<<PF<<" "<<PD<<endl;
I++;F++;D++;PI++;PF++,PD++;
cout<<I<<" "<<F<<" "<<D<<" "<<PI<<" "<<PF<<" "<<PD<<endl;
cout<<I<<" "<<F<<" "<<D<<" "<<(int)PI<<" "<<(int)PF<<" "<<(int)PD<<endl;
If you run it see the output you would see would look something like this (depending on your architecture and compiler)
0 0 0 0 0 0
1 1 1 0x4 0x4 0x8
1 1 1 4 4 8
As you can see the pointer arithmetic is handled depending on the type of the variable it points to.
So keep in mind which type of variable you are accessing when working with pointer arithmetic.
Just for the sake of example consider this code too:
void* V = 0;
int* IV = (int*)V;
float* FV = (float*)V;
double* DV = (double*)V;
IV++;FV++;DV++;
cout<<IV<<" "<<FV<<" "<<DV<<endl;
You will get the output (again depending on your architecture and compiler)
0x4 0x4 0x8
Remember that the code snippets above are just for demonstration purposes. There are a lot of things NOT to use from here.
In 3["XoePhoenix"], array index is of type array of characters. Can we do this in C? Isn't it true that an array index must be an integer?
What does 3["XeoPhoenix"] mean?
3["XoePhoenix"] is the same as "XoePhoenix"[3], so it will evaluate to the char 'P'.
The array syntax in C is not more than a different way of writing *( x + y ), where x and y are the sub expressions before and inside the brackets. Due to the commutativity of the addition these sub expressions can be exchanged without changing the meaning of the expression.
So 3["XeoPhoenix"] is compiled as *( 3 + "XeoPhoenix" ) where the string decays to a pointer and 3 is added to this pointer which in turn results in a pointer to the 4th char in the string. The * dereferences this pointer and so this expression evaluates to 'P'.
"XeoPhoenix"[ 3 ] would be compiled as *( "XeoPhoenix" + 3 ) and you can see that would lead to the same result.
3["XeoPhoenix"] is equivalent to "XeoPhoenix"[3] and would evaluate to the 4th character i.e 'P'.
In general a[i] and i[a] are equivalent.
a[i] = *(a + i) = *(i + a) = i[a]
In C, arrays are very simple data structures with consecutive blocks of memory. They therefore need to be integers as these indices are nothing more than offsets to addresses in memory.