int main(void)
{
short arr[3][2]={3,5,11,14,17,20};
printf("%d %d",*(arr+1)[1],**(arr+2));
return 0;
}
Hi. In above code as per my understanding ,*(arr+1)[1] is equivalent to *(*(arr+sizeof(1D array)*1)+sizeof(short)*1)=>arr[1][1] i.e 14. But the program output is arr[2][0]. can someone please explain how dereferencing the array second time adds sizeof(1Darray) i.e *(*(arr+sizeof(1D array)*1)+sizeof(1D array)*1)=>arr[2][0]
From the C Standard (6.5.2.1 Array subscripting)
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).
So the expression
*(arr+1)[1]
can be rewritten like
* ( *( arr + 1 + 1 ) )
that is the same as
*( *( arr + 2 ) )
arr + 2 points to the third "row" of the array. Dereferencing the pointer expression you will get the "row" itself of the type short[2] that used in the expression *( arr[2] ) is converted to pointer to its first element. So the expression equivalent to arr[2][0] yields the value 17.
Thus these two expressions
*(arr+1)[1],
and
**(arr+2)
are equivalent each other.
Note: pay attention to that there is a typo in your code
printf("%d %d",*(arr+1)[1],**(arr+2);
You need one more parenthesis
printf("%d %d",*(arr+1)[1],**(arr+2) );
Related
I'm trying to solve this question that has a vector and there is a pointer that leads to an negative index. I'd like to know if C ignores it and considers it as a positive number or it goes backwards in the vector, like -1 being the last element of the array.
Yes, you can have negative indexes. This follows from the definition of pointer arithmetic in C, namely that p[i] is exactly equivalent to *(p + i), and it's perfectly legal for i to be a negative number.
So, no, the sign is not ignored. Also, if you're dealing directly with an array, there's no rule that says that a negative index is computed with respect to the end of the array. If you're dealing directly with an array, a negative index ends up trying to access memory "off to the left" of the beginning of the array, and is quite against the rules, leading to undefined behavior.
You might wish to try this illustrative program:
#include <stdio.h>
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int *p = &arr[4];
int *endp = &arr[8];
printf("%d %d %d\n", arr[0], arr[4], arr[8]); /* prints "1 5 9" */
printf("%d %d %d\n", p[-1], p[0], p[1]); /* prints "4 5 6" */
printf("%d %d %d\n", endp[-2], endp[-1], endp[0]); /* prints "7 8 9" */
printf("x %d %d x\n", arr[-1], endp[1]); /* WRONG, undefined */
}
The last line is, as the comment indicates, wrong. It attempts to access memory outside of the array arr. It will print "random" garbage values, or theoretically it might crash.
Is *(arr -1) the same as arr[-1]?
Yes. Strictly speaking, the array subscripting [] operator is defined as arr[n] being equivalent to *( (arr) + (n) ). In this case: *( (arr) + (-1) ).
or it goes backwards in the vector
Well it goes backwards until it reaches array item 0, from there on you access the array out of bounds. It doesn't "wrap around" the index or anything like that. Example:
int arr[3] = ...;
int* ptr=arr+1;
printf("%d\n", ptr[-1]); // ok, prints item 0
int* ptr=arr;
printf("%d\n", ptr[-1]); // not ok, access out-of-bounds.
According to the C Standard (6.5.2.1 Array subscripting)
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).
So the expression arr[-1] is evaluated as *( arr + -1 ) that is the same as *( arr - 1).
However pay attention to that arr[-1] is not the same as -1[arr] because before the square brackets there must be a postfix expression
postfix-expression [ expression ]
and the unary minus operator has less precedence than a protfix expression.
But you may write ( -1 )[arr] that is the same as arr[-1].
I've read about it and I mostly get it, but this situation confused me a bit. Why don't we use arrow operator -> in scanf? I understand that dot is used for objects and arrow for pointers but here, g is a pointer to structure.
DOCUMENT *take(int *pn){
DOCUMENT *g;
printf("How much documents? ");
scanf("%d", pn);
g = (DOCUMENT *)calloc(*pn, sizeof(DOCUMENT));
for (int i = 0; i < *pn; i++)
{
printf("Type in author, name of document and number of pages: ");
scanf("%s %s %d", g[i].author, g[i].name, &g[i].s );
}
return g;
}
The array index operator [] has a dereference built into it.
g[i] is exactly the same as *(g + i). So g[i] refers to a DOCUMENT, not a DOCUMENT * and thus you use the member access operator . instead of the pointer-to-member operator ->.
If the left operand of the . or -> is a pointer, then you use ->. Otherwise if it is an object (plain variable), then you use ..
In this case g[i] is taking a pointer and doing array subscripting on it. The result of that is an object ("lvalue"), not a pointer, hence g[i]..
Also note operator precedence in expressions like &g[i].s. The array subscripting operator [] and the struct member operator . have same operator precedence - they belong to the same group of postfix operators. But that group has left-to-right operator associativity so [] takes precedence over the .. Then after those two follow &, the unary address operator, with lowest operator precedence in the expression. So operator precedence guarantees that the expression is equivalent to &((g[i]).s).
The subscript operator used with a pointer yields the object pointed to by the pointer expression. That is, for example, this expression
g[i].author
(where g[i] is the i-th element of the array of structures) is equivalent to
( g + i )->author
where g + i is a pointer that points to the i-th element of the array of structures.
The subscript operator g[i] is equivalent to the expression *( g + i ).
You may write
g[i].author
or like
( *( g + i ) ).author
or like
( g + i )->author
Thus this call of scanf
scanf("%s %s %d", g[i].author, g[i].name, &g[i].s );
can be rewritten like
scanf("%s %s %d", ( g + i )->author, ( g + i )->name, &( g + i )->s );
When you write g[i], this is actually equal to *(g + i). That is, you dereference the pointer and it becomes an object again, so you can access it with dot (.) instead of arrow (->).
This question already has answers here:
With arrays, why is it the case that a[5] == 5[a]?
(20 answers)
Closed 3 years ago.
Today I stumbled over a C riddle that got a new surprise for me.
I didn't think that -1[p] in the example below would compile, but it did. In fact, x ends up to be -3.
int x;
int array[] = {1, 2, 3};
int *p = &array[1];
x = -1[p];
I searched the internet for something like -1[pointer] but couldn't find anything. Okay, it is difficult to enter the correct search query, I admit. Who knows why -1[p] compiles and X becomes -3?
I'm the person that made this "riddle" (see my Twitter post)
So! What's up with -1[p]?
ISO C actually defines [] to be symmetrical, meaning x[y] is the same as y[x], where x and y are both expressions.
Naively, we could jump to the conclusion that -1[p] is therefore p[-1] and so x = 1,
However, -1 is actually the unary minus operator applied to the constant 1, and unary minus has a lower precedence than []
So, -1[p] is -(p[1]), which yields -3.
This can lead to funky looking snippets like this one, too:
sizeof(char)["abc"] /* yields 'b' */
First thing to figure out is the precedence. Namely [] has higher precedence than unary operators, so -1[p] is equal to -(1[p]), not (-1)[p]. So we're taking the result of 1[p] and negating it.
x[y] is equal to *(x+y), so 1[p] is equal to *(1+p), which is equal to *(p+1), which is equal to p[1].
So we're taking the element one after where p points, so the third element of array, i.e. 3, and then negating it, which gives us -3.
According to the C Standard (6.5.2 Postfix operators) the subscript operator is defined the following way
postfix-expression [ expression ]
So before the square brackets there shall be a postfix expression.
In this expression statement
x = -1[p];
there is used the postfix expression 1 (that is at the same time a primary expression), the postfix expression 1[p] (that is the subscript operator) and the unary operator - Take into account that when the compiler splits a program into tokens then integer constants are considered as tokens themselves without the minus. minus is a separate token.
So the statement can be rewritten like
x = -( 1[p] );
because a postfix expression has a higher priority than an unary expression.
Let's consider at first the postfix sub-expression 1[p]
According to the C Standard (6.5.2.1 Array subscripting)
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).
So this sub-expression evaluates like *( ( 1 ) + ( p ) ) and is the same as *( ( p ) + ( 1 ) ).
Thus the above statement
x = -1[p];
is equivalent to
x = -p[1];
and will yield -3, because the pointer p points to the second element of the array due to the statement
int *p = &array[1];
and then the expression p[1] yields the value of the element after the second element of the array. Then the unary operator - is applied.
This
int array[] = {1, 2, 3};
looks like
array[0] array[1] array[2]
--------------------------
| 1 | 2 | 3 |
--------------------------
0x100 0x104 0x108 <-- lets assume 0x100 is base address of array
array
Next when you do like
int *p = &array[1];
the integer pointer p points to address of array[1] i.e 0x104. It looks like
array[0] array[1] array[2]
--------------------------
| 1 | 2 | 3 |
--------------------------
0x100 0x104 0x108 <-- lets assume 0x100 is base address of array
|
p holds 0x104
And when you do like
x = -1[p]
-1[p] is equivalent to -(1[p]) i.e -(p[1]). it looks like
-(p[1]) ==> -(*(p + 1*4)) /* p holds points to array[1] i.e 0x104 */
==> -(*(0x104 + 4))
==> -(*(0x108)) ==> value at 0x108 is 3
==> prints -3
What happens here is really interesting.
p[n] means *(p+n). Thats why you see 3, because "p" points to array[1] which is 2, and -p[1] is interpreted as -(*(p+1)) which is -3.
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].
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.