I am a student who will be a freshman in an university after this summer vacation.I want to learn about computer programing in advance but I run into some problems. Why when I run the program in devc ,the result is -1 and 44? When I read the book called Pointers On C,In the chapter on functions,the book says that the name of array is a pointer,and in C language arr[m]=*(arr m),and arr[0] is composed by a pointer and a [number],so can I come to a conclusion that (arr 2),which is a pointer,and[0],can compose (arr 2)[0] equaling to *(arr 2 0)?
int main(void)
{
int arr[10];
for(int i=0;i<10;i )
{
arr[i]=i 1;
}
int b=*(arr 1);
int c=(arr 2)[0];//Is this true?
printf("%d\n",b);
printf("%d",c);
return 0;
}
CodePudding user response：
Is the use of (arr 2)[0] right?
Linguistically yes. But very questionable practice.
The beautiful invention of C is pointer arithmetic and that is exemplified by the definition of the subscript operator [] is that E1[E2] is identical to (*((E1) (E2))).
So by definition (arr 2)[0] is the same as (*((arr 2) (0))) and simplifies to *(arr 2) and arr[2] simplifies to the same thing.
Array subscripting in C is syntactic sugar on pointer arithmetic. The term 'syntatic sugar' is sometimes overused. But in this case it really is just syntactic sugar.
The example demonstrates a critically important fact about how C manages arrays. Pointer arithmetic works in units of the type (measured in bytes they occupy) and array subscripting the same. So adding to a pointer (arr 2) and subscripting into an array (arr[2]) have a deep equivalence.
When you understand that and realise the that array subscripts start at 0 because they're offsets you get "C".
Please never write code like that. Writing code like 7[A] gets funny looks.
CodePudding user response：
C allows you to do all sorts of clever things with arrays and pointers.
This is either a good thing (if you like writing clever code) or very frustrating (if later you have to maintain some of that clever code)
All of the following are equivalent:
(arr 2)[0]arr[2]2[arr]*( ( arr 2 ) 0 )*( arr 2 )
While they are equivalent, that doesn't necessarily mean you should use them interchangeably - my general recommendation is to use array notation when accessing an array, and pointer arithmetic when indexing into any other structures. Decaying an array into a pointer is prone to errors!
In noting that you have defined your data structure as an array int arr[10]; my advice would be to access it as arr[2] - then everyone is clear what you are trying to do, both now and in teh future.
As an aside (and see profile for affiliation) MISRA C dedicates a whole chapter (8.11) to pointer type conversion, and another (8.18) to the pitfalls of arrays and pointers - a fair bit of guidance and a total of 17 Rules between them...
CodePudding user response：
There is a typo in your code
for(int i=0;i<10;i )
{
arr[0]=i 1;
}
It seems you mean
for(int i=0;i<10;i )
{
arr[i]=i 1;
}
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 2)[0]
evaluates like
*( ( arr 2 ) 0 )
That is the same as
*( arr 2 )
and in turn is the same as
arr[2]
You even may write
2[arr]
Pay attention to that the expression
(arr 2)[0]
will be more clear if to introduce an intermediate pointer like
int *p = arr 2;
and then to use the expression
p[0]
CodePudding user response：
#include <stdio.h> // printf
int main(void)
{
int arr[10];
printf("fill and print the array\n");
for(int i=0;i<10;i )
{
arr[i]=i 1;
printf("%d -th element of the array = arr[%d] = %d\n",i 1, i, arr[i]);
}
printf("pointer arithmetic in c \n");
int b=*(arr 1); // arr[m]=*(arr m)
int c=(arr 2)[0];//(arr 2)[0] = *(arr 2 0)
int n = 0;
printf("%d -th element of the array = arr[%d] = %d\n",n 1, n , *(arr n));
printf("%d -th element of the array = %d\n",2,b);
printf("%d -th element of the array = %d\n",3,c);
return 0;
}
compile
gcc t.c -Wall -Wextra
run
./a.out
result:
fill and print the array
1 -th element of the array = arr[0] = 1
2 -th element of the array = arr[1] = 2
3 -th element of the array = arr[2] = 3
4 -th element of the array = arr[3] = 4
5 -th element of the array = arr[4] = 5
6 -th element of the array = arr[5] = 6
7 -th element of the array = arr[6] = 7
8 -th element of the array = arr[7] = 8
9 -th element of the array = arr[8] = 9
10 -th element of the array = arr[9] = 10
pointer arithmetic in c
1 -th element of the array = arr[0] = 1
2 -th element of the array = 2
3 -th element of the array = 3
HTH