What does (*ptr)[10] mean? - c

Question

void main()
{
int (*d)[10];
d[0] = 7;
d[1]=10;
printf("%d\n",*d);
}
It should print 10 but compiler is showing error such as follows:
test.c:4:7: error: incompatible types when assigning to type ‘int[10]’ from type ‘int’
Note that I have included some errors , not all.

Answer 1

As noted by chris, d is a pointer to an array. This means you use the variable improperly when you access it, but also that you will access random memory unless you assign d to point to a valid array.
Change your program as follows:
int main(void)
{
int (*d)[10]; /* A pointer to an array */
int a[10]; /* The actual array */
d = &a; /* Make `d` point to `a` */
/* Use the pointer dereference operator (unary prefix `*`)
to access the actual array `d` points to */
(*d)[0] = 7;
(*d)[1] = 10;
/* Double dereference is okay to access the first element of the
arrat `d` points to */
printf("%d\n", **d);
return 0;
}

Answer 2

In C, [] is the same as *, the pointer syntax. Thus the following lines are the same:
int** array2d1;
int* array2d2[];
int array2d3[][];
To relate to a closer example, the main function has the following popular forms:
int main(int argc, char** argv){ ... }
or
int main(int argc, char* argv[]){ ... }
Thus
int (*d)[10]
is the same as
int* d[10]
which is the same as
int** d;
int firstArray[10];
d = &firstArray;
Effectively, you are creating a pointer to a pointer (which is a pointer to an array) and allocating the first pointer to an array that 10 elements. Therefore, when you run the following lines:
d[0] = 7;
d[1] = 10;
You are assigning the 1st array's address to 7 and the second array's address to 10. So as Joachim has mentioned, to assign values, you need to deference twice:
(*d)[0] = 7
(*d)[1] = 10
Which says "Assign 7 to the 0th index at the value pointed by d". I hope that makes sense?

Answer 3

d is a pointer to an array of 10 ints.
int (*d)[10] is the declaration for a point to an array of 10 ints.
vs.
int *d[10], which is an array of 10 int pointers.
For more complex syntax like this (usually involving pointers), I use cdecl to help me decode it.

Answer 4

It's used in this form
int d[10]
I guess you are mistaken that d must be a "kind of pointer" and therfor you put an * before the d.
But that's not what you want. You wan to name an array of integer and the notation for that is seen above.

Answer 5

Concept of pointer can get confusing sometimes in C.
Consider an array int d[6] = {0,1,2,3,4,5}
Then, *d is equivalent to d[0]. d is itself an pointer to an array and *d dereferences that pointer and gives us the value.
Hence, following code would print the same values:
int main()
{
int (d)[10];
*d = 7;
*(d + 1)=10;
printf("%d\n",*d);
printf("%d\n",d[0]);
return 0;
}
result:
7
7
Please see http://codepad.org/LYY9ig1i.
If you change your code as follows:
#include<malloc.h>
int main()
{
int *d[10]; //not (*d)[10]
d[0] = (int *)malloc(sizeof(int *) * 10);
d[0][0] = 7;
printf("%d\n",d[0][0]);
return 0;
}
Hope this helps you!