In linux system why this?

[smartbeginner]

I am not able to use pointer to array in Linux .It always shows error
when I try to read from keyboard or print to screen.
main()
{
int i,j;
int (*a)[2];
a=(int *)calloc(2,sizeof(int));
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",(*(a+i)+j));
}
This shows segmentation fault?Why is this happening
But array of pointers works.

[Scorpio]

smartbeginner wrote:

I am not able to use pointer to array in Linux .It always shows error
when I try to read from keyboard or print to screen.
main()
{
int i,j;
int (*a)[2];
a=(int *)calloc(2,sizeof(int));
casting is redundant for calloc because it returns void* type.
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",(*(a+i)+j));
}
This shows segmentation fault?Why is this happening
Because you are accessing memory location that does
not belong to your program. You are allocating only enough memory
for 2 integers and scanning into 4 integers.
But array of pointers works.

I don't know how it worked, show us the code.

[John Bode]

smartbeginner wrote:

I am not able to use pointer to array in Linux .It always shows error
when I try to read from keyboard or print to screen.
main()
{
int i,j;
int (*a)[2];
a=(int *)calloc(2,sizeof(int));
This cast is both wrong and unnecessary. Remove it from the statement.
Actually, it would be better just to write

a = malloc(sizeof *a);

Much simpler.
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",(*(a+i)+j));
}
This shows segmentation fault?Why is this happening
Because you are attempting to access memory outside of what you've
allocated. You have allocated an array of 2 ints, (*a)[0] and (*a)[1].
When both i and j are 1, you are trying to write to (*a)[2], which is
outside of the memory you allocated.
But array of pointers works.

[August Karlstrom]

smartbeginner wrote:

I am not able to use pointer to array in Linux .It always shows error
when I try to read from keyboard or print to screen.
main()
{
int i,j;
int (*a)[2];
a=(int *)calloc(2,sizeof(int));
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",(*(a+i)+j));
}
This shows segmentation fault?Why is this happening
But array of pointers works.

Your code contains lots of errors. What sources have you used when
learning C, I'm just curious?
August

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[smartbeginner]

Hello Sir,

Can you please point out my errors.So that I can study C better.Thank
you for your speedy reply

[Keith Thompson]

"smartbeginner" <rr******@gmail.com> writes:

Hello Sir,

Can you please point out my errors.So that I can study C better.Thank
you for your speedy reply

Many of your errors are documented in <http://cfaj.freeshell.org/google/>.
Fix those, and then we can talk about errors in C.

--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[Barry Schwarz]

On 26 Dec 2005 05:06:24 -0800, "smartbeginner" <rr******@gmail.com>
wrote:

I am not able to use pointer to array in Linux .It always shows error
when I try to read from keyboard or print to screen.
main()
{
int i,j;
int (*a)[2];
a=(int *)calloc(2,sizeof(int));
If your compiler did not issue a diagnostic for this statement, you
need to correct your configuration parameters or get one that deals
with the C language. The right hand side of your statement has type
pointer to int (courtesy of the cast). The left hand side has type
pointer to array of two int (from the definition of a). There is no
implicit conversion between these types. This is a constraint
violation requiring a diagnostic.

While we recommend not casting the return from the allocation
functions, in this case your cast serves to indicate that you do not
understand the type of a.

If you remove the cast (and remember to include stdlib.h), the
statement should allocate space for one array of two int. If you had
really intended to allocate space for two arrays, you should follow
the recommended practice of
a = calloc(2, sizeof *a);
In this case, malloc may be more appropriate since you never use the
zero values that calloc stores in the allocated area.
for(i=0;i<2;i++)
for(j=0;j<2;j++)
scanf("%d",(*(a+i)+j));
Here is where you introduce undefined behavior. You should consider
yourself fortunate that the UB manifested itself in an immediately
obvious manner. This discussion group is full of queries from
confused people who were not so fortunate.

Variable a is a pointer to an object (in this case an array of two
int.)

The expression a+i involves pointer arithmetic. (Pointer
arithmetic is different than "normal" arithmetic because the size of
the object pointed to participates. If p is a pointer to an object of
size 5, then p+i evaluates to an address that is 5*i bytes greater
than the value in p.) The expression evaluates to the address of the
i-th object after the one pointed to. Note that expression a+0, or
the equivalent a, evaluates to the address of the object pointed to,
which is exactly what we mean when we say a is a pointer to an object.

*a and the equivalent *(a+0) evaluate to the object, in this case
an array of two int.

*(a+i) evaluates to the i-th object after the one pointed to by a.
Unfortunately, in this case there is no such object whenever i is
greater that 0. You only allocated space for one array.
}
This shows segmentation fault?Why is this happening
But array of pointers works.

An array of pointers can work. A pointer to an array can work. A
dynamic 2-dimensional array pointed to by an int** can work. There
are probably several more approaches that can work. All you need is
the correct code for the approach you choose.
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