C Programming - Arrays - Discussion

Discussion Forum : Arrays - General Questions (Q.No. 2)
2.
What does the following declaration mean?
int (*ptr)[10];
ptr is array of pointers to 10 integers
ptr is a pointer to an array of 10 integers
ptr is an array of 10 integers
ptr is an pointer to array
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
64 comments Page 5 of 7.

Neeraj awasthi said:   1 decade ago
If () is not there then int *ptr[] start with [] and it means array of pointers of type integers.

Nagaraj said:   1 decade ago
The below program explains tat the starting address of the array. so it is [int (*ptr)[5];] is a pointer to array not array pointer which can be declared by int *ptr[5];
#include<stdio.h>
int main()
{
int arr[5];
int (*ptr)[5];
ptr=&arr[10];

printf("%ld",ptr[5]);
return 0;
}

Subaidha said:   1 decade ago
* means its represent as a pointer variable and here within a array bracket they give us 10 integers.

So that the ans is B.

Balaji said:   1 decade ago
Can you explain me the concept?

Purna chandra said:   1 decade ago
Yes p is pointer of 10 integers.

Ganesh said:   1 decade ago
*ptr is a pointer and[10] is array declarations.

LOL said:   1 decade ago
The brackets are the key to this problem. Always read C declarations from right to left (backwards) and start will stuff inside brackets.

[] = array * = pointer

int *ptr [10] backwards is [10] ptr * int

so this is Array of 10 pointers to int

int (*ptr)[10] bracket first and backwards is * ptr [10] int

so this is pointer to an array of 10 integers

Chris said:   1 decade ago
#include<stdio.h>
int main()
{
int a=5;
printf("%d %d %d %d %d", a++, a--, ++a, --a, a);
return 0;
}

// Ans : 4 5 5 4 5

Can any one explain the above code please ?

Sagar Said said:   2 decades ago
Can you explain me the concept

Apurva Nigam said:   1 decade ago
@Chris and @Sundar:

"printf("%d %d %d %d %d", a++, a--, ++a, --a, a); " evaluates the value passed in it from right to left(int turbo C).
That is
first "a" will get its value as 5
then --a will store 4
then ++a will store 5 (as "--a" had decremented a's value n ++a incremented it)
then "a--" will have 5 ( as a-- is post increment therefore a's value will be affected after evaluation of the expression "a--")
then "a++" will have value 4 (b'coz above "a--" has made its value 4)

Therefore output is: 45545
Hope u know preincrement increments the value first and then use it in the expression, whereas postincrement uses the previous value of the variable and then modifies it thet is increments by 1.

eg:
int x=5 ,y,z;
y= ++x; //first x will become 6 then y will get assigned 6
z= x++; //first x will get assigned in z that is z=6 then x gets incremented.

Good night :) :)
Hope u understood....


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