C Programming - Complicated Declarations - Discussion

Discussion Forum : Complicated Declarations - Find Output of Program (Q.No. 11)
11.
What will be the output of the program?
#include<stdio.h>

int main()
{
    char huge *near *ptr1;
    char huge *far *ptr2;
    char huge *huge *ptr3;
    printf("%d, %d, %d\n", sizeof(ptr1), sizeof(ptr2), sizeof(ptr3));
    return 0;
}
4, 4, 8
2, 4, 4
4, 4, 2
2, 4, 8
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Arvind said:   6 years ago
I am not getting this. Please explain me.

Nachiket R Rajput said:   7 years ago
Near is of 2 byte and far and huge are of 4 byte.

Kavya said:   7 years ago
I can't understand it. Please explain it.

Prarthana said:   7 years ago
I didn't understand. It gives an error message.

D:\class1\1\main.c|7|error: expected '=', ',', ';', 'asm' or '__attribute__' before '*' token|
D:\class1\1\main.c|5|error: 'near' undeclared (first use in this function)|

Please help me.

Anunita said:   8 years ago
Three types of pointer is there. Near, far and huge which consists of size 2, 4, 4 respectively.

Sindhu said:   9 years ago
How can you say that near contains only 2 bytes and far, huge contains 4 bytes?
(1)

Rajani sain said:   9 years ago
I didn't understand, please clarify the answer.

Mt_takeshi said:   10 years ago
This won't even compile according to C89, C99 or C11.

Anu said:   1 decade ago
I didn't understand, please explain again?

PowerStar said:   1 decade ago
*ptr1 is of type near pointer hence the size is 2 bytes but whereas the *ptr2 and *ptr3 is of type far and huge pointers the size of huge and far pointers are 4 bytes.


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