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What will be the output of the below code? the answer given is E)0 but I am not getting it.

#include <stdio.h>
void fun(short int *a,char *b)
    b += 2;
    short int *p = (short int*)b;
    *p = *a;

int main()
    void (*fptr)(short int *,char *)
    short int a = 101;
    char arr[] = {'a','b','c','d'};
    fptr = fun;
    printf("%d", arr[3]);

    return 0;

$A)$ Compilation error.

$B) 100$

$C)$ Garbage Value

$D)$ Segmentation Fault.

$E) 0$

in Programming by (41 points)
edited by | 190 views

1 Answer

+1 vote
Best answer

go through these links before reading answer if these concepts are not clear




A[0](1000+0) A[1](1000+1) A[2](1000+2) A[3](1000+3)
a b c d

short int *p = (short int*)b;

what this line does is assigns address 1002 in "p" in such a way that it will take 2 bytes 1002 and 1003(since it takes 2 bytes)

Now, when we assign 101 to it because of little-endian format, the higher order byte having 0 will be placed in right (A[3]) and lower order byte having value 101 at A[2], What you can do to test this is run the same program with 300 O/P would be 1 because higher order byte has 1(requires 256) 



by Active (2.1k points)
edited by

Thanks @Anuj Mishra but, is it syntactically correct to assign fun to fptr?


Yes it is correct (function name itself is like an array name - so you can remove & and it still works - definitely go through this for reading and examples-

Also if  times, search wiki for function pointer - there you'll get a good example and some important usage too


@Anuj Mishra

Please Separate both the links in the post. They are showing as one url.


Thanks @logan1x , edited now.


@Anuj Mishra

Can u plz explain this line

fptr = fun;

Here fptr is a pointer, which should take address of 'fun'. Is it not?

Yes it is, read my above comment and go through link in that comment

How a function can be used as an array?

Even in array we use sign. isnot it?


No. We don't use & for referring whole array. Go through the first answer, same is case for functions ( please read above link if you haven't )

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