C programming

Question

What is the value of the following expression?

 

main()

{  

 int i = 257;  int *iPtr = &i;  

 printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) ); }  

 

 (A) 1 1   (B)1 2     (C) 2 3  

Ans given option a pls explain.

Comments

I think it'll depend upon whether the machine supports big endian or little endian format.

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@gauravkc , yes o/p is dependent upon big endian or little endian format support by that machine

i hope, he is not interested in answer... he is interested in what happening with those statements.

 

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By default we use little indian.

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Is the ans 11....

Answer 1

Option a is right

Comments

@abhisekh

Please make me understand below points

1.How you have taken values of 1000-> 00000001 and 1001 also same.?

2.what are big endian and little endian and how they affect the output

Please suggest

Thanks

Answer 2

let assume integer takes 2 Bytes, Character takes 1 Byte and Pointer (Any type) = 4 Bytes

i=257 = 256+1 ===> 00000001 00000001 ====> For this example, O/p can not effect by endienes of system.

Values	00000001	00000001
Memory Address	101	100

 

 *( (char*)iPtr ) ===> 1) iPtr converted in Character pointer (note that character pointer points only 1 Byte Memory where integer Pointer points 2 Bytes of Memory, actually depends upon size of the datatype)

 ===============> 2) returns the value at that Memory Address which is pointed by the iPtr

 

 *( (char*)iPtr +1 ) ===> 1) iPtr converted in Character pointer (note that character pointer points only 1 Byte Memory where integer Pointer points 2 Bytes of Memory, actually depends upon size of the datatype)

 ===============> 2)increment the address by one 

 ===============> 3) returns the value at that Memory Address which is pointed by the iPtr

 

o/p is 1 1

 

more details https://www.geeksforgeeks.org/little-and-big-endian-mystery/