Pointers

Question

Please explain this question

 

void main()

{
    int a =300;

    char *ptr = (char*) &a ;

    ptr++;

    *ptr=2;

    printf("%d" , a);

}

 

Answer 1

Assuming char = 1 byte and int = 4 bytes.

This question requires you to know the concept of little endian and pointer arithmetic.

Data which requires multiple bytes are stored in most modern computers in memory using the little endian system.

In little endian machines, last byte of binary representation of the multibyte data-type is stored first.

So let’s assume that the integer a with the value 300 is stored at the memory location 1000.

300 = 0b100101100

That memory location will look like this:

1000	1001	1002	1003
0010 1100	0000 0001	0000 0000	0000 0000

Now you are storing the address of an integer to a character pointer, ptr.

So ptr = 1000.

ptr++ means the pointer ptr will start pointing to the next character and not the next int (since it is a character pointer).

So, ptr = 1001.

You can use the formula ptr + 1 = ptr + 1*sizeof(*ptr) = 1000 + sizeof(char) = 1000 + 1 = 1001.

In the next line when you are de-referencing the pointer ptr, you are actually de-referencing the value stored at the memory address 1001,

So after this assignment the memory location will look like this:

1000	1001	1002	1003
0010 1100	0000 0010	0000 0000	0000 0000

 

Now when you print the value(a) that is stored at this memory address, the program will first reverse the byte order and then evaluate the binary representation as decimal.

(00000000 00000000 00000010 00101100)b in decimal = 556.

Therefore, this program will print 556.

Comments

Thank you so much for this detailed explanation.

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no worries dude ;)