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GATE | GATE MOCK 2017 | Question 44

  • Difficulty Level : Hard
  • Last Updated : 28 Dec, 2020

Consider the following C code

int main()
   int a = 300;    
   char *b = (char *)&a;
   *++b = 2;
   printf("%d ",a);
   return 0;

Consider the size of int as two bytes and size of char as one byte. Predict the output of the following code .
Assume that the machine is little-endian.
(A) 556
(B) 300
(C) Runtime Error
(D) Compile Time Error

Answer: (A)

Explanation: The binary equivalent of 300 is 00000001 00101100. Since the machine is little endian, 00000001 00101100 is stored as 00101100 00000001 in the memory in the form of an array.

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Pointer b is then given the address of a. It means that the address of first byte of 00101100 00000001 is allotted to b (hence b is allotted 00101100).

*++b = 2 increments the value of b by 1 so that b now points to the next memory address which is 00000001 and the binary equivalent of 2 (00000010) will be stored here.

Finally, a becomes 00101100 00000010, which in decimal is basically 556.

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