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What is the output of the following code?Assume that int is 32 bits, short is 16 bits, and the representation is two’s complement.unsigned int x = 0xDEADBEEF;unsigned short y = 0xFFFF;signed int z = -1;if (x > (signed short) y)printf("Hello");if (x > z)printf("World");(a) Prints nothing.(b) Prints ”Hello”(c) Prints ”World”(d) Prints ”HelloWorld

Question

What is the output of the following code?Assume that int is 32 bits, short is 16 bits, and the representation is two’s complement.unsigned int x = 0xDEADBEEF;unsigned short y = 0xFFFF;signed int z = -1;if (x > (signed short) y)printf("Hello");if (x > z)printf("World");(a) Prints nothing.(b) Prints ”Hello”(c) Prints ”World”(d) Prints ”HelloWorld

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Solution

The output of the code will be "HelloWorld". Here's why:

  1. The variable x is an unsigned integer with a value of 0xDEADBEEF. In decimal, this is a large positive number.

  2. The variable y is an unsigned short with a value of 0xFFFF. When cast to a signed short, this becomes -1 because the most significant bit is interpreted as the sign bit in a two's complement representation.

  3. The variable z is a signed integer with a value of -1.

  4. The first if statement checks if x is greater than (signed short) y. Since x is a large positive number and (signed short) y is -1, this condition is true. Therefore, "Hello" is printed.

  5. The second if statement checks if x is greater than z. Again, since x is a large positive number and z is -1, this condition is true. Therefore, "World" is printed.

So, the output of the code is "HelloWorld", which corresponds to option (d).

This problem has been solved

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