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
Solution
The output of the code will be "HelloWorld". Here's why:
-
The variable
xis an unsigned integer with a value of0xDEADBEEF. In decimal, this is a large positive number. -
The variable
yis an unsigned short with a value of0xFFFF. When cast to a signed short, this becomes-1because the most significant bit is interpreted as the sign bit in a two's complement representation. -
The variable
zis a signed integer with a value of-1. -
The first
ifstatement checks ifxis greater than(signed short) y. Sincexis a large positive number and(signed short) yis-1, this condition is true. Therefore, "Hello" is printed. -
The second
ifstatement checks ifxis greater thanz. Again, sincexis a large positive number andzis-1, this condition is true. Therefore, "World" is printed.
So, the output of the code is "HelloWorld", which corresponds to option (d).
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