In C, what does this code print when comparing only the first two bytes?\n\n#include<stdio.h>\n#include<string.h>\n\nint main()\n{\n char dest[] = {97, 97, 0}; // "aa\0"\n char src[] = "aaa"; // "aaa\0"\n int i;\n if ((i = memcmp(dest, src, 2)) == 0)\n printf("Got it");\n else\n printf("Missed");\n return 0;\n}\n\nChoose the observed output.

Introduction / Context:
memcmp compares raw memory byte-by-byte for a specified count. Here we compare the first two bytes of two arrays that start with the characters 'a' and 'a'.

Given Data / Assumptions:

• dest = {97, 97, 0} which is "aa\0".
• src = "aaa" which is {97, 97, 97, 0}.
• memcmp(dest, src, 2) compares exactly two bytes.
• memcmp returns 0 when the compared ranges are equal.

Concept / Approach:
Only the requested byte count matters. Since the first two bytes of both arrays are 97 and 97 ('a' and 'a'), memcmp will find them equal for the first two bytes and return 0.

Step-by-Step Solution:
First byte: dest[0] = 97, src[0] = 97 → equal.Second byte: dest[1] = 97, src[1] = 97 → equal.memcmp returns 0 → condition is true → prints "Got it".

Verification / Alternative check:
If we changed the count to 3, the third byte would be 0 vs 97, and memcmp would return a negative value; the program would then print "Missed".

Why Other Options Are Wrong:
Missed: would require a mismatch in the first two bytes, which is not the case.Error in memcmp statement: The call is valid.None of above: Not applicable.

Common Pitfalls:
Assuming C-strings must be compared up to '\0' with memcmp; memcmp does not stop at null bytes and respects the explicit length.

Final Answer:
Got it