Comparing two distinct C arrays of identical contents: what does this program print? #include<stdio.h> int main() { char str1[] = "Hello"; char str2[] = "Hello"; if (str1 == str2) printf("Equal "); else printf("Unequal "); return 0; }

Introduction / Context:In C, arrays are not directly comparable by value with ==. When used in expressions, array names decay to pointers to their first elements. This program compares the addresses of two separate arrays, not their string contents.

Given Data / Assumptions:

• Two separate arrays: str1 and str2, each initialized with "Hello".
• The expression str1 == str2 compares the addresses of the first elements of each array.
• They are distinct objects in memory.

Concept / Approach:Since str1 and str2 denote different storage, their addresses are different, so pointer equality is false. Value comparison of strings requires strcmp (or memcmp for raw bytes) rather than ==. Therefore the else branch executes and prints "Unequal".

Step-by-Step Solution:str1 and str2 each hold "Hello" and terminate with '\0'.When compared with ==, they decay to pointers (char*).The pointers do not reference the same memory location, so comparison is false.Program prints "Unequal".

Verification / Alternative check:Using if (strcmp(str1, str2) == 0) would print "Equal". Or assign one pointer to reference the other’s buffer (char *p = str1; char *q = str1;), then p == q would be true.

Why Other Options Are Wrong:“Equal” mistakes value equality for pointer equality. “Error” is incorrect; code compiles. The “depends on compiler” option is wrong; the objects are distinct by definition.

Common Pitfalls:Confusing array names with pointers; relying on == for string content comparison instead of strcmp.

Final Answer:Unequal