Pointer arithmetic on string literals with printf: what is printed? #include<stdio.h> int main() { printf(5 + 'Good Morning '); return 0; }

Introduction / Context:
String literals are arrays of characters that decay to pointers when used in expressions. Adding an integer to such a pointer advances the pointer by that many characters. Passing the offset pointer to printf prints from that position onward.

Given Data / Assumptions:

• Literal is 'Good Morning\n'.
• Characters 0–4 are 'G o o d ␠' (space at index 4).
• 5 + literal points at the character 'M' of 'Morning'.

Concept / Approach:
printf with a single pointer argument and no format specifiers treats the argument as a format string. Here the pointed-to substring is 'Morning\n'. It contains no % directives, so it prints as plain text starting at 'M'.

Step-by-Step Solution:

Verification / Alternative check:
Change the offset to 0 to print the full phrase; change to 6 to print 'orning\n'.

Why Other Options Are Wrong:

Common Pitfalls:
Forgetting that pointer addition steps by characters; misunderstanding that printf without format specifiers still prints the string when the data contains no % symbols.

Final Answer:
Morning.