Difficulty: Easy
Correct Answer: C:\TC\MYPROC.EXE
Explanation:
Introduction / Context: The problem reinforces that argv[0] is conventionally the program name or path with which the program was invoked. Under DOS/Turbo C, it often includes the full path and .EXE name when launched from the shell.
Given Data / Assumptions:
Concept / Approach: argv[0] is populated by the host environment. Although not strictly standardized to be a full path, in this DOS/Turbo C context it typically is something like C:\\TC\\MYPROC.EXE when executed directly from that location.
Step-by-Step Solution:
At launch, argv[0] receives the path of the executable. printf("%s", argv[0]) prints that exact string. Hence, output is C:\\TC\\MYPROC.EXE.Verification / Alternative check: If you placed the EXE elsewhere or invoked via PATH without full path, argv[0] might differ, but the question specifies the DOS/Turbo C scenario yielding the full path.
Why Other Options Are Wrong:
Common Pitfalls: Assuming argv[0] is always just the program name; environment conventions vary across systems and shells.
Final Answer: C:\\TC\\MYPROC.EXE
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