In C, printing the first character of each user argument. For the program below executed as: cmd> myprog one two three, what does it print (characters are contiguous, no spaces)? /* myprog.c */ #include<stdio.h> int main(int argc, char *argv[]) { int i; for(i = 1; i < argc; i++) printf("%c", argv[i][0]); return 0; }

Introduction / Context: This problem checks array-of-strings indexing through argv and character indexing within each string. The program prints the first letter of each user-supplied argument in order.

Given Data / Assumptions:

• Arguments: one, two, three.
• Loop: i from 1 to argc - 1.
• Character printed: argv[i][0] (first character of each argument).

Concept / Approach: argv is an array of char*; argv[i] is a pointer to the i-th C string. Index zero selects the first character of that string. Concatenation results from printing with %c and no separators.

Step-by-Step Solution:

Verification / Alternative check: Changing the format to "%c " would show "o t t " with spaces; the characters are the same in order.

Why Other Options Are Wrong:

• oot: That would require "o", "o", "t".
• nwh / eoe: These pick other letters (e.g., second or third letters).
• tto: Wrong order.

Common Pitfalls: Accidentally starting the loop at i=0 prints the first letter of argv[0] (program name) too; here we intentionally start at the first user argument.

Final Answer: ott