In C printf, assume the intent is to print a decimal integer with a minimal field width of 1. What does this program output (ignoring a stray non-printing control glyph in the source)? #include <stdio.h> int main() { int a = 250; printf("%1d", a); /* treat the strange trailing glyph as a typo and consider %1d */ return 0; }

Introduction / Context:
The original snippet shows "%1d" followed by an unusual control character. Applying the Recovery-First Policy, we interpret the intent as printing the integer with the standard format "%1d". Field width 1 is the minimum width, so any multi-digit number prints fully.

Given Data / Assumptions:

• a = 250.
• Format is effectively "%1d" (minimum width 1).
• No additional flags or precision are set.

Concept / Approach:
In printf, the field width is a minimum. If the value needs more columns, it expands to fit. Therefore, a three-digit number like 250 prints as "250" without truncation.

Step-by-Step Solution:
Compute textual representation of 250 → "250".Compare with field width 1 → width is satisfied; no padding necessary.Printed result is exactly "250".

Verification / Alternative check:
Testing with widths smaller or larger (e.g., %5d) changes only padding, not the digits themselves.

Why Other Options Are Wrong:
"2" or "50" would imply truncation, which printf does not do for integers. "1250" would require a different argument, not present here.

Common Pitfalls:
Misunderstanding field width as a directive to clip digits rather than a minimum width for padding.

Final Answer:
250