[Solved] Identify the problem in this code:#include <stdio.h> int main() { int x; printf("Enter a number: "); scanf("%d", &x); printf("You entered %d", x); return 0;... | C

C - Input and Output

Identify the problem in this code:

#include <stdio.h>
int main() {
  int x;
  printf("Enter a number: ");
  scanf("%d", &x);
  printf("You entered %d", x);
  return 0;
}

When run, the prompt does not appear before input is requested. Why?

AVariable x is not declared

Bscanf syntax is wrong

Cprintf format specifier is incorrect

DOutput is buffered; missing fflush(stdout)

Step-by-Step Solution

Solution:

Step 1: Understand output buffering
printf output may be buffered and not shown immediately before scanf waits for input.
Step 2: Fix buffering issue
Using fflush(stdout) after printf forces output to display before input.
Final Answer:
Output is buffered; missing fflush(stdout) -> Option D
Quick Check:
Use fflush to show prompt before input [OK]

Quick Trick: Use fflush(stdout) to flush output buffer [OK]

Common Mistakes:

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