[Solved] Identify the error in the following code snippet:char name[10];scanf("%c", name); — Ans: Use %s instead of %c to read string | C

C - Input and Output

Identify the error in the following code snippet:

char name[10];
scanf("%c", name);

ANo error, code is correct

BUse &name instead of name

CUse %d instead of %c

DUse %s instead of %c to read string

Step-by-Step Solution

Solution:

Step 1: Understand scanf format specifiers
%c reads a single character, but name is an array for a string.
Step 2: Correct specifier for string input
To read a string into name, use %s which reads until whitespace.
Final Answer:
Use %s instead of %c to read string -> Option D
Quick Check:
String input needs %s in scanf [OK]

Quick Trick: Use %s to read strings with scanf [OK]

Common Mistakes:

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