[Solved] What is wrong with this code? char name[20]; printf("Enter name: "); scanf("%c", &name); — Ans: Using %c instead of %s for string input | C

C - Input and Output

What is wrong with this code?

char name[20];
printf("Enter name: ");
scanf("%c", &name);

AUsing %c instead of %s for string input

BMissing & before name

CArray size too small

DNo error

Step-by-Step Solution

Solution:

Step 1: Check format specifier for string input
To read a string, %s should be used, not %c.
Step 2: Confirm pointer usage
Passing &name is incorrect; for arrays, just name is used with %s.
Final Answer:
Using %c instead of %s for string input -> Option A
Quick Check:
Use %s for strings, not %c [OK]

Quick Trick: Use %s with array name (no &) for strings [OK]

Common Mistakes:

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