[Solved] The following code tries to print the size of a short variable but causes a compilation error:short s;printf("%d", sizeof(s));What is the main reason for the error? | C

C - Variables and Data Types

The following code tries to print the size of a short variable but causes a compilation error:

short s;
printf("%d", sizeof(s));

What is the main reason for the error?

AVariable <code>s</code> is not initialized

BMissing semicolon after <code>short s</code>

CCannot use <code>sizeof</code> on variables

DUsing wrong format specifier for <code>sizeof</code> result

Step-by-Step Solution

Solution:

Step 1: Identify the type returned by sizeof
The sizeof operator returns a value of type size_t, which is an unsigned integer type.
Step 2: Check printf format specifier compatibility
The format specifier %d expects an int, but sizeof returns size_t. The correct specifier is %zu.
Final Answer:
Using wrong format specifier for sizeof result -> Option D
Quick Check:
Use %zu for sizeof output [OK]

Quick Trick: Use %zu to print sizeof results safely [OK]

Common Mistakes:

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