[Solved] Identify the problem in this code:int x = 7; int y = 2; int z = x / y; printf("%f", z); — Ans: Using %f to print an int variable | C

C - Operators and Expressions

Identify the problem in this code:

int x = 7;
int y = 2;
int z = x / y;
printf("%f", z);

AIncorrect variable declaration

BDivision operator is incorrect

CUsing %f to print an int variable

DMissing variable initialization

Step-by-Step Solution

Solution:

Step 1: Check variable types and printf format
Variable z is int but printed with %f (float format).
Step 2: Understand format specifier mismatch
Using %f for int causes undefined or wrong output.
Final Answer:
Using %f to print an int variable -> Option C
Quick Check:
Match printf format with variable type [OK]

Quick Trick: Use %d for int, %f for float in printf [OK]

Common Mistakes:

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