C Program to Precedence Puzzle with Explanation
C
Easy
Operators
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1 min read
175 words
This problem helps you practice core C fundamentals in a practical way. It builds intuition around result, precedence, operator. Let’s break it down step by step so you can implement it confidently.
Problem Statement
Write the expression: a = 5, b = 10, c = 15; result = a + b * c / 5 - 2 % 3; calculate manually, following operator precedence. Then verify with the program.
Input Format
a = 5
b = 10
c = 15
result = a + b * c / 5 - 2 % 3
Output Format
Value of result.
Constraints
• Integer arithmetic
• Standard C operator precedence rules apply
Concept Explanation
The expression is evaluated using operator precedence and left-to-right associativity.
Step-by-Step Logic
1.Substitute values:
result = 5 + 10 * 15 / 5 - 2 % 3
2.Evaluate *, /, % first (left to right):
• 10 * 15 = 150
• 150 / 5 = 30
• 2 % 3 = 2
3.Expression becomes:
5 + 30 - 2
4.Evaluate + and -:
• 5 + 30 = 35
• 35 - 2 = 33
5.Final value stored in result is 33.
Code Solution
This explanation is written for learning purposes and to help beginners understand the concept clearly.
#include <stdio.h>
int main() {
int a = 5, b = 10, c = 15;
int result;
result = a + b * c / 5 - 2 % 3;
printf("Result = %dn", result);
return 0;
}
Output Example
Input:
a = 5, b = 10, c = 15
Common Mistakes
- Misreading input/output format.
- Not handling constraints and edge cases.
- Off-by-one errors in loops.
- Forgetting to reset variables between test cases (if any).
Solution Guide
Problem
Write the expression: a = 5, b = 10, c = 15; result = a + b * c / 5 - 2 % 3; calculate manually, following operator precedence. Then verify with the program.
Input / Output
Input
a = 5
b = 10
c = 15
result = a + b * c / 5 - 2 % 3
Constraints
• Integer arithmetic
• Standard C operator precedence rules apply
Examples
Input:
a = 5, b = 10, c = 15
Explanation
Concept Explanation
The expression is evaluated using operator precedence and left-to-right associativity.
Step-by-Step Explanation
1.Substitute values:
result = 5 + 10 * 15 / 5 - 2 % 3
2.Evaluate *, /, % first (left to right):
• 10 * 15 = 150
• 150 / 5 = 30
• 2 % 3 = 2
3.Expression becomes:
5 + 30 - 2
4.Evaluate + and -:
• 5 + 30 = 35
• 35 - 2 = 33
5.Final value stored in result is 33.
Details
Common Mistakes
- Misreading input/output format.
- Not handling constraints and edge cases.
- Off-by-one errors in loops.
- Forgetting to reset variables between test cases (if any).
Official Solution
#include <stdio.h>
int main() {
int a = 5, b = 10, c = 15;
int result;
result = a + b * c / 5 - 2 % 3;
printf("Result = %dn", result);
return 0;
}
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