luc104.c¶
Problem Statement
Rewrite expressions using bitwise compound assignment operators.
Metadata¶
| Property | Detail |
|---|---|
| Author | Amit Dutta amitdutta4255@gmail.com |
| Date | 08 Feb 2026 |
| License | MIT License (See the LICENSE file for details) |
| Difficulty | Beginner (index: 1 / 10) |
Actions¶
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Source Code¶
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a = 10, b = 20, c = 5;
printf("Original values: a=%d, b=%d, c=%d\n", a, b, c);
// a = a | 3
a |= 3;
printf("a |= 3 -> %d\n", a);
// a = a & 0x48
a &= 0x48;
printf("a &= 0x48 -> %d\n", a);
// b = b ^ 0x22
b ^= 0x22;
printf("b ^= 0x22 -> %d\n", b);
// c = c << 2
c <<= 2;
printf("c <<= 2 -> %d\n", c);
return 0;
}
Explanation¶
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[START]
## What it does
(Explain the overall purpose in 1-2 sentences)
## Step-by-step
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## Notes
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[END]
CODE (luc104.c):
#include <stdio.h>
#include <stdlib.h>
int main()
{
int a = 10, b = 20, c = 5;
printf("Original values: a=%d, b=%d, c=%d\n", a, b, c);
// a = a | 3
a |= 3;
printf("a |= 3 -> %d\n", a);
// a = a & 0x48
a &= 0x48;
printf("a &= 0x48 -> %d\n", a);
// b = b ^ 0x22
b ^= 0x22;
printf("b ^= 0x22 -> %d\n", b);
// c = c << 2
c <<= 2;
printf("c <<= 2 -> %d\n", c);
return 0;
}