Bitwise NOT in C - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this bitwise NOT operation?

Consider the following C code snippet. What will it print?

C

#include <stdio.h>

int main() {
    unsigned char x = 0b00001111; // 15 in decimal
    unsigned char y = ~x;
    printf("%u\n", y);
    return 0;
}

A 240

B 15

C 0

D 255

💡 Hint

Remember that bitwise NOT flips all bits. For an 8-bit unsigned char, 0b00001111 becomes 0b11110000.

The bitwise NOT operator (~) flips every bit. For 0b00001111 (decimal 15), flipping bits gives 0b11110000, which is 240 in decimal.

❓ Predict Output

intermediate

What is the output of bitwise NOT on a signed integer?

What will this C program print?

C

#include <stdio.h>

int main() {
    int x = 0;
    int y = ~x;
    printf("%d\n", y);
    return 0;
}

A 0

B -1

C 1

D 2147483647

💡 Hint

Bitwise NOT flips all bits. For zero, all bits flipped become all ones, which is -1 in two's complement.

In two's complement, ~0 flips all bits to 1, which represents -1 as a signed integer.

🔧 Debug

advanced

Why does this bitwise NOT code produce unexpected output?

This code is intended to print the maximum unsigned integer value after applying bitwise NOT to 0, but it prints -1 instead. Why?

C

#include <stdio.h>

int main() {
    unsigned int x = 0;
    unsigned int y = ~x;
    printf("%d\n", y);
    return 0;
}

A The bitwise NOT operator (~) doesn't work correctly on unsigned integers.

B There is integer overflow when applying ~ to 0.

C The printf format specifier "%d" expects a signed int, and the bit pattern of UINT_MAX (all 1s) is interpreted as -1 in two's complement.

D unsigned int y = ~x promotes x to signed before applying ~.

💡 Hint

The issue is with how printf interprets the unsigned value when using the wrong format specifier.

The ~ operator on unsigned int 0 produces UINT_MAX (all bits 1). However, printf("%d", y) passes an unsigned int to a format expecting signed int, leading to undefined behavior. In practice, on two's complement systems, it prints -1.

❓ Predict Output

advanced

What is the output of this bitwise NOT on a 16-bit unsigned short?

Given this C code, what will be printed?

C

#include <stdio.h>

int main() {
    unsigned short x = 0x00FF; // 255 decimal
    unsigned short y = ~x;
    printf("%u\n", y);
    return 0;
}

A 65280

B 255

C 65535

D 0

💡 Hint

Bitwise NOT flips all bits in 16 bits. 0x00FF flipped becomes 0xFF00.

0x00FF in binary is 0000000011111111. Flipping all bits gives 1111111100000000, which is 65280 in decimal.

❓ Predict Output

expert

What is the output of this bitwise NOT on a signed char with negative value?

Analyze this C code and determine what it prints.

C

#include <stdio.h>

int main() {
    signed char x = -1;
    signed char y = ~x;
    printf("%d\n", y);
    return 0;
}

A 127

B -1

C -128

D 0

💡 Hint

Bitwise NOT flips all bits. For -1 in two's complement, all bits are 1, flipping gives all 0 bits.

In two's complement, -1 is represented by all bits set to 1. Bitwise NOT flips all bits to 0, which is 0 as a signed char.