Decimal to Binary

[ornate copper]

This code is supposed to convert a decimal to a 16-bit binary. Can you guys please help me understand lines 10-11 and 22-23, thanks

#include <stdio.h>

int main() {
    // Declare variables
    unsigned int decimal, binary = 0;
    int i;

    // Ask for user's input
    printf("Enter non-negative decimal integer to convert: ");
    if (scanf("%u", &decimal) != 1) {
        printf("Error occured\n");
        return 1; // Exit with an error code
    }

    // Check if the input number can be represented in 16 bits
    if (decimal > 65535) {
        printf("Error occured\n");
        return 1; // Exit with an error code
    }

    // Convert decimal to binary
    for (i = 15; i >= 0; i--) {
        int bit = (decimal >> i) & 1;
        printf("%d", bit);
    }

    printf("\n");

    return 0; // Exit without an error code
}

[slate tartan]

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[fair kayak]

What are the lines 10-11 and 22-23?

[ornate copper]

if (scanf("%u", &decimal) != 1) and c for (i = 15; i >= 0; i--) { int bit = (decimal >> i) & 1; basically

[fair kayak]

scanf returns how many conversions it successfully did (should be 1) and %u is the format specifier for an unsigned integer.

The loop just checks bit by bit (from most signficiant to least significant) whether the current bit is set

[ornate copper]

What do you mean by "how many conversions scanf did"

[dull ice]

how many % formatters it matched

[fair kayak]

Take the input This is a sentence.
Now let's test it out:

;compile | This is a sentence longer than five words

#include <stdio.h>

char a[10] = { 0 }, b[10] = { 0 }, c[10] = { 0 }, d[10] = { 0 }, e[10] = { 0 };
int conversions = scanf("%s %s %s %s %s", a, b, c, d, e);
printf("Got %d conversions for input %s %s %s %s %s\n", conversions, a, b, c, d, e);

[neat nebula]

Program Output

Got 5 conversions for input This is a sentence longer

elmonkeking#0000 | 68ms | c | x86-64 gcc 13.2 | godbolt.org

[ornate copper]

So basically, in our code, scanf must perform EXACTLY ONE conversion of the user's input

For the rest of the program to work

[fair kayak]

Yeah, but because it accepts at most 1 conversion it needs to perform at least one

[ornate copper]

Would the program NOT work without ```c
if (scanf("%u", &decimal) != 1) {
printf("Error occured\n");
return 1; // Exit with an error code
}

[fair kayak]

it would run into problems

[dull ice]

it's useful here, since %u only matches unsigned integers, so if the user types in e.g. "abc" it won't match , and it'll return 0

[fair kayak]

;compile | a

#include <stdio.h>

unsigned x;
int conversions = scanf("%u", &x);
printf("got %u with %d conversions\n", x, conversions);

[neat nebula]

Program Output

got 0 with 0 conversions

elmonkeking#0000 | 77ms | c | x86-64 gcc 13.2 | godbolt.org

[fair kayak]

So yeah, it catches the case where you enter e.g. just an a

or a negative number

or anything other that is not an unsigned integer

[ornate copper]

if (scanf("%u", &decimal) != 1) { // Checks if user's input is an unsigned integer is this comment appropriate here?

[dull ice]

I generally prefer commenting the conditional block as a whole, rather than the if itself, so I'd probably word it closer to "if the input successfully matched"

but the best comments are ones that make the most sense to you, since they're notes for yourself

[fair kayak]

I think that comment is true, but tbh I try to avoid scanf whenever I can. That thing's a nasty bitch

[ornate copper]

Got y'all

All in all, does this code have any flaws?

Or does it successfully convert decimal to binary

[dull ice]

try it and find out

[ornate copper]

It seems to work perfectly fine for me

No unintended errors

Conversions are accurate

[dull ice]

any warnings?

[fair kayak]

Are you even compiling with warnings enabled?
-Wall -Wextra -Wpedantic?

[ornate copper]

I'm new to C, not sure what these are

Just compiling through Terminal

[fair kayak]

hmmm

[ornate copper]

That's intended though

[fair kayak]

Ah, not bigger than 16 bits

[ornate copper]

Can't be represented as a 16-bit number

[fair kayak]

I see

[ornate copper]

Right

[fair kayak]

;compile -Wall -Wextra -Wpedantic | 8191

#include <stdio.h>

int main() {
    // Declare variables
    unsigned int decimal, binary = 0;
    int i;

    // Ask for user's input
    printf("Enter non-negative decimal integer to convert: ");
    if (scanf("%u", &decimal) != 1) {
        printf("Error occured\n");
        return 1; // Exit with an error code
    }

    // Check if the input number can be represented in 16 bits
    if (decimal > 65535) {
        printf("Error occured\n");
        return 1; // Exit with an error code
    }

    // Convert decimal to binary
    for (i = 15; i >= 0; i--) {
        int bit = (decimal >> i) & 1;
        printf("%d", bit);
    }

    printf("\n");

    return 0; // Exit without an error code
}

[neat nebula]

Program Output

Enter non-negative decimal integer to convert: 0001111111111111

Compiler Output

<source>: In function 'main':
<source>:5:27: warning: unused variable 'binary' [-Wunused-variable]
    5 |     unsigned int decimal, binary = 0;
      |                           ^~~~~~

elmonkeking#0000 | 71ms | c | x86-64 gcc 13.2 | godbolt.org

[fair kayak]

@ornate copper There is one warning for you

[ornate copper]

Build failed?

[fair kayak]

Because of -Werror all warnings are being treated as errors, so it fails to build even if there's just a warning

Remove that and as you can see it will still work, but will also throw you the warning

[ornate copper]

So it’s just a warning?

[fair kayak]

Yes

[ornate copper]

Thank you

Completely unrelated, but can you please help me understand another program

int main() {

    char UserInput;
    int ShiftAmount;

    printf("Enter lower-case letter to encrypt: ");
    scanf("%c", &UserInput);

    if (UserInput < 'a' || UserInput > 'z') {
        printf("Error: user did not enter lower-case letter, exiting\n");
        return 1;
    }

    printf("Enter the shift amount for Caesar cipher: ");
    scanf("%d", &ShiftAmount);

    // Ensure the shift amount is within (1, 25)
    ShiftAmount = ShiftAmount % 26;
    if (ShiftAmount < 0) {
        ShiftAmount = ShiftAmount + 26;
    }

    // Perform the Caesar shift
    char Shifted = 'a' + (UserInput - 'a' + ShiftAmount) % 26;

    printf("Ciphertext is %c\n", Shifted);

    return 0;
}```

Specifically this part char Shifted = 'a' + (UserInput - 'a' + ShiftAmount) % 26;

[dull ice]

what about it

[ornate copper]

I don't understand what the line does

[dull ice]

based on the variable name, it's calculating what the shifted character corresponding to your input is

[ornate copper]

Right, I get that

I just don't understand the code itself

[dull ice]

what about it?

[ornate copper]

'a' + (UserInput - 'a' + ShiftAmount) % 26

[dull ice]

what part of that

[ornate copper]

Why are there 'a's

And why does mod 26 intervene here

[dull ice]

to shift all the values to start at 0

'b' - 'a' = 1 (with ASCII / the encoding scheme you posted)

[dull ice]

for this: #1153814340395356302 message

[slate tartan]

@ornate copper Has your question been resolved? If so, run !solved :)

[ornate copper]

Thank you guys

!solved