C Converting 4 digit numbers to binary - c

I want to print the binary equivalent of the numbers from a file in a 20 bit field with spaces after every 4 bits but when the number exceeds 1024 (I've tested this), it no longer outputs in binary but in a base I don't know (i.e. 1234 = 0000 0014 1006 5408)
Here is my code
#include <stdio.h>
long convertToBinary(long);
int main()
{
char binNum[20]; //binary number char array
int i;
long b, decNum; //b is output binary number, decNum is each input number from file
FILE *filePtr;
filePtr = fopen("numbers.txt", "r");
while (!feof(filePtr))
{
fscanf(filePtr, "%li", &decNum);
if (decNum < 0 || decNum > 65535)
{
printf("Error, %5li is out of range.\n", decNum);
continue;
}
b = convertToBinary(decNum); //function call for binary conversion
i = 18;
while (i >= 0)
{
if (i == 4 || i == 9 || i == 14)
{
binNum[i] = ' ';
}
else if (b != 0)
{
binNum[i] = b % 10 + '0';
b /= 10;
}
else
{
binNum[i] = '0';
}
i--;
}
binNum[19] = '.';
printf("The binary representation of %5li is: ", decNum);
for (i = 0; i<20; i++)
{
printf("%c", binNum[i]);
}
printf("\n");
}
return 0;
}
//Recursion function
long convertToBinary(long number)
{
if (number == 0)
{
return 0;
}
else
{
return (number % 2 + 10 * convertToBinary(number / 2));
}
}
The file numbers.txt has numbers like : 0 1 2 3 16 17 1234 5678 65535

Your convertToBinary function is producing a decimal-coded-binary (as opposed to a binary-coded-decimal) which uses a decimal digit per bit.
It looks like long has the same range as int on your system, i.e. 231, which covers ten decimal digits. Once you get eleventh digit, you get an overflow.
Switching to unsigned long long, which has 64 bits, will fix the problem. However, this is sub-optimal: you would be better off writing a to-binary conversion that writes zeros and ones into an array of chars that you provide. You can do it recursively if you'd like, but an iterative solution will do just fine.
Note that the to-binary conversion does not need to insert spaces into char array. This can be done by your main function.

Related

why this code isn't converting decimals to binary from decimals 0 to 31

So I was trying to convert a decimal number to binary using c. In this code every thing seems to be working well. Also this code does work for decimals from 32, 33 and go on. But this code doesn't work for decimals from 0 to 31. What's the bug in here.
#include <stdio.h>
#include <math.h>
int main(void)
{
int decimal;
printf("Enter the decimal value: ");
scanf("%i", &decimal);
int n, remainder;
int i, j, k;
for (int i = 0; i < decimal; i++)
{
if (pow(2, i) <= decimal)
{
n = i;
}
}
char index[n];
int quotient[n];
quotient[0] = decimal;
for (i = 0; i <= n; i++)
{
quotient[i + 1] = quotient[i] / 2;
remainder = quotient[i] % 2;
if (remainder == 1)
{
index[i] = '1';
}
else
{
index[i] = '0';
}
}
for (int k = n; k >= 0; k--)
{
printf("%c", index[k]);
}
return 0;
}
The number returned by scanf() is a bit pattern (binary) representation of the base10 value you enter.
You've got the right idea (trying to 'pick apart' those bits), but the method used is dubious and confusing.
Below is a loop of some familiar values being converted to strings of 1's and 0's.
Consider what it is doing at each step...
int main() {
for( int i = 253; i <= 258; i++ ) {
printf( "Decimal %d: ", i );
unsigned int bitmask = 0;
bitmask = ~bitmask;
bitmask &= ~(bitmask >> 1); // High bitmask ready
// skip over leading 0's (optional)
while( bitmask && (bitmask & i) == 0 ) bitmask >>= 1;
// loop using bitmask to output 1/0, then shift mask
do {
putchar( (bitmask & i) ? '1' : '0' );
} while( (bitmask >>= 1) != 0 );
putchar( '\n' );
}
return 0;
}
Output:
Decimal 253: 11111101
Decimal 254: 11111110
Decimal 255: 11111111
Decimal 256: 100000000
Decimal 257: 100000001
Decimal 258: 100000010
I don't like this, but you seem to want to use pow() to find the highest set bit in the incoming integer. Perhaps this will lead you to the solution you are looking for.
int main() {
int n = 0;
// generate a series of numbers as input
for( int decimal = 1; decimal < 1000*1000*1000; decimal = decimal * 2 + 1 ) {
// Limited of 32 bit integers INCLUDING sign bit
for (int i = 0; i < 31; i++) {
int guess = (int)pow( 2, i ); // make a guess with this value
printf( "Guess %d\n", guess );
if( guess > decimal ) { // guess now encompasses set bits
n = i; // Found what is needed.
break;
}
}
printf( "decimal input = %d: examining %d bits\n", decimal, n );
getchar();
}
return 0;
}
Here's one sample of the output of the above.
Guess 1
Guess 2
Guess 4
Guess 8
Guess 16
Guess 32
Guess 64
decimal input = 63: examining 6 bits
Note: this sort of thing will only work for positive values. It'll probably blow up if you want the bit pattern of a negative integer.
And, because you want to store an array of quotient, you need to dimension it to have +1 elements to avoid stepping out of bounds as you peel-off each bit through division...

finding how many times number2 is showing in number1

I am solving an exercise in C and I got stuck. I don't know the logic of the code to get to my solution. For example we enter 2 numbers from input let the numbers be 123451289 and 12 and I want to see how many times number 2 is showing at number 1 (if this is confusing let me know). For the numbers earlier the program outputs 2. I tried solving it here is my code:
#include <stdio.h>
int main() {
int num1, num2, counter = 0;
scanf("%d%d", num1, num2);
if (num1 < num2) {
int temp = num1;
num1 = num2;
num2 = temp;
}
int copy1 = num1;
int copy2 = num2;
while (copy2 > 0) {
counter++; // GETTING THE LENGHT OF THE SECOND NUMBER
copy2 /= 10;
// lastdigits = copy1 % counter //HERE I WANT TO GET THE LAST DIGITS OF THE FIRST NUMBER
// But it does not work
}
}
My question is how can I get the last digits of the first number according to the second one for example if the second number have 3 digits I want to get the last 3 digits of the first number. For the other part I think I can figure it out.
I must solve this problem WITHOUT USING ARRAYS.
The problem: find all the needles (e.g. 12) in a haystack (e.g. 123451289).
This can be done simply without arrays using a modulus of the needle. For 12, this is 100. That is, 12 is two digits wide. Using the modulus, we can
isolate the rightmost N digits of the haystack and compare them against the needle.
We "scan" haystack repeatedly by dividing by 10 until we reach zero.
Here is the code:
#include <stdio.h>
int
main(void)
{
int need, hay, counter = 0;
scanf(" %d %d", &hay, &need);
// ensure that the numbers are _not_ reversed
if (hay < need) {
int temp = need;
need = hay;
hay = temp;
}
// get modulus for needle (similar to number of digits)
int mod = 1;
for (int copy = need; copy != 0; copy /= 10)
mod *= 10;
// search haystack for occurences of needle
// examine the rightmost "mod" digits of haystack and check for match
// reduce haystack digit by digit
for (int copy = hay; copy != 0; copy /= 10) {
if ((copy % mod) == need)
++counter;
}
printf("%d appears in %d exactly %d times\n",need,hay,counter);
return 0;
}
UPDATE:
I'm afraid this does not work for 10 0. –
chqrlie
A one line fix for to the modulus calculation for the 10/0 case. But, I've had to add a special case for the 0/0 input.
Also, I've added a fix for negative numbers and allowed multiple lines of input:
#include <stdio.h>
int
main(void)
{
int need, hay, counter;
while (scanf(" %d %d", &hay, &need) == 2) {
counter = 0;
// we can scan for -12 in -1237812
if (hay < 0)
hay = -hay;
if (need < 0)
need = -need;
// ensure that the numbers are _not_ reversed
if (hay < need) {
int temp = need;
need = hay;
hay = temp;
}
// get modulus for needle (similar to number of digits)
int mod = need ? 1 : 10;
for (int copy = need; copy != 0; copy /= 10)
mod *= 10;
// search haystack for occurences of needle
// examine the rightmost "mod" digits of haystack and check for match
// reduce haystack digit by digit
for (int copy = hay; copy != 0; copy /= 10) {
if ((copy % mod) == need)
++counter;
}
// special case for 0/0 [yecch]
if ((hay == 0) && (need == 0))
counter = 1;
printf("%d appears in %d exactly %d times\n", need, hay, counter);
}
return 0;
}
Here is the program output:
12 appears in 123451289 exactly 2 times
0 appears in 10 exactly 1 times
0 appears in 0 exactly 1 times
UPDATE #2:
Good fixes, including tests for negative numbers... but I'm afraid large numbers still pose a problem, such as 2000000000 2000000000 and -2147483648 8 –
chqrlie
Since OP has already posted an answer, this is bit like beating a dead horse, but I'll take one last attempt.
I've changed from calculating a modulus of needle into calculating the number of digits in needle. This is similar to the approach of some of the other answers.
Then, the comparison is now done digit by digit from the right.
I've also switched to unsigned and allow for the number to be __int128 if desired/supported with a compile option.
I've added functions to decode and print numbers so it works even without libc support for 128 bit numbers.
I may be ignoring [yet] another edge case, but this is an academic problem (e.g. we can't use arrays) and my solution is to just use larger types for the numbers. If we could use arrays, we'd keep things as strings and this would be similar to using strstr.
Anyway, here's the code:
#include <stdio.h>
#ifndef NUM
#define NUM long long
#endif
typedef unsigned NUM num_t;
FILE *xfin;
int
numget(num_t *ret)
{
int chr;
num_t acc = 0;
int found = 0;
while (1) {
chr = fgetc(xfin);
if (chr == EOF)
break;
if ((chr == '\n') || (chr == ' ')) {
if (found)
break;
}
if ((chr >= '0') && (chr <= '9')) {
found = 1;
acc *= 10;
chr -= '0';
acc += chr;
}
}
*ret = acc;
return found;
}
#define STRMAX 16
#define STRLEN 100
const char *
numprt(num_t val)
{
static char strbuf[STRMAX][STRLEN];
static int stridx = 0;
int dig;
char *buf;
buf = strbuf[stridx++];
stridx %= STRMAX;
char *rhs = buf;
do {
if (val == 0) {
*rhs++ = '0';
break;
}
for (; val != 0; val /= 10, ++rhs) {
dig = val % 10;
*rhs = dig + '0';
}
} while (0);
*rhs = 0;
if (rhs > buf)
--rhs;
for (char *lhs = buf; lhs < rhs; ++lhs, --rhs) {
char tmp = *lhs;
*lhs = *rhs;
*rhs = tmp;
}
return buf;
}
int
main(int argc,char **argv)
{
num_t need, hay, counter;
--argc;
++argv;
if (argc > 0)
xfin = fopen(*argv,"r");
else
xfin = stdin;
while (1) {
if (! numget(&hay))
break;
if (! numget(&need))
break;
counter = 0;
// we can scan for -12 in -1237812
if (hay < 0)
hay = -hay;
if (need < 0)
need = -need;
// ensure that the numbers are _not_ reversed
if (hay < need) {
num_t temp = need;
need = hay;
hay = temp;
}
// get number of digits in needle (zero has one digit)
int ndig = 0;
for (num_t copy = need; copy != 0; copy /= 10)
ndig += 1;
if (ndig == 0)
ndig = 1;
// search haystack for occurences of needle
// starting from the right compare digit-by-digit
// "shift" haystack right on each iteration
num_t hay2 = hay;
for (; hay2 != 0; hay2 /= 10) {
num_t hcopy = hay2;
// do the rightmost ndig digits match in both numbers?
int idig = ndig;
int match = 0;
for (num_t need2 = need; idig != 0;
--idig, need2 /= 10, hcopy /= 10) {
// get single current digits from each number
int hdig = hcopy % 10;
int ndig = need2 % 10;
// do they match
match = (hdig == ndig);
if (! match)
break;
}
counter += match;
}
// special case for 0/0 et. al. [yecch]
if (hay == need)
counter = 1;
printf("%s appears in %s exactly %s times\n",
numprt(need), numprt(hay), numprt(counter));
}
return 0;
}
Here's the program output:
12 appears in 123451289 exactly 2 times
123 appears in 123451289 exactly 1 times
1234 appears in 123451289 exactly 1 times
1 appears in 123451289 exactly 2 times
0 appears in 10 exactly 1 times
0 appears in 0 exactly 1 times
1000000000 appears in 1000000000 exactly 1 times
2000000000 appears in 2000000000 exactly 1 times
This looks along the lines of what you're attempting.
You can use the pow() function from math.h to raise 10 to the power of how many digits you need for your modulus operation.
Compile with -lm or make your own function to calculate 10^num_digits
#include <stdio.h>
#include <math.h>
int main() {
int x = 123456789;
double num_digits = 3.0;
int last_digits = x % (int)pow(10.0, num_digits);
printf("x = %d\nLast %d Digits of x = %d\n", x, (int)num_digits, last_digits);
return 0;
}
Outputs:
x = 123456789
Last 3 Digits of x = 789
I think you are trying to ask :- if number1 = 1234567 and number2 = 673, then, length of number2 or number2 has 3 digits, so, you now want the last 3 digits in number1, i.e, '456', if I'm not wrong.
If that is the case, then, what you did to find the number of digits in num2 is correct, i.e,
while (copy2>0) {
counter++; // GETTING THE LENGHT OF THE SECOND NUMBER
copy2/=10;
}
you can do the same for number1 and find out its number of digits, then you can compare whether the number of digits in number2 is less than that in number1. Ex, 3 is less than number of digits in number1, so you can proceed further. Let's say number of digits in number1 is 7 and you want the last 3 digits, so you can do iterate over the digits in number1 till count of digits in number2 and pop out each last digit and store them in an array.
The code:
#include <stdio.h>
int main()
{
int num1,num2;
int count1 = 0, count2 = 0;
scanf("%d",&num1);
scanf("%d",&num2);
if(num1<num2){
int temp = num1;
num1 = num2;
num2 = temp;
}
int copy1 = num1;
int copy2 = num2;
while (copy1>0)
{
count1++;
copy1/=10;
}
while (copy2>0)
{
count2++;
copy2/=10;
}
// printf("num1 has %d digits and num2 has %d digits\n", count1, count2);
if (count1 >= count2)
{
int arr[count2];
int x = count2;
int p = num1;
int i = 0;
while (x > 0)
{
arr[i++] = p%10;
x --;
p/=10;
}
for (int j = 0; j < i; j++)
{
printf("%d ", arr[j]);
}
}
return 0;
}
output : 8 7 6
let's say, num1 = 12345678, num2 = 158, then arr = {8,7,6}.
You must determine the number of digits N of num2 and test if num1 ends with num2 modulo 10N.
Note these tricky issues:
you should not sort num1 and num2: If num2 is greater than num1, the count is obviously 0.
num2 has at least 1 digit even if it is 0.
if num1 and num2 are both 0, the count is 1.
if num2 is greater then INT_MAX / 10, the computation for mod would overflow, but there can only be one match, if num1 == num2.
it is unclear whether the count for 1111 11 should be 2 or 3. We will consider all matches, including overlapping ones.
to handle larger numbers, we shall use unsigned long long instead of int type.
Here is a modified version:
#include <limits.h>
#include <stdio.h>
int main() {
int counter = 0;
unsigned long long num1, num2;
if (scanf("%llu%llu", &num1, &num2) != 2) {
printf("invalid input\n");
return 1;
}
if (num1 == num2) {
/* special case for "0 0" */
counter = 1;
} else
if (num1 > num2 && num2 <= ULLONG_MAX / 10) {
unsigned long long copy1 = num1;
unsigned long long mod = 10;
while (mod < num2) {
mod *= 10;
}
while (copy1 > 0) {
if (copy1 % mod == num2)
counter++;
copy1 /= 10;
}
}
printf("count=%d\n", counter);
return 0;
}
Note that leading zeroes are not supported in either number: 101 01 should produce a count of 1 but after conversion by scanf(), the numbers are 101 and 1 leading to a count of 2. It is non trivial to handle leading zeroes as well as numbers larger than ULLONG_MAX without arrays.
This was the answer that i was looking for but thank you all for helping :)
#include <stdio.h>
#include <math.h>
int main(){
int num1,counter1,counter2,num2,temp,digit,copy1,copy2;
scanf("%d%d",&num1,&num2);
if(num1<num2){
temp = num1;
num1 = num2;
num2 = temp;
}
copy1 = num1;
copy2 = num2;
counter1 = counter2 = 0;
while (copy2>0) {
counter1++;
copy2/=10;
}
counter1 = pow(10,counter1);
if(num1> 1 && num2>1)
while (copy1>0) {
digit = copy1%counter1;
if(digit==num2){
counter2++;
}
copy1/=10;
} else{
if(num2<1){
while (copy1>0) {
digit = copy1%10;
if(digit==copy2){
counter2++;
}
copy1/=10;
}
}
}
printf("%d",counter2);
}

Printing a copy of a given 4 digit number but every prime digit is followed by the number which is one greater than it

So I've been trying to create a program which asks the user for a 4 digit number, and prints it's copy but every prime digit is just followed by a number just greater than it. (Eg. 2345 becomes 2334456)
So what I tried was first finding storing all the digits and then printing them out as strings, followed by the next number if any of them were prime using if statements. Now this seems to give a different output than I'd expect. 2345 gives 23235656 for example. Where did I go wrong?
#include <stdio.h>
int main() {
unsigned int userinput;
unsigned int onesplace;
unsigned int tensplace;
unsigned int hundredsplace;
unsigned int thousandsplace;
printf("Please print your number /n");
scanf("%u", &userinput);
onesplace = userinput%10;
tensplace = ((userinput-onesplace/10))%10;
hundredsplace = ((userinput - onesplace -10*tensplace)/100)%10;
thousandsplace = ((userinput - onesplace - 10*tensplace - 100*hundredsplace)/1000);
printf("%u", thousandsplace);
if ((thousandsplace == 2)||(thousandsplace == 3)||(thousandsplace == 5)||(thousandsplace == 7)) {
unsigned int newnum = thousandsplace + 1;
printf("%u", newnum);
}
printf("%u", hundredsplace);
if ((hundredsplace == 2)||(hundredsplace == 3)||(hundredsplace == 5)||(hundredsplace == 7)) {
unsigned int newnum2 = hundredsplace + 1;
printf("%u", newnum2);
}
printf("%u", tensplace);
if ((tensplace == 2)||(tensplace == 3)||(tensplace == 5)||(tensplace == 7)) {
unsigned int newnum3 = tensplace + 1;
printf("%u", newnum3);
}
printf("%u", onesplace);
if ((onesplace == 2)||(onesplace == 3)||(onesplace == 5)||(onesplace == 7)) {
unsigned int newnum4 = onesplace + 1;
printf("%u", newnum4);
}
}
You are miscalculating the tensplace, essentially forgetting to divide the tens value by 10.
Change (userinput-onesplace/10) to (userinput/10).
Then with an input of 2345 you get output
Please print your number /n2334456
The probably unwanted "/n", visible here in the single output line because I used https://www.tutorialspoint.com/compile_c_online.php (with is non-interactively provided standard input), is a separate unrelated problem.

C variable specified as a long long but recognized as an int

I'm working on a program that checks the validity of credit card numbers for the CS50 class I'm taking (it's legal I swear haha) and I'm currently working on correctly getting the first two numbers of each CC# to check what company it is from. I've commented what each part does for clarity and also commented where my problem arises.
#include <stdio.h>
#include <stdlib.h>
#include <cs50.h>
#include <math.h>
#include <string.h>
int main(void)
{
long long ccn = get_long_long("Enter CCN: \n");
int count = 0;
long long ccn1 = ccn;
// finds the amount of digits entered and stores that in int count.
while (ccn1 != 0)
{
ccn1 /= 10;
+count;
}
printf("%i \n", count);
// ln 17- 19 should take int count, subtract two, put that # as the power of 10,
// then divide the CC# by that number to get the first two numbers of the CC#.
long long power = count - 2;
// here is where i get the error. its a long long so it
// should hold up to 19 digits and im only storing 14 max
// but it says that 10^14th is too large for type 'int'
long long divide = pow(10,power);
long long ft = ccn / divide;
printf("power: %i \n", power); //ln 20-22 prints the above ints for debug
printf("Divide: %lli \n", divide);
printf("First two: %lli \n", ft);
string CCC;
// ln 24-35 cross references the amount of digits in the CC#
// and the first two digits to find the comapany of the credit card
if ((count == 15) && (ft = 34|37))
{
CCC = "American Express";
}
else if ((count == 16) && (ft = 51|52|53|54|55))
{
CCC = "MasterCard";
}
else if ((count = 13|16) && (ft <=49 && ft >= 40))
{
CCC = "Visa";
}
printf("Company: %s\n", CCC);
}
The first issue is the +count in the loop. This should be ++count. Because of this, count stays at 0 and power = -2. You can avoid all that power stuff. You already have the loop, you can use it to get the first two digits:
int ft = 0;
while (ccn1 != 0)
{
// When you are down to 10 <= ccn1 < 100, store it
if (ccn1 < 100 && ccn1 > 9) ft = ccn1;
ccn1 /= 10;
++count;
}
Your second issue is how you do your comparisons.
if ((count == 15) && (ft = 34|37))
First, = is assignment, and == tests equality. Second, | is bitwise OR, || is logical OR. Third, you can't test multiple values like that. Correct way:
if ((count == 15) && (ft == 34 || ft == 37))

Extracting individual digits from a long in C

I'm doing a homework assignment for my course in C (first programming course).
Part of the assignment is to write code so that a user inputs a number up to 9 digits long, and the program needs to determine whether this number is "increasing"/"truly increasing"/"decreasing"/"truly decreasing"/"increasing and decreasing"/"truly decreasing and truly increasing"/"not decreasing and not increasing". (7 options in total)
Since this is our first assignment we're not allowed to use anything besides what was taught in class:
do-while, for, while loops, else-if, if,
break,continue
scanf, printf ,modulo, and the basic operators
(We can't use any library besides for stdio.h)
That's it. I can't use arrays or getchar or any of that stuff. The only function I can use to receive input from the user is scanf.
So far I've already written the algorithm with a flowchart and everything, but I need to separate the user's input into it's distinct digits.
For example, if the user inputs "1234..." i want to save 1 in a, 2 in b, and so on, and then make comparisons between all the digits to determine for example whether they are all equal (increasing and decreasing) or whether a > b >c ... (decreasing) and so on.
I know how to separate each digit by using the % and / operator, but I can't figure out how to "save" these values in a variable that I can later use for the comparisons.
This is what I have so far:
printf("Enter a positive number : ");
do {
scanf ("%ld", &number);
if (number < 0) {
printf ("invalid input...enter a positive integer: ");
continue;
}
else break;
} while (1);
while (number < 0) {
a = number % 10;
number = number - a;
number = number / 10;
b = a;
}
Why not scan them as characters (string)? Then you can access them via an array offset, by subtracting the offset of 48 from the ASCII character code. You can verify that the character is a digit using isdigit from ctype.h.
EDIT
Because of the incredibly absent-minded limitations that your professor put in place:
#include <stdio.h>
int main()
{
int number;
printf("Enter a positive number: ");
do
{
scanf ("%ld", &number);
if (number < 0)
{
printf ("invalid input...enter a positive integer: ");
continue;
}
else break;
} while (1);
int a = -1;
int b = -1;
int c = -1;
int d = -1;
int e = -1;
int f = -1;
int g = -1;
int h = -1;
int i = -1;
while (number > 0)
{
if (a < 0) a = number % 10;
else if (b < 0) b = number % 10;
else if (c < 0) c = number % 10;
else if (d < 0) d = number % 10;
else if (e < 0) e = number % 10;
else if (f < 0) f = number % 10;
else if (g < 0) g = number % 10;
else if (h < 0) h = number % 10;
else if (i < 0) i = number % 10;
number /= 10;
}
/* Printing for verification. */
printf("%i", a);
printf("%i", b);
printf("%i", c);
printf("%i", d);
printf("%i", e);
printf("%i", f);
printf("%i", g);
printf("%i", h);
printf("%i", i);
return 0;
}
The valid numbers at the end will be positive, so those are the ones you validate to meet your different conditions.
Since you only need to compare consecutive digits, there is an elegant way to do this without arrays:
int decreasing = 2;
int increasing = 2;
while(number > 9)
{
int a = number % 10;
int b = (number / 10) % 10;
if(a == b)
{
decreasing = min(1, decreasing);
increasing = min(1, increasing);
}
else if(a > b)
decreasing = 0;
else if(a < b)
increasing = 0;
number /= 10;
}
Here, we walk through the number (by dividing by 10) until only one digit remains. We store info about the number up to this point in decreasing and increasing - a 2 means truly increasing/decreasing, a 1 means increasing/decreasing, and a 0 means not increasing/decreasing.
At each step, a is the ones digit and b is the tens. Then, we change increasing and decreasing based on a comparison between a and b.
At the end, it should be easy to turn the values of increasing and decreasing into the final answer you want.
Note: The function min returns the smaller of its 2 arguments. You should be able to write your own, or replace those lines with if statements or conditionals.
It's stupid to ask you to do loops without arrays --- but that's your teacher's fault, not yours.
That being said, I would do something like this:
char c;
while (1) {
scanf("%c", &c);
if (c == '\n') /* encountered newline (end of input) */
break;
if (c < '0' || c > '9')
break; /* do something to handle bad characters? */
c -= '0';
/*
* At this point you've got 0 <= c < 9. This is
* where you do your homework :)
*/
}
The trick here is that when you type numbers into a program, you send the buffer all at once, not one character at a time. That means the first scanf will block until the entire string (i.e. "123823" or whatever) arrives all at once, along with the newline character ( '\n' ). Then this loop parses that string at its leisure.
Edit For testing the increasing/decreasing-ness of the digits, you may think you need to store the entire string, but that's not true. Just define some additional variables to remember the important information, such as:
int largest_digit_ive_seen, smallest_digit_ive_seen, strict_increasing_thus_far;
etc. etc.
Let us suppose you have this number 23654
23654 % 10000 = 2 and 3654
3654 % 1000 = 3 and 654
654 % 100 = 6 and 54
54 % 10 = 5 and 4
4
This way you can get all the digits. Of course, you have to know if the number is greater than 10000, 1000, 100 or 10, in order to know the first divisor.
Play with sizeof to get the size of the integer, in order to avoid a huge if...else statement
EDIT:
Let us see
if (number>0) {
// Well, whe have the first and only digit
} else if (number>10) {
int first_digit = number/10;
int second_digit = number % 10;
} else if (number>100) {
int first_digit = number/100;
int second_digit = (number % 100)/10;
int third_digit = (number % 100) % 10;
} ...
and so on, I suppose
// u_i is the user input, My homework asked me to extract a long long, however, this should also be effective for a long.
int digits = 0;
long long d_base = 1;
int d_arr[20];
while (u_i / d_base > 0)
{
d_arr[digits] = (u_i - u_i / (d_base * 10) * (d_base * 10)) / d_base;
u_i -= d_arr[digits] * d_base;
d_base *= 10;
digits++;
}
EDIT: the extracted individual digit now lives in the int array d_arr. I'm not good at C, so I think the array declaration can be optimized.
Here's a working example in plain C :
#include <stdio.h>
unsigned long alePow (unsigned long int x, unsigned long int y);
int main( int argc, const char* argv[] )
{
int enter_num, temp_num, sum = 0;
int divisor, digit, count = 0;
printf("Please enter number\n");
scanf("%d", &enter_num);
temp_num = enter_num;
// Counting the number of digits in the entered integer
while (temp_num != 0)
{
temp_num = temp_num/10;
count++;
}
temp_num = enter_num;
// Extracting the digits
printf("Individual digits in the entered number are ");
do
{
divisor = (int)(alePow(10.0, --count));
digit = temp_num / divisor;
temp_num = temp_num % divisor;
printf(" %d",digit);
sum = sum + digit;
}
while(count != 0);
printf("\nSum of the digits is = %d\n",sum);
return 0;
}
unsigned long alePow(unsigned long int x, unsigned long int y) {
if (x==0) { return 0; }
if (y==0||x==1) { return 1; }
if (y==1) { return x; }
return alePow(x*x, y/2) * ((y%2==0) ? 1 : x);
}
I would suggest loop-unrolling.
int a=-1, b=-1, c=-1, d=-1, e=1, f=-1, g=-1, h=-1, i=-1; // for holding 9 digits
int count = 0; //for number of digits in the given number
if(number>0) {
i=number%10;
number/=10;
count++;
}
if(number>0) {
h=number%10;
number/=10;
count++;
}
if(number>0) {
g=number%10;
number/=10;
count++;
}
....
....
/* All the way down to the storing variable a */
Now, you know the number of digits (variable count) and they are stored in which of the variables. Now you have all digits and you can check their "decreasing", "increasing" etc with lots of if's !
I can't really think of a better soltion given all your conditions.

Resources