when i call my multi() it will calculate the binary number just fine, but won't calculate the adding(). If i switch the two in order, it will calculate the adding() and not calculate the multi(). Is there a reason why the second function call isn't displaying its value? It will print the printf, but not the value. Any help would be great.
#include <stdio.h>
long binary1, binary2, binary3, binary4, multiply = 0;
int binaryproduct(int, int);
int digit, factor = 1;
int multi();
int adding();
int main()
{
binary3 = binary1;
binary4 = binary1;
printf("Enter the first binary number: ");
scanf("%ld", &binary1);
printf("Enter the second binary number: ");
scanf("%ld", &binary2);
multi();
printf("\n");
adding();
}
int adding()
{
int i = 0, remainder = 0, sum[20];
while (binary3 != 0 || binary4 != 0) {
sum[i++] =(binary3 % 10 + binary4 % 10 + remainder) % 2;
remainder =(binary3 % 10 + binary4 % 10 + remainder) / 2;
binary1 = binary3 / 10;
binary2 = binary4 / 10;
}
if (remainder != 0) {
sum[i++] = remainder;
}
--i;
printf("Sum of two binary numbers: ");
while (i >= 0) {
printf("%d", sum[i--]);
}
return 0;
}
int multi()
{
while (binary2 != 0) {
digit = binary2 % 10;
if (digit == 1) {
binary1 = binary1 * factor;
multiply = binaryproduct(binary1, multiply);
} else {
binary1 = binary1 * factor;
}
binary2 = binary2 / 10;
factor = 10;
}
printf("Product of two binary numbers: %ld", multiply);
return 0;
}
int binaryproduct(int binary3, int binary4)
{
int i = 0, remainder = 0, sum[20];
int binaryprod = 0;
while (binary1 != 0 || binary2 != 0) {
sum[i++] =(binary1 % 10 + binary2 % 10 + remainder) % 2;
remainder =(binary1 % 10 + binary2 % 10 + remainder) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (remainder != 0) {
sum[i++] = remainder;
}
--i;
while (i >= 0) {
binaryprod = binaryprod * 10 + sum[i--];
}
return binaryprod;
}
Your globals aren't being initialized. In you adding() function you use:
binary1 = binary3/10
binary2 = binary4/10
and in the binaryproduct function you have two local ints named binary3, binary4 that you don't use.
You should be using unique names, initializing you variables, use local variables in places where globals are unnecessary.
Related
I'm very new to C language much less programming in general and am working through the cs50 edx course. I encountered this problem in week 1 where I'm supposed to validate credit cards given their characteristics (they're in the comments).
After hours worth of research and trials, I managed to solve it though I think it's way too long and I am 90% sure it can be shortened with loops. I added the long version that works and an iteration of it which uses loops that I just can't get to work. Any advice would help and be very much appreciated!
#include <stdio.h>
#include <cs50.h>
int main(void)
{
//user input for credit card number
long creditCard = get_long("Credit Card: ");
//initialize array for finding digits
int digit[16];
int oddDigits[8];
//separate all digits as variables in array digits[]
digit[0] = ((creditCard / 1000000000000000) % 10);
digit[1] = ((creditCard / 100000000000000) % 10);
digit[2] = ((creditCard / 10000000000000) % 10);
digit[3] = ((creditCard / 1000000000000) % 10);
digit[4] = ((creditCard / 100000000000) % 10);
digit[5] = ((creditCard / 10000000000) % 10);
digit[6] = ((creditCard / 1000000000) % 10);
digit[7] = ((creditCard / 100000000) % 10);
digit[8] = ((creditCard / 10000000) % 10);
digit[9] = ((creditCard / 1000000) % 10);
digit[10] = ((creditCard / 100000) % 10);
digit[11] = ((creditCard / 10000) % 10);
digit[12] = ((creditCard / 1000) % 10);
digit[13] = ((creditCard / 100) % 10);
digit[14] = ((creditCard / 10) % 10);
digit[15] = (creditCard % 10);
//double the value of every other digit as oddDigits[]
oddDigits[0] = digit[0] * 2;
oddDigits[1] = digit[2] * 2;
oddDigits[2] = digit[4] * 2;
oddDigits[3] = digit[6] * 2;
oddDigits[4] = digit[8] * 2;
oddDigits[5] = digit[10] * 2;
oddDigits[6] = digit[12] * 2;
oddDigits[7] = digit[14] * 2;
//the first part in finding the sum of every other digit
for (int o = 0; o < 8; o++) {
if (oddDigits[o] > 9) {
oddDigits[o] -= 9;
}
}
//part 2 of finding the sum of every other digit
int sum = 0;
for (int s = 0; s < 8; s++) {
sum += oddDigits[s];
}
//sum of everything else
for (int s2 = 1; s2 < 16; s2 += 2) {
sum += digit[s2];
}
//check if last digit is 0
int check = sum % 10;
//last part of validation
if (check != 0)
printf("INVALID\n");
//type of card
else {
if (digit[0] == 0 && digit[1] == 3 && (digit[2] == 4 || digit[2] == 7)) //15 digits starting with either 34 or 37
{
printf("AMAX\n");
}
else if (digit[0] == 5 && (digit[1] == 1 || digit[1] == 2 || digit[1] == 3 || digit[1] == 4 || digit[1] == 5)) //16 digits starting with either 51 - 54
{
printf("MASTERCARD\n");
}
else if (digit[0] == 4) //16 digits starting with 4
{
printf("VISA\n");
}
else if (digit[0] == 0 && digit[1] == 0 && digit[2] == 0 && digit[3] == 4) //13 digits starting with 4
{
printf("VISA\n");
}
else
{
printf("INVALID\n");
}
}
}
The only part I changed is where I try to get every digit of the number using for loops and the part where I try to double every other digit... I did a test where after the loop it will print all the digits separately and it always prints 1 no matter what and I'm wondering what I'm doing wrong here...
//initialize array for finding digits
int digit[16];
int oddDigits[8];
//separate all digits as variables in array digits[]
for (int d = 0; d < 16; d++) {
for (long i = 1000000000000000; i > 9; i /= 10) {
digit[d] = ((creditCard / i) % 10);
}
}
//double the value of every other digit as oddDigits[]
for (int n = 0; n < 8; n++) {
for (int t = 0; t < 15; t += 2) {
oddDigits[n] = digit[t]*2;
}
}
While you are separating all digits as variables in array digits[], the for loop runs till.
digit[14] = ((creditCard / 10) % 10);
to fix this change it as
for (long i = 1000000000000000; i > 0; i /= 10)
{
digit[d] = ((creditCard / i) % 10);
}
ie change 9 to 0. in this way, the last iteration of the loop will run
digit[15] = ((creditCard/1) % 10);
which is same as
digit[15] = (creditCard % 10);
now see, the outer loop is running 16 times. the inner loop is also running 16 times and the inner loop is running only once per iteration of the outer loop. So one of the loops is redundant. remove the outer loop. keep the d variable as 0 and increment it by 1 after dividing the digit. i.e.
int d = 0;
//separate all digits as variables in array digits[]
for (long i = 1000000000000000; i > 9; i /= 10)
{
digit[d] = ((creditCard / i) % 10);
d++;
}
You can also do it like this
long i = 1000000000000000;
for (int d = 0; d < 16; d++)
{
digit[d] = ((creditCard / i) % 10);
i /= 10;
}
Either one will work.
Now this part //double the value of every other digit as oddDigits[]
It can also be done with 1 loop. In your code, the variable t is just double of the variable n in each step. So, you don't need the t variable too. (just replace t by 2*n)
//double the value of every other digit as oddDigits[]
for (int n = 0; n < 8; n++)
{
oddDigits[n] = digit[2*n]*2;
}
I am trying to solve this tutorial practice question that doesn't have an answer that I can check my code against. The goal is to write a program to display numbers whose digits are 2 greater than the corresponding digits of the entered number. So if the number input is 5656 then the output number should be 7878. I have figured out how to separate each number and add them, but I can't seem to get them to print in a four-digit sequence.
#include <stdio.h>
int main ()
{
int n, one, two, three, four, final;
scanf("%d", &n);
one = (n / 1000);
n = (n % 1000) + 2;
two = (n / 100) + 2;
n = (n % 100) + 2;
three = (n / 10) + 2;
n = (n % 10) + 2;
four = (n / 1) + 2;
n = (n % 1) + 2;
final = (one * 1000) + (two * 100) + (three * 10) + four;
printf("%d", final);
return 0;
}
#include <stdio.h>
int main()
{
int n,a[4], final;
scanf("%d", &n);
for(int i=3;i>=0;i--)
{
a[i]=n%10+2;
n/=10;
}
final = (a[0] * 1000) + (a[1] * 100) + (a[2] * 10) + a[3];
printf("%d", final);
return 0;
}
Below function works with N number of digits.
Idea is to extract each digit from the input number and add its decimal position.
#include <stdio.h>
int power(int x, int y)
{
int res = 1;
for (;y>0;y--)
{
res *=x;
}
return res;
}
int main ()
{
int n;
scanf("%d", &n);
int sum = 0;
int i=0;
while(n>0)
{
sum += ((n%10) +2)*power(10,i);
i++;
n /=10;
}
printf("%d", sum);
return 0;
}
Another idea:
char str[10]; // enough to contain an int as string + 1
char *s = str+sizeof(str); // points to last char + 1
int n;
scanf("%d", &n);
*--s = 0; // terminate the string
while(n) {
*--s = (((n % 10)+2)%10) + '0'; // write a char from the end
n /= 10;
}
printf("%s\n", s);
int a = 5696;
// output is 7818
// ---------Java-----------
// --------solution--------
int first = a/1000+2;
int b = a%1000;
int second = b/100+2;
int c = b%100;
int d = c/10+2;
int third = d/10;
int e = c%10;
int fourth = e+2;
String result = Integer.toString(first)+Integer.toString(second)+Integer.toString(third)+Integer.toString(fourth);
System.out.println(result);
Lets say I have an integer called SIN and the scanf input receives 193456787.
so SIN = 193456787;
What I want to do is add up all the other numbers after the first digit.
So 9 + 4 + 6 + 8 = 27
Can somebody please explain to a beginner how to do this?
Print the number and then sum every other digit
int sum_every_other_digit_after_first(unsigned long long x) {
char buf[sizeof x * CHAR_BIT];
sprintf(buf, "%llu", x);
char *p = buf;
int sum = 0;
while (*p) {
p++; // Skip digit
if (*p) {
sum += *p++ - '0';
}
}
return sum;
}
or as inspired by #PageNotFound
int sum_every_other_digit_after_first(unsigned long long x) {
int esum = 0;
int osum = 0;
while (x > 0) {
esum += x%10;
x /= 10;
if (x == 0) {
return osum;
}
osum += x%10;
x /= 10;
}
return esum;
}
or for fun, a recursive solution
int sum_every_other_digit_after_first_r(unsigned long long x, int esum, int osum) {
if (x >= 100) {
int digit2 = x % 100;
esum += digit2 % 10;
osum += digit2 / 10
return sum_every_other_digit_after_first_r(x / 100, esum, osum);
}
if (x >= 10) {
return esum + x % 10;
}
return osum;
}
sum_every_other_digit_after_first_r(1234567,0,0) --> 12
My solution
#include <stdio.h>
int main()
{
int SIN = 193456787;
int a = 0, b = 0, cnt = 0;
while (SIN > 0) {
if (cnt % 2) b += SIN % 10;
else a += SIN % 10;
cnt++;
SIN /= 10;
}
printf("%d\n", cnt%2 ? b : a);
return 0;
}
Note: Please comment if this is not what you intended, as your question is a little ambigous.
#include <stdio.h>
int main() {
unsigned number;
scanf("%u\n", &number);
unsigned result = 0;
unsigned tmp = number;
unsigned numberOfDigits = 0;
do
numberOfDigits++;
while((tmp /= 10) != 0);
if(numberOfDigits % 2 != 0)
number /= 10;
while(number >= 10) {
result += number % 10;
number /= 100; // Skip two digits
}
printf("%u\n", result);
}
Digits Sum function
I have to write a function - digits sum,
for ex.: the number : 9887, 9+8+8+7= 32 , 3+2= 5
Will this function work well?
int sum_digits(int num){
if (num < 10)
return num;
int a = sum_digits (num / 10) + num % 10;
if (a > 9)
a = sum_digits (a / 10) + a % 10;
return a;
}
int sum_digits(int num){
if (num < 10)
return num;
int a = sum_digits (num / 10) + num % 10;
if (a > 10)
a += sum_digits (a / 10) + a % 10;
return a;
}
This will work.
Note that a= is replaced by a+= and a>9 is replaced by a>10
if the user inputs the "value" as 1223445 the output should read as follows:
After change #1: 12235
After change #2: 135
the code is meant to take out two consecutive numbers with the same value. the first loop works but then it stops and I cannot figure why. here is the code:
{
int count;
int y;
int z;
int b;
int c;
int d;
int a;
int ct;
y = 0;
z = 1;
d = 0;
count = 0;
if (value > 0)
while ((z * 10 + z) != (y % 100))
{
y = value % 10 + y * 10;
z = y % 10;
value /= 10;
count = count + 1;
}
value = value * pow(10, count - 2);
y = y / 100;
count = count - 3;
while(y > 0)
{
b = y % 10;
c = pow(10, count);
d = d + c * b;
y = y / 10;
count = count - 1;
}
value = value + d;
ct = 1;
printf("After change #%d: %d\n", ct, value);
a = value;
while (a > 1)
{
if((a % 100) - (a % 10) - (10 * (a % 10)) == 0)
Change(value);
else
a = a / 10;
}
return;
}
Here's a simple solution: (not sure how you want to handle odd numbered repeats e.g. '111', '11111')
public static long strip(long value, long sum) {
if(value==0) return sum;
long tens = value % 100;
long ones = value % 10;
if(ones*10 == tens-ones) {
return strip(value /100, sum);
}
if(sum==0) sum +=(value%10);
else {
long x = (long)Math.ceil((Math.log10(sum)));
sum =(long) (Math.pow(10,x) * ones + sum);
}
return strip(value /10, sum);
}