Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 3 years ago.
Improve this question
I'm trying to write a program that calculates 1+2+3...+n with recursion. I'm just starting to learn C so I'm pretty bad...
Here's what I have right now:
#include <stdio.h>
int adder;
adder=1;
int sum(int numba) {
if (adder==numba) {
return(sum);
}
return(sum+adder);
++adder;
}
int main() {
char line[100];
int nummba;
printf("Enter in a number: ");
fgets(line, sizeof(line), stdin);
sscanf(line, "%d", nummba);
printf("The sum of numbers from 1 to %d is %d.", nummba, sum(nummba));
return(0);
}
Obviously doesn't work...
How do I use recursion to solve 1+2+3...n?
Thanks.
For starters there is no need to use global variables.
Secondly in these statements
return(sum);
and
return(sum+adder);
sum is a pointer to function. So the expressions of the return statements do not make sense.
Moreover this code snippet
int adder;
adder=1;
will not even compile because you may not use statements in the file scope. At least you should write
int adder = 1;
The function can look the following way
unsigned long long int sum( unsigned int n )
{
return n == 0 ? 0 : n + sum( n - 1 );
}
Here is a demonstrative program.
#include <stdio.h>
unsigned long long int sum( unsigned int n )
{
return n == 0 ? 0 : n + sum( n - 1 );
}
int main(void)
{
printf( "Enter in a non-negative number: " );
unsigned int n = 0;
scanf( "%u", &n );
printf( "The sum of numbers from 0 to %u is %llu.\n", n, sum( n ) );
return 0;
}
Its output might look like
Enter in a non-negative number: 10
The sum of numbers from 0 to 10 is 55.
Recursion requires that a function calls itself. You are not doing this, so you do not have a recursive function in the first place. Consider the following code:
#include <stdio.h>
int sum(int n) {
if (n == 0)
return n;
else
return n + sum(n - 1);
}
int main() {
int n;
char line[100];
printf("Enter in a number: ");
fgets(line, sizeof(line), stdin);
sscanf(line, "%d", &n);
printf("The sum of numbers from 1 to %d is %d.\n", n, sum(n));
return(0);
}
If you find recursion a bit difficult for now, you can opt for a less difficult solution using a while loop:
int sum(int n)
{
int i = 0;
int j = 1;
while (j <= n)
{
i += j++;
}
return i;
}
change your code to:
#include <stdio.h>
int sum_val=0;
void sum(int numba)
{
if(numba>=1)
{
sum_val=sum_val+numba;
sum(numba-1);
}
}
int main()
{
int nummba;
printf("Enter in a number: ");
scanf(" %d",&nummba);
sum(nummba);
printf("The sum of numbers from 1 to %d is %d.", nummba, sum_val);
return(0);
}
Did you compile your code?
Related
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 4 years ago.
Improve this question
The program should have the following guidelines:
The numbers are read from the standard input.
The first number is the length of the sequence (n) followed by n numbers.(ie, if you put '54321' the length of the sequence is 5 numbers)
If n is 0 or negative, the program displays the message “Error_1” followed
by a new line on the standard input.
If the length is shorter than n, it displays “Error_2” followed by a new line and quits.
I'm finding point number 2 difficult
My code is:
#include <stdio.h>
int main() {
int i,j,k;
printf("Enter a Number:\n");
scanf("%d", &i);
if (i <= 0 ) {
printf("Error_1\n");
} else if(){
printF("Error_2\n")
}
}
#include <stdio.h>
int main() {
int i,j,k;
printf("Enter a Number:\n");
scanf("%d", &i);
if (i <= 0 ) {
printf("Error_1\n");
} else{
scanf("%d",&j);
k=0;
while(j>0)
{
k++;
j=j/10;
}
if(k<i)
printf("Error_2\n");
}
}
so what i did was, i found out the length of number entered and if the length does not match with provided length , it prints error2. i found out the length by continously dividing the number by 10 till it becomes 0.
Check this out. You can use log base 10 to compute the length of the sequence. Easier and clener.
To compile the code use -lm flag.See the following command:
gcc sampleFilename.c -lm
#include <stdio.h>
#include<math.h>
int main() {
int i,num, length;
printf("Enter a Number:\n");
scanf("%d", &i);
if (i <= 0 ) {
printf("Error_1\n");
}
else{
printf("Enter the number: ");
scanf("%d",&num);
length=(int) log10(num)+1;// compute the length of the number .. read about log10
if (length < i ) //length of the sequence is less than the number 'i'
printf("Error_2\n");
}
}
there is something you have to understand
your code:
"if (i <= 0 ) {
printf("Error_1\n");
} else if () {/* I'm talking about this*/
printF("Error_2\n")
}"
commend: /* its mean else if (its empty!!) you should write condition in the if, what you need to do its just use if and not else or else if
https://www.programiz.com/c-programming/c-if-else-statement*/
currect code below:
#include <stdio.h>
int main()
{
int num,cnt=0;
printf("Enter length: "\n);
scanf("%d", &num);
while(num > 10) //checking first number
{
cnt++;
num /= 10;
}
if(num <= 0)
{
printf("Error_01\n");
}
if(num == (count + 1)!)
{
printf("Error_02");
}
return 0;
}
Try this below code,
Hope this will work
#include <stdio.h>
int main()
{
int i,j,k;
printf("Enter any number: \n");
scanf("%d", &i);
k = 0;
if(i <= 0)
{
printf("Error_01\n");
}
printf("Enter value number: \n");
scanf("%d", &j);
while(j != 0) // check till first digit
{
k++;
j /= 10;
}
if(k != i)
{
printf("Error_02\n");
}
return 0;
}
by the way this was referred from
https://codeforwin.org/2016/10/c-program-to-count-number-of-digits-in-number.html
I have a problem with C program. The idea of it is similar to Armstrong number checking. Say if the input number is 123. Program needs to check if condition, for example 123=1^1+2^2+3^3 is true. I know how to add digits,but have a problem with powers. It is obvious that I need a loop for powers from 1 to the number of digits. In Armstrong number algorithm you have similar power on every digit. For example 153=1^3+5^3+3^3. Here is what I have so far:
#include<stdio.h>
int main()
{
int n,d,s=0,o,i,k;
printf("n=");scanf("%d",&n);
d=n;
while(d!=0)
{
o=d%10;
s=s+o;
d=d/10;
k++
}
printf("sum:%d",s);
printf("number of digits:%d",k);
return 0;
}
Thanks for the answers.
You need first get the lenth of number, which is used to determine how many times you need to get into loop to calculate each bit.
For example, number 123, you first need to know the number is 3 bits len, then you can mutilply number 3 three times, number 2 twice, and number 1 once.
I use a temporary string to achieve this
here is code， a little bit alteration on yours
#include <stdio.h>
#include <string.h>
#define MAX_NUM_LEN 16
int main()
{
char tmp_num[MAX_NUM_LEN] = {0};
int len,n,d,s=0,o,i,tmp_len, tmp_o;
printf("n=");scanf("%d",&n);
sprintf(tmp_num, "%d", n);
len = strlen(tmp_num);
tmp_len = len;
d=n;
while(d!=0)
{
o=d%10;
for (tmp_o = 1, i = tmp_len; i > 0; i--)
tmp_o *= o;
s=s+tmp_o;
d=d/10;
tmp_len--;
}
printf("sum:%d\n",s);
printf("number of digits:%d\n",len);
return 0;
}
results:
According of what I've understood I think this is what the OP is looking for:
int power(int base, int exp)
{
if (base == 0) return 0;
int result=1;
while (exp-- > 0) result*=base;
return result;
}
void calculate(int number)
{
int d=number;
int tmpnumber=number;
int n=0;
while (d > 0)
{
n++;
d /=10;
}
printf("Number of digits: %d\n", n);
int k=0;
int sum=0;
while (n--)
{
// get digits from left to right
d=number / power(10, n);
k++;
sum+=power(d, k);
number %= power(10, n);
printf("%d^%d=%d\n", d, k, power(d, k));
}
printf("\n%5d %5d", tmpnumber, sum);
}
int main(int argc,char *argv[])
{
int value;
while (TRUE)
{
printf("Enter value (0 = Quit): ");
scanf("%d", &value);
if (value <= 0) return 0;
calculate(value);
printf("\n");
}
}
it's me again. I deleted my previous question because it was very poorly asked and I didn't even include any code (i'm new at this site, and new at C). So I need to write a program that prints out the digits smaller than 5 out of a given number, and the number of the digits.
For example: 5427891 should be 421 - 3
The assignment also states that i need to print the numbers smaller than 5 in a recursive function, using void.
This is what I've written so far
#include<stdio.h>
void countNum(int n){
//no idea how to start here
}
int main()
{
int num, count = 0;
scanf("%d", &num);
while(num != 0){
num /= 10;
++count;
}
printf(" - %d\n", count);
}
I've written the main function that counts the number of digits, the idea is that i'll assign (not sure i'm using the right word here) the num integer to CountNum to count the number of digits in the result. However, this is where I got stuck. I don't know how to extract and print the digits <5 in my void function. Any tips?
Edit:
I've tried a different method (without using void and starting all over again), but now i get the digits I need, except in reverse. For example, instead of printing out 1324 i get 4231.
Here is the code
#include <stdio.h>
int rec(int num){
if (num==0) {
return 0;
}
int dg=0;
if(num%10<5){
printf("%d", num%10);
dg++;
}
return rec(num/10);
}
int main(){
int n;
scanf("%d", &n);
int i,a;
for(i=0;i<n;i++)
{
scanf("%d", &a);
rec(a);
printf(" \n");
}
return 0;
}
Why is this happening and how should I fix it?
There is nothing in your question that specifies the digits being input are part of an actual int. Rather, its just a sequence of chars that happen to (hopefully) be somewhere in { 0..9 } and in so being, represent some non-bounded number.
That said, you can send as many digit-chars as you like to the following, be it one or a million, makes no difference. As soon as a non-digit or EOF from stdin is encountered, the algorithm will unwind and accumulate the total you seek.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int countDigitsLessThanFive()
{
int c = fgetc(stdin);
if (c == EOF || !isdigit((unsigned char)c))
return 0;
if (c < '5')
{
fputc(c, stdout);
return 1 + countDigitsLessThanFive();
}
return countDigitsLessThanFive();
}
int main()
{
printf(" - %d\n", countDigitsLessThanFive());
return EXIT_SUCCESS;
}
Sample Input/Output
1239872462934800192830823978492387428012983
1232423400123023423420123 - 25
12398724629348001928308239784923874280129831239872462934800192830823978492387428012983
12324234001230234234201231232423400123023423420123 - 50
I somewhat suspect this is not what you're looking for, but I'll leave it here long enough to have you take a peek before dropping it. This algorithm is fairly pointless for a useful demonstration of recursion, to be honest, but at least demonstrates recursion none-the-less.
Modified to print values from most significant to least.
Use the remainder operator %.
"The result of the / operator is the quotient from the division of the first operand by the second; the result of the % operator is the remainder. In both operations, if the value of the second operand is zero, the behavior is undefined" C11dr §6.5.5
On each recursion, find the least significant digit and test it. then divide the number by 10 and recurse if needed. Print this value, if any, after the recursive call.
static int PrintSmallDigit_r(int num) {
int count = 0;
int digit = abs(num % 10);
num /= 10;
if (num) {
count = PrintSmallDigit_r(num);
}
if (digit < 5) {
count++;
putc(digit + '0', stdout);
}
return count;
}
void PrintSmallDigits(int num) {
printf(" - %d\n", PrintSmallDigit_r(num));
}
int main(void) {
PrintSmallDigits(5427891);
PrintSmallDigits(-5427891);
PrintSmallDigits(0);
return 0;
}
Output
421 - 3
421 - 3
0 - 1
Notes:
This approach works for 0 and negative numbers.
First of all, what you wrote is not a recursion. The idea is that the function will call itself with the less number of digits every time until it'll check them all.
Here is a snippet which might help you to understand the idea:
int countNum(int val)
{
if(!val) return 0;
return countNum(val/10) + ((val % 10) < 5);
}
void countNum(int n, int *c){
if(n != 0){
int num = n % 10;
countNum(n / 10, c);
if(num < 5){
printf("%d", num);
++*c;
}
}
}
int main(){
int num, count = 0;
scanf("%d", &num);
countNum(num, &count);
printf(" - %d\n", count);
return 0;
}
for UPDATE
int rec(int num){
if (num==0) {
return 0;
}
int dg;
dg = rec(num/10);//The order in which you call.
if(num%10<5){
printf("%d", num%10);
dg++;
}
return dg;
}
int main(){
int n;
scanf("%d", &n);
int i,a;
for(i=0;i<n;i++){
scanf("%d", &a);
printf(" - %d\n", rec(a));
}
return 0;
}
This is an activity given by my instructor.
Create a program that accepts numeric input from the user. If the user enters an even number, store it to an array for even numbers. If the user enters an odd number, store it to another array for odd numbers. Input terminates if the user entered 10 numbers already. Display the size of each array and their elements.
Example:
Input: 5, 6, 12, 10, 0, 3, 4, 100, -1, 7
Even numbers (6): 6 12 10 0 4 100
Odd numbers (4): 5 3 -1 7
and this is the code I've come up with.
#include <stdio.h>
int sort(int);
int main(){
int input, count;
for(count=0;count!=10;count++){
printf("Enter 10 digits: ");
scanf("%d", &input);
sort(input);
}
printf("%d", input);
return 0;
}
int sort(int inp){
int odd[10];
int even[10];
if(inp%2==0){
odd[]=inp;
}
else
even[]=inp;
return 0;
}
Please help me on how to store the numbers into two separate arrays. Any tips will be greatly appreciated.
Check the below code. It's self-explaining.
#include <stdio.h>
#include <stdlib.h>
#define NUM 10
int main()
{
int input, i;
int oddcounter = 0, evencounter =0;
int oddarr[NUM];
int evenarr[NUM];
printf("Enter 10 integers\n");
for (i = 0; i < NUM; i++)
{
if ( scanf("%d", &input) == 1 )
{
if ((input % 2) == 0)
{
evenarr[evencounter++] = input;
}
else
{
oddarr[oddcounter++] = input;
}
}
}
printf("Number of elem in oddarray : %d, evenarray : %d\n\n", oddcounter, evencounter);
printf("Odd elements are :");
for (i = 0; i < oddcounter ; i++) printf("%d\t", oddarr[i]);
printf("\n");
printf("Even elements are :");
for (i = 0; i < evencounter; i++) printf("%d\t", evenarr[i]);
printf("\n");
return 0;
}
In addition to Sourav's comment which indicates that you shouldn't have the int[] arrays be local to sort, this syntax isn't correct for assigning to arrays in C:
odd[]=inp;
On my compiler, it generates the following error:
24:9: error: expected expression
odd[]=inp;
To store to odd, you need to indicate the index at which you'd like to store, for example:
odd[1]=inp;
which also means you'll need to keep track the latest index you wrote to for each array!
You need to tell the compiler which index of the array you are storing your data to. In sort:
if(inp%2==0){
odd[]=inp;
}
else
even[]=inp;
return 0;
}
Should look something like:
if(inp%2==0){
odd[endofoddindex]=inp;
}
else
even[endofevenindex]=inp;
return 0;
}
That said, you won't get much use out of the arrays being local variables, since they are deallocated on each call. Your best bet is to declare the arrays in main and pass them in.
Your even and odd arrays are both local. This means that they exist as long as the function exists. So you won't be retrieve the data you have stored(You also don't store it correctly).
So you need both the arrays in main and also two other variables for using as the index of both the array(i and j in the below program). The modified program is given below:
#include <stdio.h>
int sort(int);
int main(){
int input, count,i=0,j=0; //i and j to be used as array indices
int odd[10];
int even[10]; //arrays in main
for(count=0;count!=10;count++){
printf("Enter 10 digits: ");
scanf("%d", &input);
if(sort(input)) //if odd number was found
odd[i++]=input;
else //even number found
even[j++]=input;
}
printf("%d", input);
//print even and odd arrays here
return 0;
}
int sort(int inp){
if(inp%2==0)
return 0; //if number is even,return 0
return 1; //else return 1
}
You need to either have your arrays as globals or pass them into your sort function. Where they are they currently they get recreated every time the sort function is called and are inaccessible to the rest of your program.
You will also need to keep track of the max number of ints in each array and the current number.
Your test in sort would be something like this:
if( inp % 2 == 0)
{
//TODO check that currentEvenCount < maxEvenCount
even[ currentEvenCount ] = inp;
currentEvenCount++
}
else
{
//TODO check that currentOddCount < maxOddCount
odd[ currentOddCount ] = inp;
currentOddCount++;
}
To declare your arrays as globals just move the declaration outside of any function above anywhere they are referenced
int even[10];
int odd[10];
int main() ...
To pass them as parameters to sort function you could declare sort like this:
sort( int inp, int even[], int maxEvenCount, int* currentEvenCount, int odd[]. int maxOddCount, int* currentOddCount)
{
...
if( inp % 2 == 0)
{
//TODO check that currentEvenCount < maxEvenCount
even[ *currentEvenCount ] = inp;
(*currentEvenCount)++
}
}
The * in front of currentEventCount is dereferencing the pointer and getting/setting the actual value pointed to.
You would then call sort like so:
int main()
{
int evenArray[10];
int oddArray[10];
int currentEvenCount = 0;
int currentOddCount = 0;
...
sort( input, evenArray, 10, ¤tEvenCount, oddArray, 10, ¤tOddCount);
}
There is no any sense to define the arrays as local variables of function sort because each time the function is called the arrays are created anew.
The program could look the following way
#include <stdio.h>
#define N 10
enum Type { Even, Odd };
enum Type sort( int x )
{
return x % 2 == 0 ? Even : Odd;
}
int main( void )
{
int odd[N];
int even[N];
int odd_count = 0;
int even_count = 0;
int i;
printf( "Enter %d numbers: ", N );
for( i = 0; i < N; i++ )
{
int num;
scanf( "%d", &num );
switch ( sort( num ) )
{
case Even:
even[even_count++] = num;
break;
case Odd:
odd[odd_count++] = num;
break;
}
}
printf( "Even numbers (%d):", even_count );
for ( i = 0; i < even_count; i++ ) printf( " %d", even[i] );
printf( "\n" );
printf( "Odd numbers (%d):", odd_count );
for ( i = 0; i < odd_count; i++ ) printf( " %d", odd[i] );
printf( "\n" );
return 0;
}
If to enter
5 6 12 10 0 3 4 100 -1 7
then the output will be
Even numbers (6): 6 12 10 0 4 100
Odd numbers (4): 5 3 -1 7
Simply copy, paste and investigate the program.:)
Hope this program will solve you issue. Here is the working code.
#include <stdio.h>
int sort(int[]);
int main(){
int input[10], count;
printf("Enter 10 digits: ");
for(count=0;count<10;count++){
scanf("%d", &input[count]);
}
sort(input);
return 0;
}
int sort(int inp[]){
int odd[10];
int even[10];
int oddCount=0, evenCount=0;
int i;
for(i=0; i<10;i++)
{
if(inp[i]%2==0){
even[evenCount]=inp[i];
evenCount++;
}
else
{
odd[oddCount]=inp[i];
oddCount++;
}
}
printf("ODD COUNT is %d \n", oddCount);
for(i=0; i<oddCount;i++)
{
printf("ODD VALUE %d \n", odd[i]);
}
printf("EVEN COUNT is %d \n", evenCount);
for(i=0; i<evenCount;i++)
{
printf("EVEN VALUE %d \n", even[i]);
}
return 0;
}
Your variable input is just an int, it's not an array. You need two arrays:
int even[10], odd[10];
Then you need to keep track of the number of numbers you've read so far, and for each number check which array to store it in.
I don't see a need to do sorting, so not sure why you have a function called sort().
It should just be something like:
int even[10], odd[10];
int oddindex = 0, evenindex = 0;
while(scanf(" %d", &x) == 1)
{
if(x % 2 == 0)
even[evenindex++] = x;
else
odd[oddindex++] = x;
if((evenindex + oddindex) >= 10)
break;
}
/* Loop here to print numbers. */
An answer suitable for an assignment question:
int main()
{
int i,c,o[10],e[10];int oc=0;int ec=0;int*pc;for(c=0;c<10;c++){scanf("%d",&i);pc=(i&1)?&o[oc++]:&e[ec++];*pc=i;}
// Now print out the values as requested in oc, o, ec and e.
}
This question already has answers here:
Check if a value from scanf is a number?
(2 answers)
Closed 9 years ago.
my program adds numbers entered by the user. It runs and works great until a character is entered instead of an integer. Is there a simple way to make sure only integers are entered from the keyboard?
Here is my code.
#include<stdio.h>
#include <stdlib.h>
int main(int argc, char **argv)
{
int n, sum = 0, i, TotalOfNumbers;
printf("Enter the number of integers you want to add\n");
scanf("%d", &n);
printf("Enter %d integers\n",n);
for (i = 1; i <= n; i++)
{
scanf("%d",&TotalOfNumbers);
sum = sum + TotalOfNumbers;
}
printf("Sum of entered integers = %d\n",sum);
return 0;
}
You need to check the return value of scanf. If the input was a valid number, it will return 1. If the input was not a valid number, it will return something else. Here is your code modified to put the checks in.
#include<stdio.h>
#include <stdlib.h>
int get_number()
{
int num;
int ret;
ret = scanf("%d", &num);
if (ret != 1) {
printf("bad number\n");
exit(EXIT_FAILURE);
}
return num;
}
int main(int argc, char **argv)
{
int n, sum = 0, i, TotalOfNumbers;
printf("Enter the number of integers you want to add\n");
n = get_number();
printf("Enter %d integers\n",n);
for (i = 1; i <= n; i++)
{
TotalOfNumbers = get_number();
sum = sum + TotalOfNumbers;
}
printf("Sum of entered integers = %d\n",sum);
return 0;
}
Check the ferror state on the input stream
scanf("%d",&TotalOfNumbers);
if(!ferror(stdin)){
sum = sum + TotalOfNumbers;
}
In addition to posted answer, there options not general as posted, but quicker.
First if you want to skip some final set of characters.In following example all letters,! and + will be skiped
int n;
scanf("%*[a-zA-Z!+]%d",&n);
printf("\n%d",n);
for input
weweqewqQQWWW!!!!+++3332
the output is
3332
Next option is to use buffer wich allowed to read everything untill number is met, and then read the number. The disadvantage is that buffer size is limited
char buf[25];
int n;
scanf("%[^0-9]%d",buf,&n);
printf("\n%d",n);
For input
fgfuf#$#^^#^##4565
Output
4565