I need help solving this problem, if anyone had a similar problem it would help me.
The task is: Write a program that loads an array a
of n
natural numbers and then forms an array b
so that the elements of the array b
are equal to the sum the digits of the array element a
.
Example:
n=5
a[0]=1
a[1]=2
a[2]=3
a[3]=32
a[4]=54
Array b is:
b[0]=1
b[1]=2
b[2]=3
b[3]=5
b[4]=9
I did like this:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAX 100
void form(int [], int [], int);
int sum_digit(int);
void print (int [], int);
int main() {
int n;
printf("Enter n:");
scanf("%d", &n);
while (n < 1 || n > MAX) {
int n;
printf("Enter n:");
scanf("%d", &n);
}
int a[MAX], b[MAX];
printf("Enter element of array a:\n");
for (int i = 0; i < n; i ) {
printf("%d. element:",i 1);
scanf("%d", &a[i]);
}
form(a, b, n);
print(b, n);
return 0;
}
void form(int a[MAX], int b[MAX], int n) {
for (int i = 0; i < n; i ) {
while (a[i] > 0) {
int pom = a[i] % 10;
b[i] = sum_digit(pom);
a[i] = a[i] / 10;
}
}
}
int sum_digit(int s) {
int sum = 0;
sum = sum s;
return sum;
}
void print(int b[MAX], int n) {
for (int i = 0; i < n; i ) {
printf("%d ", b[i]);
}
}
The problem arises for multi-digit array elements.
Each time the sum_digit
function is called the sum is set to zero.
So for example for element 32
, I can't get 5 because I can't add 2
and 3
. If anyone has an idea how to solve this I would be grateful.
CodePudding user response:
There are multiple problems in your code:
you do not test the return value of
scanf()
so any invalid or missing input will cause undefined behavior because the destination variables will not be updated.There is a new variable
n
defined in thewhile
loop, so the original variable in the outer scope is never modified byscanf()
.form
andsum_digit
are both incorrect: you should compute the sum insum_digit()
with a loop and store all sums in a loop inform
.
Here is a modified version:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAX 100
void form(int [], int [], int);
int sum_digits(int);
void print (int [], int);
int main() {
int n;
for (;;) {
printf("Enter n:");
if (scanf("%d", &n) != 1)
return 1;
if (n >= 1 && n <= MAX)
break;
printf("Incorrect value: %d\n", n);
}
int a[MAX], b[MAX];
printf("Enter elements of array a:\n");
for (int i = 0; i < n; i ) {
printf("%d. element:", i 1);
if (scanf("%d", &a[i]) != 1)
return 1;
}
form(a, b, n);
print(b, n);
return 0;
}
void form(int a[MAX], int b[MAX], int n) {
for (int i = 0; i < n; i ) {
b[i] = sum_digits(a[i]);
}
}
int sum_digits(int s) {
int sum = 0;
/* this loop handles positive and negative numbers alike */
while (s != 0) {
sum = sum s % 10;
s = s / 10;
}
/* return the positive sum of digits */
return abs(sum);
}
void print(int b[MAX], int n) {
for (int i = 0; i < n; i ) {
printf("%d ", b[i]);
}
printf("\n");
}
CodePudding user response:
For starters pay attention to that as the type of elements of the array a is int
then it means that the user can enter negative numbers in the array. But the sum of digits should be a non-negative value.
And the array a
should not be changed. The task is just to form the array b
.
Now here you are.
void form( const int a[], int b[], int n )
{
for ( int i = 0; i < n; i )
{
b[i] = sum_digit( a[i] );
}
}
and
int sum_digit( int num )
{
enum { Base = 10 };
int sum = 0;
do
{
int digit = num % Base;
if ( digit < 0 ) digit = -digit;
sum = digit;
} while ( num /= Base );
return sum;
}
Here is a demonstration program.
#include <stdio.h>
int sum_digit( int );
void form( const int a[], int b[], int n )
{
for (int i = 0; i < n; i )
{
b[i] = sum_digit( a[i] );
}
}
int sum_digit( int num )
{
enum { Base = 10 };
int sum = 0;
do
{
int digit = num % Base;
if (digit < 0) digit = -digit;
sum = digit;
} while (num /= Base);
return sum;
}
void print( const int a[], int n )
{
for (int i = 0; i < n; i )
{
printf( "%d ", a[i] );
}
putchar( '\n' );
}
int main( void )
{
enum { MAX = 5 };
int a[MAX] = { 1, -22, 333, -4444, 55555 };
int b[MAX];
print( a, MAX );
form( a, b, MAX );
print( b, MAX );
}
The program output is
1 -22 333 -4444 55555
1 4 9 16 25
CodePudding user response:
sum_digit()
should calculate the sum of all the digits, not just one digit. So you need to move the loop from form()
to sum_digit()
.
void form(int a[MAX],int b[MAX],int n)
{
for(int i=0;i<n;i )
{
b[i]=sum_digit(a[i]);
}
}
int sum_digit(int s)
{
int sum=0;
while (s > 0) {
sum = s % 10;
s /= 10;
}
return sum;
}
Since the assignment says that the array contains natural numbers, you could also change all your int
declarations to unsigned
.