Issue in detecting Armstrong Numbers - c

Question

I have this problem where the number isn't being identified as an Armstrong number. I tried entering 153 which is an Armstrong number which isn't being identified as one. I am learning C. Any help would be nice.
#include <stdio.h>
int main()
{
int x[3],a,b,c,temp;
int temp2 = 0;
printf("Enter a supposed Armstrong number: ");
scanf("%d",&x);
for(int i =0;i<3;i++)
{
temp = x[i]*x[i]*x[i];
temp2 = temp2 + temp;
}
if(temp2 == x)
{
printf("This is an Armstrong number!");
}
else
{
printf("This is not an Armstrong number!");
}
}

Answer 1

TL;DR - To store a 3-digit integer value, you don't need an array with 3 ints.
In your code, x is an array type, and the way you use it, only x[0] get a valid value, x[1] and x[2] are left uninitialized. So, your code produces undefined behavior.
I believe, you don't need to have an array, instead make use of the modulo operator to extract each digit one by one and carry on with the calculation.

Answer 2

with respect to your code x is an integer type array.so,integer type array doesn't work like a char type array.
char type array takes base address and it is still open util '\0'(null) char is found.it is defined for scanf().but for int type array you need to manually define the particular position of the array in the memory to store the inputs.
so,u have to use like this =>
#include <stdio.h>
int main(){
int x[3],n,r,c,temp,i;
int temp2 = 0;
printf("Enter a supposed Armstrong number: ");
scanf("%d",&n);
c=n;
for(i =0;i<3;i++){
x[i]=n%10;
n=n/10;
temp = x[i] * x[i] * x[i];
temp2 = temp2 + temp;
}
if(temp2 == c){
printf("This is an Armstrong number!");
}
else{
printf("This is not an Armstrong number!");
}
return 0;
}

Answer 3

Firstly take input into a single variable of type int and then store it in array after breaking them by % operator then you will be able to find Armstrong number.
E.g
int x=153;
int y[3];
for(int i=2;i>=;i--)
{
y[i]=x%10;
x=x/10;
}
After this step your loop will work properly

Answer 4

You were no need to use an array for getting all digits from given integer input.
#include <stdio.h>
int main() {
int input, temp, rmdr, sum = 0;
printf("Enter a supposed Armstrong number: ");
scanf("%d", &input);
temp = input; // store value of given input (backUp)
for ( ; input>0; ) { // loop till input has data
rmdr = input % 10; // get last digit from input
sum += (rmdr * rmdr * rmdr); // collect cube of remainders
input /= 10; // remove last digit from input
}
// collected sum will compare with temp (backUp)
// because given input will be 0 after loop
if (temp == sum)
printf("This is an Armstrong number!");
else
printf("This is not an Armstrong number!");
}

Answer 5

Thanks all, really helped. I needed to do this for a mini-project.I decided to do this because it seemed the easiest way.
#include <stdio.h>
int main()
{
int x,y,z,temp;
int temp2 = 0;
printf("Enter a supposed Armstrong number: ");
scanf("%d",&x);
z = x;
for(int i =0;i<3;i++)
{
y = x%10;
temp = y*y*y;
temp2 = temp2 + temp;
x = x/10;
}
if(temp2 == z)
{
printf("This is an Armstrong number!");
}
else
{
printf("This is not an Armstrong number!");
}
}