finding the sum of digits? [closed] - c

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I tried this problem using for loop but i did not got the correct output as specified . i do not know what is the problem in my code ?
#include <stdio.h>
int main()
{
int T,N;
scanf("%d %d" ,&T,&N);
for(int i=N;i>0;i=i/10)
{
int r=N%10;
int sum=0;
sum=sum+r;
printf("The sum is : %d" ,sum);
}
}
The sum is : 5The sum is : 5The sum is : 5The sum is : 5The sum is : 5
The output is coming like this while we just need sum of digits printed

Answer 1

for(int i = N; i > 0 ; i = i/10)
{
int r = N % 10; // calculating remainder of UNMODIFIED input, so will
// ALWAYS be last digit
int sum = 0; // you are initializing the sum to 0 for every single iteration
sum = sum + r; // so this will *always* result in 0 + N % 10
printf("The sum is : %d", sum);
}
To fix, you need to initialise the sum just once to collect all of the single digits. Additionally, you need to use the modified value:
int sum = 0;
for(int i = N; i > 0 ; i = i/10)
{
int r = i % 10;
// ^ (!)
sum += r; // alternative variant...
printf("The sum is : %d\n", sum);
// ^^ for better output formatting
}
Until now we are still printing the sum with every iteration as well. That might be useful, if you want to follow how the sum evolves (assuming input was 1210):
The sum is 0
The sum is 1
The sum is 3
The sum is 4
But actually, you'd rather want to print only the result, wouldn't you? So you'd move the printing out of the loop as well:
for(...)
{
...
}
printf("The sum is : %d\n", sum);
Alternative variant: If you don't need the value of N afterwards any more anyway, you can iterate directly on it:
for( ; N > 0; N /= 10)
// ^ empty initialization, nothing to be done...
{
int r = N % 10; // NOW using N is fine...
...
}
Finally: if you compare with != instead of >, you can cover negative intput (as you use signed integers...) as well.
Edit according to question:
it asked input and output like this. Input 3 12345 31203 2123 Output 15 9 8
Well, in this case, you need a double loop:
int t;
// well, actually, you should check if you did get correct input:
if(scanf("%d", &t) != 1))
{
// invalid input
// appropriate error handling, e. g. printing a message and:
return -1;
}
for( ; t > 0; --t) // handles the number of tasks to solve
{
int n; // inside loop: read in a new value with every task
scanf("%d", &n); // TODO: check input, see above
int sum = 0;
for(...) { ... } // loop handling the input value, see above
printf(...);
}