double array output in C program

Question

I seem to stuck in a extremely simple question, but I cannot fix it and understand why. Below is my code:

#include <stdio.h>
#define SIZE 8

int main(void){
  int i, j;
  double nums[SIZE];
  double input;
  
  printf("Enter 8 doubles: ");
  for (i = 0; i < SIZE; i++){
    nums[i] = scanf("%lf", &input);
  }

  for (i = 0; i < SIZE; i++){
    printf("%lf ", nums[i]);
  }

  return 0;
}

Below is the result I got:

Enter 8 doubles: 1 2 3 4 5 6 7 8
1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000

What is the main problem about my code? Thank you!

scanf returns the number of items scanned. You want to assign input to your array items. — Retired Ninja
– Retired Ninja, Commented Oct 18, 2020 at 7:28

Farhad Sarvari · Accepted Answer · 2020-10-18 07:39:46Z

1

Try this one and you should read the data in variable with scanf because return from scanf is the number of items of the argument list successfully read:

 #include <stdio.h>
 #define SIZE 3

 int main(void){
   int i, j;
   double nums[SIZE];
   double input;

   printf("Enter 8 doubles: ");
   for (i = 0; i < SIZE; i++){
     scanf("%lf", &input);
     nums[i] = input;// or scanf("%lf", &num[i]); 
   }

   for (i = 0; i < SIZE; i++){
     printf("%lf ", nums[i]);
   }

   return 0;
 }

answered Oct 18, 2020 at 7:39

Farhad Sarvari

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1 Comment

Farhad Sarvari Over a year ago

@Jabou Please let me know if this answer works for you.

Juned Khan · Accepted Answer · 2020-10-18 07:32:11Z

0

scanf return number of elements scanned.

i think you meant:

for (i = 0; i < SIZE; i++){
    scanf("%lf", &nums[i]);
}

answered Oct 18, 2020 at 7:32

Juned Khan

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