I am running this program
#include <stdio.h>
int main ()
{
int n;
int a = 1;
int b = 2;
int product;
int i;
printf("How many numbers of the sequence would you like \n");
scanf("%d",&n);
for (i=0;i<n;i++)
{
printf("%d\n",a);
product = a * b;
a = b;
b = product;
}
return 0;
}
When I enter n = 3, the result is 1 2 2
Why is it ? I meant to make it so it show 1 2 4 , what have I done wrong ? And why is it print 1 2 2 .
And why is it print 1 2 2 .
Step-by-step trace at printf("%d\n",a);:
i a b product
0 1 2 ?
1 2 2 2
2 2 4 4
3 4 8 8
4 8 32 32
I meant to make it so it show 1 2 4
Then interchange the below two lines
a = b;
b = product;
It should be
for (i=0;i<n;i++)
{
printf("%d\n",a);
product = a * b;
b = product;
a = b;
}
Your code does exactly what it's supposed to do.
Maybe you wanted :
#include <stdio.h>
int main ()
{
int n;
int a = 1;
int b = 2;
int product;
int i;
printf("How many numbers of the sequence would you like \n");
scanf("%d",&n);
printf("%d\n",a);
for (i=1;i<n;i++)
{
product = a * b;
a = b;
b = product;
printf("%d\n",b);
}
return 0;
}
1 2 2 because you print a not product and value of a in each iteration is -
1st iteration a=1
2nd iteration a=2 // a=b and b=2 i.e a=2
3rd iteration a=2 // why ? because again a=b and b did not change
thus output 1 2 2
Simple solution would be -
1.Initialize product to 1 // int product =1;
2.print product instead of printf("%d\n",a); // printf("%d\n",product);
