Generate sequence using array?

Question

I'm trying to get a list of integers from 2 to N, however I am getting a weird output.

Code:

#define N 10

int main(int argc, char *argv[]) {

    int array[N]={2-N};
    for(int i=0;i<N;printf("%d ",array[i]), i++);
}

Output:

-8 0 0 0 0 0 0 0 0 0 

I'm trying to get it to read: 2,3,4,5,6,7,8,9,10

Any ideas? Thanks in advance

Answer 1

This statement

int array[N]={2-N};

Sets the first element of the array to 2-N, and as N is 8, that will set it to -8. The other elements are undefined, but happen to be zero.

You want something like

for (int i = 0; i < N; i++)
    array[i] = i+2;

Answer 2

N is 10, 2-N is -8. That is why the first array element is -8.

The other array elements are 0 because they were not explicitly initialized, so are initialized by default to 0.

You could initialize it as

int array[N]={2, 3, 4, 5, 6, 7, 8, 9, 10};

https://publib.boulder.ibm.com/infocenter/lnxpcomp/v8v101/index.jsp?topic=%2Fcom.ibm.xlcpp8l.doc%2Flanguage%2Fref%2Faryin.htm

Answer 3

Put that printf statement inside the loop itself, properly.

Answer 4

int array[N]={2-N};  

will initialize first element of array, i.e, array[0] to -8 and rest of them initialized to 0 implicitly (take a look here). So, you are getting right output.
To initialize your array from 2 to N, you need to use a for loop:

for (int i = 0; i < N; i++)
    array[i] = i+2;