~ no response on stdout ~ in C [closed]

Answer 1

Your variables are uninitialized. As a result your program invokes Undefined Behavior.

For example you do not initialize n, but you then declare int b[n]. What is the size of array b? Nobody really knows, since n has a garbage value.

Start by figuring out what the values of your variables should be, and then start coding.

---

Array indexing starts from 0, thus your for loops are not looking good.

Change this:

for(a=1;a<=query;a++)

to this:

for (a = 0; a < query; a++)

Answer 2

Your problem is in this line:

int a,query,in,n,b[n],sum[a];

What is the meaning of b[n]? And of sum[a]? Assuming that you would like to have a language with magically growing arrays, C is a bad choice.

In C language arrays are a structure with size known at compile time. Anyway, let's assume that your compiler supports the horrible hack of variable-length arrays. You can do this:

#include <stdlib.h>
#include <stdio.h>
#include <math.h>

int main(void) {
    int a, query, in , n;
    scanf("%d", &query);
    int sum[query+1];

    for (a = 1; a <= query; a++) {
        sum[a] = 0;
        scanf("%d", &in);
        int b[in+1];
        for (n = 1; n <= in ; n++) {
            b[n] = 1 + 7 * (n - 1) + 6 * (n - 1) * (n - 2) + (n - 1) * (n - 2) * (n - 3);
            sum[a] = sum[a] + b[n];
        }
    }

    for (a = 1; a <= query; a++) {
        printf("%d\n", sum[a]);
    }

    return 0;
}

Note the changes: first you need to know the size of the array, then you can allocate it. Moreover, I increased the size of the array in order to support your choice of starting from 1. Lastly I also zeroed the initial value for sum array.

Answer 3

int main()
{
    int a, query, in, n, *b, *sum;
    scanf("%d",&query);

    sum = malloc(query * sizeof(int));

    /* do some checks if the malloc was successful */ 

    for(a = 0; a < query; a++)
    {
        scanf("%d",&in) ;    /* you should check if scan has returned 1 */
        b = malloc(in * sizeof(int)); /* and again check against any allocation errors */

        for(n = 0; n < in; n++)
        {
           b[n] = 1+7*(n)+6*(n)*(n-1)+(n)*(n-1)*(n-2);
           sum[a] = sum[a] + b[n];
        }
        free(b);
    }

    /* the rest */

Answer 4

In int a,query,in,n,b[n],sum[a];, the value of a is not initialised and is having garbage value which could be anything. This value is used as the size of the variable length array sum. So the size of the array could be anything which is probably not what you want.

a could be 0 in which case sum is an array of size 0 which will in turn make sum[1] incorrect (it would be undefined behavior).

The same applies to n and b[n].

In the nested for loop, with sum[a]=sum[a]+b[n]; you are using sum[a] which has not been initialised. It has garbage value and the result is indeterminate.

If you wanted all elements of sum to be initialised with 0, you can do

int sum[20]={0};

at the time of declaring it.

The same goes for b[n]=1+7*(n-1)+6*(n-1)*(n-2)+(n-1)*(n-2)*(n-3); as well.

And array indexing starts at 0. Perhaps you could use the loops like

for (a = 0; a < query; a++) {

instead of

for (a = 1; a <= query; a++) 

If you choose the starting index as 0, the inner nested loop should be like

for(n=0;n<in;n++)
{
    //b[n]=1+7*(n-1)+6*(n-1)*(n-2)+(n-1)*(n-2)*(n-3);
    b[n]=1+7*(n)+6*(n)*(n-1)+(n)*(n-1)*(n-2);
    sum[a]=sum[a]+b[n];
}

See demo.