What will happen when the following code is executed?
void main()
{
    char arr1[] = "REGALINT";
    char arr2[10] = "REGALINT";
    printf("%d,",sizeof(arr1));
    printf("%d",sizeof(arr2));
}

What will be printed on the standard output as a result of the following code snippet? void main() { int arr[5]={1,2,3,4,5}; printf('%d\n', *(arr+4)); }

Given the following array: int a[8] = {1,2,3,4,5,6,7,0}; what would be the output of printf('%d',a[4]); ?

What will happen when the following code is executed? { int num; num =0; do {-- num; printf('%d\n', num); num++; } while (num >=0); }

Which is/are the correct signature/s of main with command line arguments?

What would be printed on the standard output as a result of the following code snippet? int Recur(int num) { if(num==1 || num==0) return 1; if(num%2==0) return Recur(num/2)+2; else return Recur(num-1)+3; } int main() { int a=9; printf('%d\n', Recur(a)); return 0; }

What would be printed on the standard output as a result of the following code snippet? char i = 'A'; char *j; j = & i; *j = *j + 32; printf('%c',i);

Which one is not a type of operator ?

Which statements are correct for the keyword register?

An array is defined with the following statement in a file, file1.c int a[ ] = { 1, 2, 3, 4, 5, 6 }; In another file, file2.c, the following code snippet is written to use the array a: extern int a[]; int size = sizeof(a); What is wrong with the above code snippet?

What would be printed on the standard output as a result of the following code snippet? int Recur(int num) { if(num==1 || num==0) return 1; if(num%2==0) return Recur(num/2)+2; else return Recur(num-1)+3; } int main() { int a=9; printf('%d\n', Recur(a)); return 0; }