C Programming

Consider the following ANSI C program

 #include < stdio.h >
int foo(int x, int y, int q) 
    {
        if ((x < = 0) && (y < = 0))
        return q;
        if (x < = 0)
        return foo(x, y-q, q);
        if (y < = 0)
        return foo(x-q, y, q);
        return foo(x, y-q, q) + foo(x-q, y, q);
    }
int main( )
{
    int r = foo(15, 15, 10);
    printf("%d", r);
    return 0;
}

The output of the program upon execution is ____

Consider the following ANSI C code segment:

 z=x + 3 + y-> f1 + y-> f2;
for (i = 0; i < 200; i = i + 2) 
{
    if (z > i) 
    {
        p = p + x + 3;
        q = q + y-> f1;
    } else 
    {
        p = p + y-> f2;
        q = q + x + 3;
    }
}

Assume that the variable y points to a struct (allocated on the heap) containing two fields f1 and f2, and the local variables x, y, z, p, q, and i are allotted registers. Common sub-expression elimination (CSE) optimization is applied on the code. The number of addition and the dereference operations (of the form y -> f1 or y -> f2) in the optimized code, respectively, are:

Consider the following ANSI C function:

int SomeFunction (int x, int y)
{
    if ((x==1) || (y==1)) return 1;
    if (x==y) return x;
    if (x > y) return SomeFunction(x-y, y);
    if (y > x) return SomeFunction (x, y-x);
 
} 

The value returned by SomeFunction(15, 255) is __________

Consider the following ANSI C program.

 #include < stdio.h >
int main( ) 
{
    int arr[4][5];
    int  i, j;
    for (i=0; i < 4; i++) 
  {
        for (j=0; j < 5; j++) 
        {
            arr[i][j] = 10 * i + j;
        }
    }
    printf("%d", *(arr[1]+9));
    return 0;
}

What is the output of the above program?

Consider the following ANSI C function:

 int SimpleFunction(int Y[], int n, int x)
{
int total = Y[0], loopIndex;
for (loopIndex=1; loopIndex<=n-1; loopIndex++)
    total=x*total +Y[loopIndex];
return total;
}

Let Z be an array of 10 elements with Z[i]=1, for all i such that 0\leq i \leq 9. The value returned by SimpleFunction(Z,10,2) is __________

Consider the following ANSI C program.

 #include < stdio.h >
int main()
{
    int i, j, count;
    count=0;
    i=0;
    for (j=-3; j < =3; j++)
    {
        if (( j > = 0) && (i++))
        count = count + j;
    }
    count = count +i;
    printf("%d", count);
    return 0;
}

Which one of the following options is correct?

In the following procedure

 Integer procedure P(X,Y);
Integer X,Y;
value x;
begin
      K=5;
      L=8;
      P=x+y;
end

X is called by value and Y is called by name. If the procedure were invoked by the following program fragment

 K=0;
L=0;
Z=P(K,L);

then the value of Z will be set equal to

Following declaration of an array of struct, assumes size of byte, short, int and long are 1,2,3 and 4 respectively. Alignment rule stipulates that n byte field must be located at an address divisible by n, the fields in the struct are not rearranged, padding is used to ensure alignment. All elements of array should be of same size.

 Struct complx
       Short s
       Byte b
       Long l
       Int i
End Complx
Complx C[10]

Assuming C is located at an address divisble by 8, what is the total size of C, in bytes?

What is the output in a 32 bit machine with 32 bit compiler?

 #include < stdio.h >
rer(int **ptr2, int **ptr1)
{
    int *ii;
    ii=*ptr2;
    *ptr2=*ptr1;
    *ptr1=ii;
    **ptr1*=**ptr2;
    **ptr2+=**ptr1;
}
void main(){
    int var1=5, var2=10;
    int *ptr1=&var1,*ptr2=&var2;
    rer(&ptr1,&ptr2);
    printf("%d %d",var2,var1);
}

What is output of the following 'C' code assuming it runs on a byte addressed little endian machine?

 #include < stdio.h >
int main()
{
    int x;
    char *ptr;
    x=622,100,101;
    printf("%d",(*(char *)&x)*(x%3));
    return 0;
}