Consider the following C-code fragment running on a $32$-bit x86 machine:

typedef struct {
union {
unsigned char a;
unsigned short b;
} U;
unsigned char c;
} S;

S B[10];
S *p = address of B[4];
S *q = address of B[5];

p->U.b = 0x1234;

If $M$ is the value of $q-p$ and $N$ is the value of address of $(p->c)$ minus address of $p$, then $(M,N)$ is:

Given that structure $S$ takes $32$ bits.

So,

$32\text{ bits}=4\text{ bytes}$

Now,

$p$ points to $B[4]$

$q$ points to $B[5]$

Since $p$ and $q$ are pointers of type $S$, pointer subtraction gives the difference in number of structures.

So,

$q-p=1$

Therefore,

$M=1$

Now, union $U$ contains $unsigned\ char\ a$ and $unsigned\ short\ b$.

Size of union $U$ is $2$ bytes.

After union $U$, variable $c$ is stored.

So, offset of $c$ from the beginning of structure is $2$ bytes.

Therefore,

$N=2$

Hence,

$(M,N)=(1,2)$