GATE IT 2008 | Question: 52

Question

C program is given below:

# include <stdio.h>
int main ()
{
        int i, j;
        char a [2] [3] = {{'a', 'b', 'c'}, {'d', 'e', 'f'}};
        char b [3] [2];
        char *p = *b;
        for (i = 0; i < 2; i++) {
              for (j = 0; j < 3; j++) {
              *(p + 2*j + i) = a [i] [j];
              }
        }
}

What should be the contents of the array b at the end of the program?

• A. $\text{a    b}$
  $\text{c    d}$
  $\text{e    f}$
   
• B. $\text{a    d}$
  $\text{b    e}$
  $\text{c    f}$
   
• C. $\text{a    c}$
  $\text{e    b}$
  $\text{d    f}$
   
• D. $\text{a    e}$
  $\text{d    c}$
  $\text{b    f}$

Comments

Yes you're wrong. We have a 2D array here not a 3D array. There's no "0th 2D array" :P

---

Sorry, I mean base address of 2D array, only.

Edit:- Got it. Thanks.

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Approach:

*b points to individual elements of the 1d array.

*(p+0+0) = b[0][0] = a

*(p+2+0) = b[1][0] = b

*(p+4+0) = b[2][0] = c

Answer 1

The correct answer is option (B).

first integer type two variables declared $i$ and $j$

then an integer type $2-d$ array $a[2][3]$ is declared and initialized and $2-d$ array $b[3][2]$ is created but not initialized. i.e 

            address     value                       address         value

$a[0]0[]$   $2000$         $a$               $b[0][0]$   $3000$          garbage value

$a[0][1]$   $2001$         $b$               $b[0][1]$   $3001$          garbage value

$a[0][2]$   $2002$         $c$               $b[1][0]$   $3002$          garbage value

$a[1][0]$   $2003$         $d$               $b[1][1]$   $3003$          garbage value

$a[1][1]$   $2004$         $e$               $b[2][0]$   $3004$          garbage value

$a[1][2]$   $2005$         $f$                $b[2][1]$   $3005$          garbage value

now the char type pointer is declared and the base address of  array  $b$ is put in it. so $p=3000$

now the for loop is started where $i$ is initialized to $0$ ,so 

$i=0 : i<2$ (true)

      $j=0; j<3$ (true)

          $*(3000+2*0+0) =a [0][0]  \Rightarrow *(3000) = a$

            $j++$ 

         $j=1; j<3$ (true)

         $ *(3000+2*1+0) =a [0][1]  \Rightarrow *(3002) = b$

           $ j++$

         $j=2; j<3$ (true)

          $*(3000+2*2+0) =a [0][2]  \Rightarrow  *(3004) = c$

           $ j++$

         $j=3; j<3$ (false)

    $i++$

$i=1 : i<2$ (true)

       $j=0; j<3$ (true)

          $*(3000+2*0+1) =a [1][0]  \Rightarrow *(3001) = d$

            $j++$ 

         $j=1; j<3$ (true)

          $*(3000+2*1+1) =a [1][1]  \Rightarrow *(3003) = e$

            $j++$

         $j=2; j<3$ (true)

         $ *(3000+2*2+1) =a [1][2]  \Rightarrow  *(3005) = f$

            $j++$

         $j=3; j<3$ (false)

    $i++$

now the values in array $b$ is 

 $b[0][0]$   $3000$           $a$

 $b[0][1]$   $3001$           $d$

 $b[1][0]$   $3002$           $b$

 $b[1][1]$   $3003$           $e$

 $b[2][0]$   $3004$           $c$

 $b[2][1]$   $3005$           $f$

Hence, the output will be (B) choice.

Note:

*(p + 2*j + i)

$p +$ size of inner dimension $* j + i$, hence is same as $p[j][i]$. Hence with this statement we can identify that the code is transposing the matrix $a$ and storing in $b$ using pointer $p$. 

Comments

Thanks for the correction :)

Added a note at the end of your answer which can be used to solve these questions faster.

---

@jeevesh kataria
char *p = *b;
--- u can also break this one statement into two statements to reduce its complexity
1.  char *p;
 2. p=*b;
p is a character pointer and now it is holding base address of array b suppose base address of array is 3000 so we can represent b[0] as *(3000+0) or first character of b and initially b[0] contains a garbage value
so char *p = *b; is like p contains base address of a " character at address 3000" and initially it is a garbage value.
----------hope this will help .....

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Amazing.... ...............

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what is sizeofinner dimension please explain ?

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how p contains base address of b becz it is containing garbage initially becz b is 3000 nd *b means at 3000 ....so how?

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b is a 2D array, or a pointer to a 1D array. *b gives 1D array or its base address. **b has garbage value.

---

 

Can someone tell me what is mistake in this output 

# include <stdio.h>
int main ()
{
int i, j;
char a [2] [3] = {{'a', 'b', 'c'}, {'d', 'e', 'f'}};
char b [3] [2];
char *p = *b;
for (i = 0; i < 2; i++) {
for (j = 0; j < 3; j++) {
*(p + 2*j + i) = a [i] [j];
}
}

for (i = 0; i < 3; i++) {
for (j = 0; j < 2; j++) {
printf("%c\n",a [i][j]) ;
}
}
}

It gives me 

a

b

d

e.

you can check here https://www.onlinegdb.com/online_c_compiler

---

@mehul  Check it ...

Code 1 : 

# include <stdio.h>
int main ()
{
        int i, j;
        char a [2] [3] = {{'a', 'b', 'c'}, {'d', 'e', 'f'}};
        char b [3] [2];
        char *p = b;     // address initialization to p of b
        for (i = 0; i < 2; i++) 
        {
              for (j = 0; j < 3; j++) 
              {
              *(p + 2*j + i) = a [i] [j];
              }
        }
        for (i = 0; i < 3; i++) 
        {
              for (j = 0; j < 2; j++)
              {
                  printf("b[%d][%d] = %c \n",i,j,b[i][j]);
              }
        }
}

 

Code 2 :

int main ()
{
        int i, j;
        char a [2] [3] = {{'a', 'b', 'c'}, {'d', 'e', 'f'}};
        char b [3] [2];
        char *p = &b[0];  // address initialization of b to p
        for (i = 0; i < 2; i++) 
        {
              for (j = 0; j < 3; j++) 
              {
              *(p + 2*j + i) = a [i] [j];
              }
        }
        for (i = 0; i < 3; i++) 
        {
              for (j = 0; j < 2; j++)
              {
                  printf("b[%d][%d] = %c \n",i,j,b[i][j]);
              }
        }
}

 

Code 3 :

int main ()
{
        int i, j;
        char a [2] [3] = {{'a', 'b', 'c'}, {'d', 'e', 'f'}};
        char b [3] [2];
        char *p = &b[0][0];   //address initialization of b to p
        for (i = 0; i < 2; i++) 
        {
              for (j = 0; j < 3; j++) 
              {
              *(p + 2*j + i) = a [i] [j];
              }
        }
        for (i = 0; i < 3; i++) 
        {
              for (j = 0; j < 2; j++)
              {
                  printf("b[%d][%d] = %c \n",i,j,b[i][j]);
              }
        }
}

 

Code 4 : 

# include <stdio.h>
int main ()
{
        int i, j;
        char a [2] [3] = {{'a', 'b', 'c'}, {'d', 'e', 'f'}};
        char b [3] [2];
        char *p = &b;     // address initialization to p of b
        for (i = 0; i < 2; i++) 
        {
              for (j = 0; j < 3; j++) 
              {
              *(p + 2*j + i) = a [i] [j];
              }
        }
        for (i = 0; i < 3; i++) 
        {
              for (j = 0; j < 2; j++)
              {
                  printf("b[%d][%d] = %c \n",i,j,b[i][j]);
              }
        }
}

All of the code snippets gives Option B as output  ...
 

---

In case of 2-d array, both b & *b will have same value(address) but to get b[0][0], we've to use **b(here both b & *b will differ concept wise not value wise).

In case of pointer to array, b & *b will have different value(address) and to get b[0][0], **b will be used as above.

 

 

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But the address calculation in column major array is column wise. Here why are you calculating in row major order.

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The solution can be b) and d) both?why there is no ambiguity in solution?As d) is column major and b) is row major.Why the solution is only row major i.e. b)? My seniors pls explain.Any help will be appreciated.

---

This might help 

https://stackoverflow.com/questions/14201679/why-does-c-language-not-support-column-major-array

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@Kalpna Bhargav When We do 

p+1 in a 2d Array it points to the next ROW!

So, p + 2 should point to the second next row

So, p+2 should be 3004?

 

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@Bibhu Pala here p is a character pointer pointing to individual cells of b 2d array. Please note that *p=*b and not **p=b.

b points to individual 1D arrays.

*b points to individual elements of the 1d array.

**b is the data present in one of the cell of 1d array.

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*(p + 2*j + i)

p+ size of inner dimension ∗j+i, hence is same as p[j][i]. Hence with this statement we can identify that the code is transposing the matrix a and storing in b using pointer p. 

Can someone pls elaborate it more?

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Say a = 2;

a + 1 = 2+1 = 3,   if a is an int

a + 1 = 2 + 1 * sizeof(*a),   if a is a pointer

Answer 2

Both will behave similarly.

$char *p=b$

$char *p=*b$

---

But $char *p=b+1\ \&\ char*p=*b+1$ will behave differently.

Comments

But char∗p=b+1 & char∗p=∗b+1char∗p=b+1 & char∗p=∗b+1 will behave differently.

Wouldn’t first part will give you the error ? as ‘b’ is an array name so it holds the address of the first element of an array , which here is a 1-d array of size 2. So first option should be like this :

char (*p)[2] = b + 1 ; 

Edit : It works but gives a warning : initialization from incompatible pointer type