Question 1 |

What is the output of the following program?

Consider the starting address of int array a is 1000 and b is 2000.

Assume size of int data type is 4 Byte.

Consider the starting address of int array a is 1000 and b is 2000.

Assume size of int data type is 4 Byte.

```
#include < stdio.h >
int main()
{
int a[10], b[3][5];
printf(" %d %d ", a, b) ;
printf("%d %d", a+2,b+2) ;
return 0;
}
```

1000 2000 1002 2002 | |

1000 2000 1008 2008 | |

1000 2000 1008 2040 | |

1000 2000 1008 2024 |

Question 1 Explanation:

Increment of 'a' is by 4-bytes, sizeof(int),
but the increment of 'b' is by 20-bytes. The
question is why?

The answer lies in the row-major memory space allocation of 1-D and 2-D array by the C compiler.

b is the starting address of the 0th row.

b+1 is the starting address of the 1st row.

b+i is the starting address of the ith row.

The size of a row is c * sizeof(int) = 5 * sizeof(int) = 5 * 4 = 20 bytes

where c is the number of columns

'b' is a pointer constant of type int [][5], a row of five int. If such a pointer is incremented, it goes up by 5 * sizeof(int) (number of bytes).

Type int [][5] is equivalent to int (*)[5]

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The answer lies in the row-major memory space allocation of 1-D and 2-D array by the C compiler.

b is the starting address of the 0th row.

b+1 is the starting address of the 1st row.

b+i is the starting address of the ith row.

The size of a row is c * sizeof(int) = 5 * sizeof(int) = 5 * 4 = 20 bytes

where c is the number of columns

'b' is a pointer constant of type int [][5], a row of five int. If such a pointer is incremented, it goes up by 5 * sizeof(int) (number of bytes).

Type int [][5] is equivalent to int (*)[5]

**OUTPUT : 1000 2000 1008 2040**Click to Join Our Telegram Group for Latest Update of MOCK TEST

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Question 2 |

What is the output of the following program?

Consider the starting address of int array b is 1000.

Assume size of int data type is 4 Byte.

Consider the starting address of int array b is 1000.

Assume size of int data type is 4 Byte.

```
#include < stdio.h >
int main()
{
int b[3][5]={{0,1,2,3,4},
{5,6,7,8,9},
{10,11,12,13,14}
}; ;
printf("%d %d ", b, *b) ;
printf("%d %d ", b+1, *b+1) ;
printf("%d %d ",b+2, *(b+2), *(b+2)+3) ;
return 0;
}
```

1000 0 1004 5 1008 10 13 | |

1000 0 1020 5 1040 1013 | |

1000 100 1004 1004 1008 1008 1012 | |

1000 1000 1020 1004 1040 1040 1052 |

Question 2 Explanation:

If b is the address of the 0th row, *b is the 0th row itself. A row may be viewed as an 1-D array, so *b is the address of the 0th element of the 0th row.

Similarly b+i is the address of the ith row, *(b+i) is the ith row, so *(b+i) is the address of the 0th element of the ith row.

If *b is the address of the 0th element of the 0th row, *b + 1 is the address of the 1st element of the 0th row.

Similarly *b + j is the address of the jth element of the 0th row.

The difference between b + 1 and b is 20 (bytes) but the difference between *b + 1 and *b is the sizeof(int), 4 (bytes).

If *(b+i) is the address of the 0th element of the ith row, *(b+i) + 1 is the address of the 1st element of the ith row.

Similarly *(b+i) + j is the address of the jth element of the ith row.

The difference between b + i and b is 20i (bytes), but the difference between *(b + i) + j and *(b+i) is 4j (bytes).

**OUTPUT : 1000 1000 1020 1004 1040 1040 1052**

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Question 3 |

What is the output of the following program?

Consider the starting address of int array b is 1000.

Assume size of int data type is 4 Byte.

Consider the starting address of int array b is 1000.

Assume size of int data type is 4 Byte.

```
#include < stdio.h >
int main()
{
int b[3][5] = {{0,1,2,3,4},
{5,6,7,8,9},
{10,11,12,13,14}
};
printf("%d ", b[2][3]);
printf("%d ", (*(b+2))[3]);
printf("%d ", *(b[2]+3));
printf("%d ", *(b+2)+3);
printf("%d ", b[2]+3);
return 0;
}
```

13 13 13 13 13 | |

13 13 13 1052 1012 | |

13 1052 13 1052 1052 | |

13 13 13 1052 1052 |

Question 3 Explanation:

We know that

b is the address of the 0th row,

b+i is the address of the ith row,

*(b+i) is the address of the 0th element of the ith row,

*(b+i)+j is the address of the jth element of the ith row,

We know that

*(b+i)+j is the address of the jth element of the ith row,

b[i][j] is the jth element of the ith row,

&b[i][j] is the address of the jth element of the ith row, so

*(b + i) + j is equivalent to &b[i][j]

We know that *(b+i)+j is the address of the jth element of the ith row, so

*(*(b + i) + j) is equivalent to b[i][j]

*(*(b + i) + j) is equivalent to b[i][j]

*(b + i) + j is equivalent to &b[i][j]

*(b[i] + j) is equivalent to b[i][j]

b[i] + j is equivalent to &b[i][j]

(*(b+i))[j] is equivalent to b[i][j]

b[2][3] is int value stored at row-2 and col-3 = 13

(*(b+2))[3] is equivalent to *(*(b+2)+3) is equivalent to b[2][3]=13

*(b[2]+3) is equivalent to *(*(b+2)+3) is equivalent to b[2][3]=13

For *(b+2)+3,

*(b+2) is the address of the 0th element of the 2nd row =1000+2*5*4=1040 and *(b+2)+3 is address of 3rd element of tht row =1040+3*4=1052.

b[2]+3 is equivalent to *(b+2)+3 = 1052

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b is the address of the 0th row,

b+i is the address of the ith row,

*(b+i) is the address of the 0th element of the ith row,

*(b+i)+j is the address of the jth element of the ith row,

We know that

*(b+i)+j is the address of the jth element of the ith row,

b[i][j] is the jth element of the ith row,

&b[i][j] is the address of the jth element of the ith row, so

*(b + i) + j is equivalent to &b[i][j]

We know that *(b+i)+j is the address of the jth element of the ith row, so

*(*(b + i) + j) is equivalent to b[i][j]

*(*(b + i) + j) is equivalent to b[i][j]

*(b + i) + j is equivalent to &b[i][j]

*(b[i] + j) is equivalent to b[i][j]

b[i] + j is equivalent to &b[i][j]

(*(b+i))[j] is equivalent to b[i][j]

b[2][3] is int value stored at row-2 and col-3 = 13

(*(b+2))[3] is equivalent to *(*(b+2)+3) is equivalent to b[2][3]=13

*(b[2]+3) is equivalent to *(*(b+2)+3) is equivalent to b[2][3]=13

For *(b+2)+3,

*(b+2) is the address of the 0th element of the 2nd row =1000+2*5*4=1040 and *(b+2)+3 is address of 3rd element of tht row =1040+3*4=1052.

b[2]+3 is equivalent to *(b+2)+3 = 1052

**OUTPUT : 13 13 13 1052 1052**Click to Join Our Telegram Group for Latest Update of MOCK TEST

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Question 4 |

What is the output of the following program?

Consider the starting address of int array b is 1000.

Assume size of int data type is 4 Byte.

Consider the starting address of int array b is 1000.

Assume size of int data type is 4 Byte.

```
#include < stdio.h >
int main()
{
int b[3][5] = {{0,1,2,3,4},
{5,6,7,8,9},
{10,11,12,13,14}
};
printf("%d ", *b+1);
printf("%d ", *(b+1)+1);
printf("%d ", *(*(b+1)+1));
return 0;
}
```

1020 1024 1024 | |

1004 1008 1008 | |

1024 1024 6 | |

1004 1024 6 |

Question 4 Explanation:

*b is the address of the 0th
element of the 0th row is 1000. So, *b + 1 is the address of the 1st
element of the 0th row is 1004

b is the starting address of the 0th row (=1000),

b+1 is the starting address of the 1st row (1020),

*(b+1) is the starting address of the 0th element of the 1st row (1020),

*(b+1)+1 is the address of the 1st element of the 1st row=1024,

*(*(b+1)+1) is equivalent to b[1][1]= 6

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b is the starting address of the 0th row (=1000),

b+1 is the starting address of the 1st row (1020),

*(b+1) is the starting address of the 0th element of the 1st row (1020),

*(b+1)+1 is the address of the 1st element of the 1st row=1024,

*(*(b+1)+1) is equivalent to b[1][1]= 6

**OUTPUT :**1004 1024 6Click to Join Our Telegram Group for Latest Update of MOCK TEST

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Question 5 |

What is the output of the following program?

Consider the starting address of int array b is 1000.

Assume size of int data type is 4 Byte.

Consider the starting address of int array b is 1000.

Assume size of int data type is 4 Byte.

```
#include < stdio.h >
int main()
{
int b[3][5] = {{0,1,2,3,4},
{5,6,7,8,9},
{10,11,12,13,14}
};
printf("%d ", *(b+2)-*(b+1));
printf("%d ", (b+2)-(b+1));
return 0;
}
```

1 1 | |

20 20 | |

5 1 | |

20 4 |

Question 5 Explanation:

For *(b+2)-*(b+1)

b is the starting address of the 0th row (=1000),

b+1 is the starting address of the 1st row (1020),

*(b+1) is the starting address of the 0th element of the 1st row (1020),

b+2 is the starting address of the 2nd row (1040),

*(b+2) is the starting address of the 0th element of the 2nd row (1040),

Pointer arithmetic is computed as (1040-1020)/sizeof(int)=20/4=5

**NOTE:***(b+1) and *(b+2) is pointer to int element. Hence, we have divided the pointer subtraction with sizeof(int).

For (b+2)-(b+1)

b is the starting address of the 0th row (=1000),

b+1 is the starting address of the 1st row (1020),

b+2 is the starting address of the 2nd row (1040),

Pointer arithmetic is computed as (1040-1020)/(5*sizeof(int))=20/20=1

**NOTE:**b is a pointer constant of type int [][5], a row of five int. If such a pointer is incremented, it goes up by 5* sizeof(int) (number of bytes). Type int [][5] is equivalent to int (*)[5].

**OUTPUT :**5 1

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There are 5 questions to complete.