# C Programming - Arrays - Discussion

Discussion Forum : Arrays - Find Output of Program (Q.No. 6)

6.

What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?

```
#include<stdio.h>
int main()
{
int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};
printf("%u, %u\n", a+1, &a+1);
return 0;
}
```

Answer: Option

Explanation:

**Step 1**: *int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};* The array a[3][4] is declared as an integer array having the 3 rows and 4 colums dimensions.

**Step 2**: *printf("%u, %u\n", a+1, &a+1);*

The base address(also the address of the first element) of array is 65472.

For a two-dimensional array like *a* reference to array has type "pointer to array of 4 ints". Therefore, *a+1* is pointing to the memory location of first element of the second row in array *a*. Hence 65472 + (4 ints * 2 bytes) = 65480

Then, *&a* has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, *&a+1* denotes "12 ints * 2 bytes * 1 = 24 bytes".

Hence, begining address 65472 + 24 = 65496. So, *&a+1* = 65496

Hence the output of the program is 65480, 65496

Discussion:

29 comments Page 1 of 3.
Dharmesh said:
2 years ago

For second part of the question &a+1 means (&a) +1 as precedence of & is higher than the precedence of the arithmetic operator, and &a means base address of the array.

After 1st part pointer is pointing to 2nd array's 1st element. In order to reach the base address ie. Address of 1st array 1 element we need to add a size of 8*2=16 bytes to the address of (a+1) address.

65480+16=65496.

Now we have reached till base address &a now in order to reach (&a) +1 we need to add base address+ 4*2 more i.e. 65496+8= 65104.

The difference between 1st part and 2nd part will be 32 bits.

Address of (&a) +1 - a = 32 bits.

Thanks!

After 1st part pointer is pointing to 2nd array's 1st element. In order to reach the base address ie. Address of 1st array 1 element we need to add a size of 8*2=16 bytes to the address of (a+1) address.

65480+16=65496.

Now we have reached till base address &a now in order to reach (&a) +1 we need to add base address+ 4*2 more i.e. 65496+8= 65104.

The difference between 1st part and 2nd part will be 32 bits.

Address of (&a) +1 - a = 32 bits.

Thanks!

(1)

Dipak said:
2 years ago

In this, the base address is 64472.

&a means the initial address of array i.e. 64472.

In this array, 12 elements & each contain 2 bytes so 12 * 2 = 24.

so &a+1= 64472 + 24 = 64496.

means it shift the address after the a[3][4].

&a means the initial address of array i.e. 64472.

In this array, 12 elements & each contain 2 bytes so 12 * 2 = 24.

so &a+1= 64472 + 24 = 64496.

means it shift the address after the a[3][4].

Omkar said:
2 years ago

@Sairam.

As you said a+1 means a[1][0] and a+2 means a[2][0].

So, if a[0][1] we have to access then what we have to add in a

is a+4 means a[0][1].

As you said a+1 means a[1][0] and a+2 means a[2][0].

So, if a[0][1] we have to access then what we have to add in a

is a+4 means a[0][1].

Tejas B said:
3 years ago

Thanks for explaining @Sairam.

SAIRAM said:
3 years ago

@All.

Here, I will explain it clearly,

#include<stdio.h>

int main()

{

int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};

printf("%u, %u\n", a+1, &a+1);

return 0;

}

a [0] [1] [2] [3]

[0] 1 2 3 4 --> a[0][0] = 1 a[0][1] = 2 a[0][2] = 3 a[0][3] = 4

[1] 4 3 2 1 --> a[1][0] = 4 a[1][1] = 3 a[1][2] = 2 a[1][3] = 1

[2] 7 8 9 0 --> a[2][0] = 7 a[2][1] = 8 a[2][2] = 9 a[2][3] = 0

In general,

'a' contains base address which is the address of the first row first element in the array i.e., a[0][0],

&a is also refers to same base address a[0][0]

a+1 points to the address of a[1][0], which is the second-row first element.

Similarly a+2 points to the address of a[2][0], which is the third-row first element.

Each element or integer occupies 2bytes.

a[3][4] means we have 3 rows and 4 columns. I each row we have 4 elements.

Question:

What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?

this means address of a[0][0] = 1 is 65472

this means address of a[0][1] = 2 is 65474

this means address of a[0][2] = 3 is 65476

this means address of a[0][3] = 4 is 65478

this means address of a[1][0] = 4 is 65480.

a+1 points the address of a[1][0], which is second row first element. so answer is 65480

&a represents the whole array but stores only the base address which is the starting element array a[0][0].

&a+1 points fourth row first element which means a[3][0]

address of a[0][0] is 65472 ------ 1st row 1st element

address of a[0][1] is 65474

address of a[0][2] is 65476

address of a[0][3] is 65478

address of a[1][0] is 65480 ------ 2nd row 1st element

address of a[1][1] is 65482

address of a[1][2] is 65484

address of a[1][3] is 65486

address of a[2][0] is 65488 ------ 3rd row 1st element

address of a[2][1] is 65490

address of a[2][2] is 65492

address of a[2][3] is 65494

address of a[3][0] is 65496 ------ 4th row 1st element.

Totally we have to cross 12 elements to reach &a+1 i.e., a[3][0].

Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes".

Hence, begining address of 4th row first element is 65472 + 24 = 65496. So, &a+1 = 65496 which is also address of a[3][0].

Here, I will explain it clearly,

#include<stdio.h>

int main()

{

int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0};

printf("%u, %u\n", a+1, &a+1);

return 0;

}

a [0] [1] [2] [3]

[0] 1 2 3 4 --> a[0][0] = 1 a[0][1] = 2 a[0][2] = 3 a[0][3] = 4

[1] 4 3 2 1 --> a[1][0] = 4 a[1][1] = 3 a[1][2] = 2 a[1][3] = 1

[2] 7 8 9 0 --> a[2][0] = 7 a[2][1] = 8 a[2][2] = 9 a[2][3] = 0

In general,

'a' contains base address which is the address of the first row first element in the array i.e., a[0][0],

&a is also refers to same base address a[0][0]

a+1 points to the address of a[1][0], which is the second-row first element.

Similarly a+2 points to the address of a[2][0], which is the third-row first element.

Each element or integer occupies 2bytes.

a[3][4] means we have 3 rows and 4 columns. I each row we have 4 elements.

Question:

What will be the output of the program if the array begins at 65472 and each integer occupies 2 bytes?

this means address of a[0][0] = 1 is 65472

this means address of a[0][1] = 2 is 65474

this means address of a[0][2] = 3 is 65476

this means address of a[0][3] = 4 is 65478

this means address of a[1][0] = 4 is 65480.

a+1 points the address of a[1][0], which is second row first element. so answer is 65480

&a represents the whole array but stores only the base address which is the starting element array a[0][0].

&a+1 points fourth row first element which means a[3][0]

address of a[0][0] is 65472 ------ 1st row 1st element

address of a[0][1] is 65474

address of a[0][2] is 65476

address of a[0][3] is 65478

address of a[1][0] is 65480 ------ 2nd row 1st element

address of a[1][1] is 65482

address of a[1][2] is 65484

address of a[1][3] is 65486

address of a[2][0] is 65488 ------ 3rd row 1st element

address of a[2][1] is 65490

address of a[2][2] is 65492

address of a[2][3] is 65494

address of a[3][0] is 65496 ------ 4th row 1st element.

Totally we have to cross 12 elements to reach &a+1 i.e., a[3][0].

Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes".

Hence, begining address of 4th row first element is 65472 + 24 = 65496. So, &a+1 = 65496 which is also address of a[3][0].

(8)

Saha said:
3 years ago

@Neha.

How 65496 came? Explain please.

How 65496 came? Explain please.

(1)

Ravi said:
4 years ago

&a should be a pointer to an array of 3 pointers points to arrays of 4 ints, Am I correct?

Mohan said:
5 years ago

a+1 means 1th one-dimensional array of a.

&a+1 means next 2D array like a.

&a+1 means next 2D array like a.

Mohan said:
5 years ago

a+1 means 1th one-dimensional array of a.

&a+1 means next 2D array like a.

&a+1 means next 2D array like a.

Abhishek said:
6 years ago

Thanks for explaining @Neha.

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