GATE Overflow Test Series | Programming | Test 1 | Question: 20

Question

What will be the output of the following C program?

#include <stdio.h>
int main()
{
    int a[10][20][30]={0};
    printf("%d %d %d",&a+1 - &a, 
       a[10][20] - a[10][10], a[10] - a[5]);
    return 0;
}

• A. 1 300 100
• B. 600 300 100
• C. 1 100 5
• D. None of the above

Comments

%d will not be their, it will result in some error, it will be %ld(long int)

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Sir i get the method but i am confuse why a[10][20] not giving index out of bound error?Index are from [0 to 9][0 to 19]

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out of bound error will only happen when you try to de-reference the pointers you don’t have access to. Here if we do *a[10][20]=5, then segmentation fault will appear.
Very nice doubt.

Answer 1

    #include <stdio.h>
    int main()
    {
        int a[10][20][30]={0};
        printf("%d %d %d",&a+1 - &a, 
           a[10][20] - a[10][10], a[10] - a[5]);
        return 0;
    }

Output : 1,300,100

Let BA=1000 & Size of each element=1 and not 2 for simplicity.

$\&a+1-\&a$

$1000+1*(sizeof(a))-1000$

$1000+1*6000-1000$

$7000-1000$

$6000$

Now divide by 3-d array size

$\frac{6000}{6000}=1$

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$a[10[20]-a[10][10]$

$*(*(a+10)+20)-*(*(a+10)+10)$

$(1000+10*20*30+20*30)-(1000+10*20*30+10*30)$

$7600-7300$

$300$

Now divide by element size

$\frac{300}{1}=300$

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$a[10]-a[5]$

$*(a+10)-*(a+5)$

$(1000+10*20*30)-(1000+5*20*30)$

$7000-4000$

$3000$

Now divide by row size

$\frac{3000}{30}=100$

Answer 2

When $p_1$ and $p_2$ are two pointers pointing to same type, $p_1-p_2 = \dfrac{\text{integer value of }p_1 - \text{integer value of }p_2}{\text{sizeof}(*p_1)}$

&a+1 - &a  

• p1 = &a+1 = address of a + 1*sizeof(a).
• p2 = address of a
• So, p1 - p2 = 1*sizeof(a) / sizeof(a) = 1.

a[10][20] - a[10][10]

• p1 = a[10][20], point to int[30]
• *p1 = int  
• p2 = a[10][10]
• So, p1 - p2 = ([a + 10*20*30*sizeof(int) ]- [a + 9*20*30*sizeof(int) + 10 * 30 * sizeof(int)  ]) / (sizeof(int)) = 600-300 = 300

a[10] - a[5])

• p1 = a[10]
• *p1 = int[30] (p1 is 2D array int[20][30])
• p2 = a[5]
• So, p1 - p2 = ([a + 10*20*30*sizeof(int) ]- [a + 5*20*30*sizeof(int) ]) / (30 * sizeof(int)) = 100

Comments

I understand that out of bound error will occur only for de-referencing inaccessible pointers. But when we calculate a[10][20] – a[10][10] , the following formula’s denominator will need access to *(a[10][20]) which should give segmentation fault but it is still running currently. How?
integer value of p1−integer value of p2 / sizeof(∗p1)

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To know the size of *p, there is no need to get the content of memory address pointed to by p. Its known straight away from the declaration of p.

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Why are &a and a the same?

Shouldn’t &a indicate the address where the array’s address is stored?