This question is based upon the figure shown below

The given program, which was supposed to give output as shown in the image, has some errors. Analyse the code and choose the correct line number(s) containing error(s).

1.int main( )

2.{ int i=1,j=1,k=1;

3. abc(i,j,k);

4. xyz(i,j,k);

5. return 0;

6.}

7.abc (int a, int b, int c)

8.{ for(;a<=5;a++)

9. { for(b=5;b>a;b--)

10.    { printf(" "); }

11.    for(c=1;c<=a;c++)

12.    { printf("* "); }

13.    print("\n");

14. } }

15.xyz(int *a, int *b, int *c)

16.{ for(a=4;a>=1,a--)

17. { for(b=4;b>=a;b--)

18.    { printf(" "); }

19.    for(c=a;c>=1;c++)

20.    { printf("* "); }

21.    printf("\n");

22. } }

a. 4

b. 9

c. 13

d. 14

e. 15

f. 16

g. 17

h. 19

Given the array:

int num[3][4]= {

                {3,6,9,12},

                {15,25,30,35},

                {66,77,88,99}

                };

what would be the output of *(*(num+1)+1)+1?

int a, b, number1, tmp_no;

printf("Enter any value between 2 and 15\n");

scanf("%d", &number1);

printf("Enter elements\n");

for (a = 0; a < number1; a++)

{

    scanf("%d", &arr1[a]);

}

for (a = 0; a < number1; a++)

{

    for (b = 0; b < (number1 – a – 1); b++)

    {

        if (arr1[b] > arr1[b + 1])

        {

            tmp_no = arr1[b];

            arr1[b] = arr1[b + 1];

            arr1[b + 1] = tmp_no;

        }

    }

}

for (a = 0; a < number1; a++)

{

    printf("%d\n", arr1[a]);

}

}

Which of the given functions will be performed by the following program?

#include <stdio.h>

void main()

{

int arr1[15];

int a, b, number1, tmp_no;

printf("Enter any value between 2 and 15\n");

scanf("%d", &number1);

printf("Enter elements\n");

for (a = 0; a < number1; a++)

{

    scanf("%d", &arr1[a]);

}

for (a = 0; a < number1; a++)

{

    for (b = 0; b < (number1 – a – 1); b++)

    {

        if (arr1[b] > arr1[b + 1])

        {

            tmp_no = arr1[b];

            arr1[b] = arr1[b + 1];

            arr1[b + 1] = tmp_no;

        }

    }

}

for (a = 0; a < number1; a++)

{

    printf("%d\n", arr1[a]);

}

}

{15,25,30,35},

{66,77,88,99}

};

what would be the output of *(*(num+1))?

Given the array:

int num[3][4]=

{

{3,6,9,12},

{15,25,30,35},

{66,77,88,99}

};

what would be the output of *(*(num+1))?

int *pt1, sm=0;

int a, b, p, z, n=9;

z = n;

pt1 = num1;

for (a = 0; a < z; a++)

{

    for (b = 0; b < z; b++)

    {

        if (a == b)

        {

            continue;

        }

        else if (*(pt1 + a) == *(pt1 + b))

        {

            p = b;

            z--;

            while (p < z)

            {

                *(pt1 + p) = *(pt1 + p + 1);

                p++;

            }

            b = 0;

        }

    }

}

for (a = 0; a < z; a++)

{

    sm=sm+num1[a];

}

printf("%d",sm);

return 0;

}

Analyse the following code and choose its correct output from the given options.

#include<stdio.h>

int main()

{

int num1[9] = {3,2,3,4,5,2,3,9,2};

int *pt1, sm=0;

int a, b, p, z, n=9;

z = n;

pt1 = num1;

for (a = 0; a < z; a++)

{

    for (b = 0; b < z; b++)

    {

        if (a == b)

        {

            continue;

        }

        else if (*(pt1 + a) == *(pt1 + b))

        {

            p = b;

            z--;

            while (p < z)

            {

                *(pt1 + p) = *(pt1 + p + 1);

                p++;

            }

            b = 0;

        }

    }

}

for (a = 0; a < z; a++)

{

    sm=sm+num1[a];

}

printf("%d",sm);

return 0;

}