Question

Write a C function to add the elements of two same-sized integer arrays and return ptr to a third array.

int *addTwoArrays(int *a1, int *b1, int size);

The function should follow the following rules:

• If the sum for any element is negative, make it zero.
• If a1 and b1 point to the same array, it returns a NULL
• If any input array is NULL, it returns a NULL.

Please call this function with the following arrays and print the sums (print sum array elements in separate lines for cases i, ii below).

i. int a1[] = {1, -15, 2, 14, 3, -13, 0};

int b1[] = {0, 16, 2, -15, -3, 10, 0};

ii. int a2[] = {100, 101, 200, -3011};

int b2[] = {1000, 1010, -300, 10000};

Hint: For the array with sum, you may have to declare a global var

Answer 1

#include<stdio.h>
int sum[];

int *addTwoArrays(int *a1, int *b1, int size)
{
if(a1==b1)
{
return NULL;
}
if((a1==NULL)||(b1==NULL))
{
return NULL;
}
for(int i=0;i<size;i++)
{
sum[i] = 0;
}
for(int i=0;i<size;i++)
{
sum[i] = a1[i] + b1[i];
}
printf("Sum is: ");
for(int i=0;i<size;i++)
{
if(sum[i]<0)
{
sum[i]=0;
}
printf("\t%d\t",sum[i]);
}
  
}
int main()
{
int i;

//for the execution of case1
int a1[]={1,-15,2,14,3,-13,0};
int b1[]={0,16,2,-15,-3,10,0};
int size=7;
printf("\ncase 1:\n");
printf("\ta1[]=\t");
for(i=0;i<size;i++)
printf("%d\t\t",a1[i]);
printf("\n\tb1[]=\t");
for(i=0;i<size;i++)
printf("%d\t\t",b1[i]);
printf("\n");
addTwoArrays(a1,b1,size);
  
//for the execution of case2
int a[]={100,101,200,-3011};
int b[]={1000,1010,-300,10000};
size=4;
printf("\n\ncase 2:\n");
printf("\ta1[]=\t");
for(i=0;i<size;i++)
printf("%d\t\t",a[i]);
printf("\t");
printf("\n\tb1[]=\t");
for(i=0;i<size;i++)
printf("%d\t\t",b[i]);
printf("\n");
addTwoArrays(a,b,size);
return 0;
}

OUTPUT GENERATED:

case 1:
   a1[]=   1       -15       2       14       3       -13       0      
   b1[]=   0       16       2       -15       -3       10       0      
Sum is:    1       1       4       0       0       0       0  

case 2:
   a1[]=   100       101       200       -3011          
   b1[]=   1000       1010       -300       10000      
Sum is:    1100       1111       0       6989