: Consider the following code int a; int *aPtr; a = 9; aPtr = &a; printf(“a is e

Question

: Consider the following code

int a; int *aPtr;

a = 9;

aPtr = &a;

printf(“a is equal to %d ”,a);

printf(“the address of a is: %p ”, &a);

printf(“Using the pointer, a is equal to %d ”, *aPtr);

printf(“Using the pointer, the address of a is: %p ”, aPtr);

return 0;

adding functions to the above to calculate a*a and return the value: one function doing call-by- value and the other function doing call by reference

Problem 11: (1) Write a code defining an integer array of size=5 and a string array to store “word”, and prints all the elements of these two arrays in two separate lines.

Write a function ArrayModify that modifies the 4rd element of the integer array by making it 3 times bigger, and in main code print the original and modified elements

Write a function named ElementDouble that doubles the value of the modified 3rd element, and prints the value doubled. In main, print the 3rd element before and after calling function ElementDouble.

Explanation / Answer

#include<stdio.h>
void mul(int a, int b);
int main()
{
int a; int *aPtr;
a = 9;
aPtr = &a;
printf(“a is equal to %d ”,a);
printf(“the address of a is: %p ”, &a);
printf(“Using the pointer, a is equal to %d ”, *aPtr);
printf(“Using the pointer, the address of a is: %p ”, aPtr);
mul(a, b);
return 0;
}
void mul(int a ,int b){
int temp;
temp=a*b;
return temp;
}

-------------------------------------------------

#include <stdio.h>

int main(void) {
int myArray [5] = {1,2,3,4,5};
char name[5]={'L','e','s','s','d'};
for (int i=0;i<=5;i++)
{
printf("%d", myArray[i]);

  
}
printf(" ");
for (int j=0;j<=5;j++)
{
printf("%c", name[j]);
}
return 0;
}

---------------------------------------------------------

#include <stdio.h>

int main(void) {
int c,d;
int myArray [5] = {1,2,3,4,5};
printf("Before modification ");
for (int i=0;i<=4;i++){
printf("%d ", myArray[i]);
}
printf("After modification ");
for (int i=0;i<=4;i++){
   if(i==3){
myArray[i]=myArray[i]*3;
   }
   printf("%d ", myArray[i]);
}
   return 0;
}

------------------------------------------------

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