Question

Run the C program below  and complete the table showing all the variables. Add printf function calls to obtain the addresses and values of all 13 variables. (Count the array (ca) as 3 variable/values and also include argc and argv in the table).

The table must show the Address, Name, Datatype, Scope, and Value of each variable on the stack. (Hint: use the sizeof function to double check your addresses.) Explain how different the actual memory allocations are from what you practiced by hand in the classes. (Don't just say like "it is totally different from what I thought." "It is randomly allocated." or any abstract impression.)

int main(int argc, char* argv[])
{
    int num1, num2 = 13;
    char c = 'H';
    float score1 = 12.5;
    char ca[3] = "Hi";

    dummy(num2, c, ca, score1);

    /* pause here */

    return 0;
}

void dummy(int x, char y, char * z, float w)
{
    x++;
    y++;
    w = w + 2.3;
}

Answer 1

C program with printf statements added

#include<stdio.h>
int main(int argc, char* argv[])
{
int num1, num2 = 13;
char c = 'H';
float score1 = 12.5;
char ca[3] = "Hi";

dummy(num2, c, ca, score1);

//printing all variable values and addresses
printf("num1=%d, address of num1=%d \n",num1,&num1); //num1
printf("num2=%d, address of num2=%d \n",num2,&num2); //num2
printf("c=%c, address of c=%d \n",c,&c); //c
printf("score1=%f, address of score1=%d \n",score1,&score1); //score1
printf("ca[0]=%c, address of ca[0]=%d \n",ca[0],&ca[0]); //ca[0]
printf("ca[1]=%c, address of ca[1]=%d \n",ca[1],&ca[1]); //ca[1]
printf("ca[2]=%c, address of ca[2]=%d \n",ca[2],&ca[2]); //ca[2]

return 0;
}

void dummy(int x, char y, char * z, float w)
{
x++;
y++;
w = w + 2.3;

//printing values of all variables and their addresses
printf("x=%d, address of x=%d \n",x,&x); //x
printf("y=%c, address of y=%d \n",y,&y); //y
printf("w=%f, address of w=%d \n",w,&w); //w
printf("z[0]=%c, address of z[0]=%d \n",z[0],&z[0]); //z[0]
printf("z[1]=%c, address of z[1]=%d \n",z[1],&z[1]); //z[1]
printf("z[2]=%c, address of z[2]=%d \n",z[2],&z[2]); //z[2]
}

The addresses of x and num2 are different because num2 is passed by value(not passed by reference). Similarly, addresses of y & c and w & score1 are different. Also, the changes made in the function dummy() are not reflected in the variables in main beacuse the changes are made in the copies of these variables and not the actual variables.The addresses of ca and z are same because arrays are always passed by reference.

variable	datatype	address	value	scope
num1	int	2686748	2	main
num2	int	2686744	13	main
c	char	2686743	H	main
score1	float	2686736	12.5	main
ca[0]	char	2686733	H	main
ca[1]	char	2686734	i	main
ca[2]	char	2686735	NULL	main
x	int	2686788	14	dummy
y	char	2686768	I	dummy
w	float	2686700	2.5	dummy
z[0]	char	2686733	H	dummy
z[1]	char	2686734	i	dummy
z[2]	char	2686735	NULL	dummy