Will these print the same thing? #include <stdio.h> void printout(int num[]); int main(void) { int num[9] = {10,20,30,40,50,60,70,80,90}; printout(num); } void printout(int num[]) int i; for (i=0;i<9;i++) printf(“num=%d”, num[i]); #include <stdio.h> int main(void) { int num[9] = {10,20,30,40,50,60,70,80,90}; int i; for (i=0;i<9;i++) printf(“num=%d”, num[i]); } YES!

What will this print out? #include <stdio.h> void change(int array[], int val); int main(void) { int val = 3; int array[1] = {3}; change(array, val); printf(“array=%d, val=%d”,array[0],val); } void change(int array[], int val) int i; array[0] = array[0]*2; val = val * 2; 1. array=6, val=6 2. array=6, val=3 3. array=3, val=3 4. array=3, val=6

In the previous example, array was passed by ____1______, while val was are passed by ___2_____? 1. reference, value 2. value, reference 4. Oh shoot, don’t know, I’d better ask Garvin’s other half as she is the one who knows everything! 3. Yo mamma, yo daddy

What will this print out? #include <stdio.h> void change(int array, int val); int main(void) { int val = 3; int array[1] = {3}; change(array[0], val); printf(“array=%d, val=%d”,array[0],val); } void change(int not_an_array, int val) not_an_array = not_an_array*2; val = val * 2; 1. array=6, val=6 2. array=6, val=3 3. array=3, val=3 4. array=3, val=6 Notice that the function no longer takes in an array, but an array element – which is passed by value

What will this print out? #include <stdio.h> int change(int array[], int val); int main(void) { int val = 3; int array[1] = {3}; val = change(array, val); printf("array=%d, val=%d",array[0],val); } int change(int array2[], int val) array2[0] = array2[0]*2; val = val * 2; return val; 1. array=6, val=6 2. array=6, val=3 3. array=3, val=3 4. array=3, val=6

The function printout does not know anything about variable size What is missing here? #include <stdio.h> void printout(int num[]); int main(void) { int size = 9; int num[9] = {10,20,30,40,50,60,70,80,90}; printout(num); } void printout(int num[]) int i; for (i=0;i<size;i++) printf(“num=%d”, num[i]); The function printout does not know anything about variable size

So the previous examples needs to look like one of these two: #include <stdio.h> void printout(int num[], int size); int main(void) { int size = 9; int num[9] = {10,20,30,40,50,60,70,80,90}; printout(num, size); } void printout(int num[], int size) int i; for (i=0;i<size;i++) printf(“num=%d”, num[i]); #include <stdio.h> #define size 9 /* include a define */ void printout(int num[]); int main(void) { int num[9] = {10,20,30,40,50,60,70,80,90}; printout(num); } void printout(int num[]) int i; for (i=0;i<size;i++) printf(“num=%d”, num[i]);

Do you need to know the array size of a character array before hand? 1. No because a character array is a string with a terminating character, ‘\0’, and thus you can just use strlen 2. Yes because it is an array and every array has a length that needs to be known by the programmer beforehand 3. Yes because it is I think I read somewhere that says I must know the array size of a character array beforehand – some blog I think. 4. What is Garvin’s wife’s number, I’ll just ask her.

Skeleton for your inclass homework: /* include all include files that you need */ void changearray( you must figure out what goes here); int main() { /* Declare variables */     /* Read in array elements from user of a specified size (arrayLength) */     for (i = 0; i < arrayLength; i++)     {         printf("Please enter an integer to be put in the array: ");         scanf("%d", &array[i]);     } /* print out array */ /* call the changearray function */ /* print out new array */ } void changearray(you must figure out this function)