Trouble with Call by Address

Hello,

I'm trying to wirte a program but I'm having trouble with call by address. When I use the * in the function prototype and in the function itself and an & in the function call in main, I don't get an error, but the value of 0 is not passed from main to the function AddData. Can anyone tell me how to make this work and where I have made my mistake?

Thanks,
Vireyda

Code:

---

#include<stdio.h> //include standard input output functions

//definitions section
//structure for travel expense account
struct TravelCosts
        {
        char employee_name[20];
        int emp_num;
        float distance_traveled;
        float mileage_rate;
        float meals;
        float total_expenses;
        };
void AddData(struct TravelCosts expenses[], int *num);
void PrintTravel(struct TravelCosts expenses[], int num);
void ChangeCosts(struct TravelCosts expenses[], int num);
void FindSpenders(struct TravelCosts expenses[], int num);

void main()
{
        //declarations section
        char response;
        static struct TravelCosts travel[20];
        int num_records = 0;

        do
        {
        printf("Please select an action:\n\n");
        printf("Add employee expenses\t\t- A or a\n"
              "Print expense database\t\t- P or p\n"
              "Change expense data\t\t- C or c\n"
              "Find over spending employees\t- F or f\n"
              "Quit Program\t\t\t- Q or q\n");
        fflush(stdin);
        scanf("%c", &response);

        //switch to case based on response value
        switch(response)
        {case 'A':case 'a':
                AddData(travel, &num_records);
                break;
        case 'P': case 'p':
                PrintTravel(travel, num_records);
                break;
        case 'C': case 'c':
                ChangeCosts(travel, num_records);
                break;
        case 'F': case 'f':
                FindSpenders(travel, num_records);
                break;
        case 'Q': case 'q':
                break;
        default:
                printf("\nThat selection is invaild.  Please try again.\n");
        }//end switch
        }while(response!='Q'&&response!='q');//end do
}//end main

void AddData(struct TravelCosts expenses[], int *num)
        {
        printf("%d", num\n);
        }

---

You need to dereference your pointer num in the function if you want the value stored there. Should be *num.

be dereference do you mean that I should drop the int in the function? Is there a way I can do this with &'s instead?

Thanks,
Vireyda

Inside AddData, num is a pointer. To get the integer value that the pointer is pointing at, you must dereference the pointer by adding the * to it. For example:

printf("%d\n", *num);

Do I need to use *num if I want to read an input into that variable?

Thanks,
Vireyda

It depends. If you are using scanf, which takes a pointer or the address of a variable, then just use num (because num is a pointer). If you are using some other method, like cin, that wants the int itself, then use *num.

Remember that when declaring a variable, the '*' means that the variable is a pointer. When not declaring a variable, the '*' means go to the memory location pointed to by this variable. Normally, the variable is a pointer, so you are telling the code to look at the value at which it is pointing. Since scanf takes a pointer as an argument, you pass num by itself. Since printf wants the actual value, you tell it to go to the memory location at which num is pointing, by adding the '*' to it.

Of course, num_records (in your main function) is just an int, it is not a pointer. So if you pass num_records to printf, you will not need the '*' - it already contains the value. If you pass num_records to scanf, you will need to pass a pointer to that value. That is why you use '&', the address of operator.

I think I'm starting to understand, but I still feel very confused about pointers. I'm trying to use num in a for loop now, but once the new value is input by the user and read it crashes before it can do the loop. I would have thought that I would leave out the '*', but then i get an error while I compile.

Thanks again,
Vireyda


[CODE]
void AddData(struct TravelCosts expenses[], int *num)
{
//declaraions section
int i;

printf("How many travel claims are ther to be entered? ");
scanf("%d", &num);

for(i=0;i<=*num-1;++i)
[/CODE}

Quote:

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Originally Posted by jlou

If you are using scanf, which takes a pointer or the address of a variable, then just use num (because num is a pointer).

---

You put &num instead of num.

Remember, '&' turns an int into a pointer to int. It turns a pointer to int into a pointer to a pointer to an int. Since scanf takes a pointer to an int, and num is a pointer to an int, all you need to do is pass in num.

Also, just to be more clear, put parentheses around the *num in your for loop, so it is i <= (*num) - 1, or even better, use:

Code:

---

for(i=0;i<*num;++i)

---

Okay, I made the reccommended change, but my program still crashes when I get to the for loop. I don't know what's going on here.

Paste your latest code

I'm not getting any errors when I compile the program or build it. It just crashes when I get to the for loop.

Code:

---

void AddData(struct TravelCosts expenses[], int *num)
        {
        //declaraions section
        int i;

        printf("How many travel claims are ther to be entered? ");
        scanf("%d", &num);

        for(i=0;i<*num;++i)
                {
                printf("/nPlease enter the following information:");
                printf("Employee %d Name:  ", i);
                gets(expenses[i].employee_name);
                printf("Employee Number:  ");
                scanf("%d", &expenses[i].emp_num);
}

---

>scanf("%d", &num);
Remove the &, num is already a pointer. And your bracing is off, but that's probably just a copying error.

Thanks so much. It works now. I get so confused about pointers.

Vireyda :)

Why are you passing num into the function as a pointer and returning nothing? If the idea is to pass num back to the calling routine, change this function to:

Code:

---

int AddData(struct TravelCosts expenses[])
{
    int num;
    ...
    return num;
}

---

But if you don't need num in the calling routine use

Code:

---

void AddData(struct TravelCosts expenses[])
{
    int num;
    ...
    return;
}

---