Passing struct into function - c

Question

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define SIZE 100000
typedef struct {
int day;
int month;
int year;
} DATE;
typedef struct {
char name[100];
int age;
float hrlyWage;
float hrsWorked;
float regPay;
float otPay;
float totalPay;
DATE payDate;
} PAYRECORD;
int newRecord(struct PAYRECORD record[], int index){
//set name to \0 so it can work as string
record[index].name = {'\0'};
index++;
return index;
}
int main(){
char menuChoice = 'X';
struct PAYRECORD record[SIZE];
int index = 0;
while (menuChoice != 'Q'){
system("pause");
system("cls");
menuChoice = runMenu();
switch (menuChoice){
case 'A':
index = newRecord(record, index);
}
}
}
main sets up an array of structs the gets passed into newRecord, and the goal is to make it so that I can input the data here and then return the new index to keep track of my array of structs. However something is going wrong where my program doesn't seem to be recognizing newRecord as a function, which ends up throwing the whole program off.
I get syntax errors for all the functions inside of newRecord, though I beleive it's because, as I mentioned, the program seems to be unable to recognize newRecord as a User defined Function.

Answer 1

Use of struct PAYRECORD is wrong since there is no such type. You only have a typedef named PAYRECORD.
If you want to be able to use struct PAYRECORD as well as just PAYRECORD, change the definition of the struct to:
typedef struct PAYRECORD {
char name[100];
int age;
float hrlyWage;
float hrsWorked;
float regPay;
float otPay;
float totalPay;
DATE payDate;
} PAYRECORD;
If that's not your goal, change the use of struct PAYRECORD by just PAYRECORD.
Also, the line:
record[index].name = {'\0'};
in newRecord is not correct. You cannot assign to an array like that. Change it to:
record[index].name[0] = '\0';

Answer 2

The struct PAYRECORD does not exist, the compiler has no idea how big that is.
Note that PAYRECORD is a typedef to an anonymous struct. So your function
should look like this:
int newRecord(PAYRECORD record[], int index){
//set name to \0 so it can work as string
record[index].name[0] = 0;
index++;
return index;
}
Also note that {'\0'}; works only when initializing a array when you declare
it:
char arr1[10] = { '\0' }; // OK
char arr2[10];
arr2 = { '\0' }; // NOT OK
// error: expected expression before ‘{’ token
// a = { '\0' };
// ^
And when writing functions that take arrays as an argument, you should also pass
the size of the array.
int newRecord(PAYRECORD record[], int index, size_t len){
if(record == NULL)
return -1; // error, NULL passed
if(index >= len)
return -1; // error, cannot access array
record[index].name[0] = 0;
index++;
return index;
}
And then you can call it from main like this:
PAYRECORD record[SIZE];
...
int index = 0;
if(newRecord(record, index, sizeof record / sizeof *record) != index)
{
// error handling
}
This makes the code more robust. You always have to check the array boundaries,
otherwise you might read/write out of bounds. And also check that NULL has not
been passed as well, if you dereference NULL, your program will crash with
segfault.

Answer 3

Also, the parameter to newRecord could be a PAYARRAY, not an array directly; based on declaring
typedef struct { } PAYRECORD, PAYARRAY[SIZE];
int newRecord(PAYARRAY record, int index) {...}
int main(){
...
PAYARRAY record;
...
case 'A':
index = newRecord(&record, index);
}
The compiler should be converting the PAYARRAY or PAYRECORD[] argument to a PAYRECORD * pointing to the first element, so use of the '&' is indicated for the function call.