Functions and returning values in C - c

Question

Im trying to print out the part at the end of this program. I enter C17 and the part comes out as 0 when it should be 1. Why is this?
Kind Regards
Dennis
# include <stdio.h>
int Part;
int getPartType(int Part);
int calcPrice(int Part);
int main(int argc, char * argv[]){
getPartType(Part);
calcPrice(Part);
return 0;
}
// Part1: Asks for input from user for part type
int getPartType(int Part) {
int nvr;
char character_one;
char character_two;
int number;
printf("Enter the part type (C17, F25, DN3, GG7 or MV4): ");
nvr = scanf("%c%c%d",&character_one,&character_two,&number);
if (number==7 && character_two=='1') {
Part=1;
}else if (number==5 && character_two=='2') {
Part=2;
}else if (number==3 && character_two=='N') {
Part=3;
}else if (number==7 && character_two=='G') {
Part=4;
}else if (number==4 && character_two=='V') {
Part=5;
}else{
printf("Wrong Part Type\n");
Part=0;
}
return Part;
}
int calcPrice(int Part) {
printf("%d\n",Part);
return 0;
}

Answer 1

getPartType(Part); returns an int, and doesn't assign to the original Part. So you must change this line:
getPartType(Part);
to
Part = getPartType(Part);
If you want to change the original value of Part you must use pointers. You can read more about this in any decent C book (I recommend K&R). For example:
// takes pointer to integer and sets it to 5
void settofive(int *someInteger) {
*someInteger = 5; // dereference someInteger and set to 5
}
int main(int argc, char *argv[]) {
int test = 0;
int *ptrTotest = &test; // take address of test and store in ptrTotest
printf("%d\n", test); // prints out zero
settofive(ptrTotest);
printf("%d\n", test); // prints out five
return 0;
}

Answer 2

You have a little misunderstanding of function argument passing.
When you call a function like
getPartType(Part);
C will create a copy of Part on the stack and all computations within the function will be made on this copy. Therefore you will not change the variable Part. This is called Call-by-value.
To change this problem, there are two ways. You can either just use:
Part = getPartType(Part);
This will create a copy of Part, the function will work on this copy, and then return something. This something will then get stored in the original Part. In your case you can actually just use int getPartType(void) as the function declaration, because you don't work an Part.
The other way would be to pass a pointer:
getPartType(&Part);
This passes a pointer to the original Part, so you can manipulate the original part (using the *-operator). This would mean that your declaration shoudl like like void getPartType(int *). But I would say the first method is preferable if you are dealing with just one basic variable

Answer 3

C is call by value. This means that the function can't change the value of a variable in the caller's context, unless the caller passes the address of that value.
Since your function doesn't really need an input argument, it should be removed. All you need is the return value.
Also, you could consider using multiple return statements, changing the if-ladder to look like so:
if (number==7 && character_two=='1') {
return 1;
}else if (number==5 && character_two=='2') {
return 2;
and so on.
Further, the use of "magical" numerical constants is generally a bad idea. It would be better to introduce an enumeration before main(), like this:
enum Part { PART_C17 = 1, PART_F25, PART_DN3, PART_GG7, PART_MV4 };
Then change the function to return a value of this new type:
enum Part getPartType(void)
{
/* ... */
}
and update the code in the if-ladder accordingly:
if (number==7 && character_two=='1') {
return PART_C17;
}else if (number==5 && character_two=='2') {
return PART_F25;
and so on.