Pass struct to a function.

Say I define the following structure globally:

Code:

struct stfirst
{
     int num;
     char let;
}Me;

then I go ahead and declare it in main as follows:

Code:

stfirst Me[3] = {
     {1,'a'},
     {2,'b'},
     {3,'c'}
};

Is there a way to pass this struct of type first to a function called:

Code:

void show(stfirst);

where the show function will then go through the struct and display the options? I cant seem to make mine work. I dont know what names to put in place of where (ie the function definition,function call from main, and function body)


so does this now mean that I am passing an array of type stfirst to the function, then I should do as if I were just passing an array to the function.. ie

Code:

void show(Me[3])
{
//do comands
}

??
and when I call the function I should use the array name which was Me.

You could do it that way. Or this way..

Code:

#include <iostream>

struct stfirst
{
     int num;
     char let;
};

void show(stfirst* me)
{
     for(int i = 0; i < 3; i++) {
             std::cout << me[i].num << std::endl;
             std::cout << me[i].let << std::endl;
     }
     
}

int main()
{
    stfirst me[3] = {
     {1,'a'},
     {2,'b'},
     {3,'c'}
     };
     
     show(me);

    std::cin.get();
}

That could be theoretically wrong (as in it works, but you shouldnt do it). I've been wrong too much when it comes to passing parameters, references, pointers, and how to treat them.

But when I try my program it doesn work. I can pass a single struct to my function, but I cant pass the array of struct type stfirst. to pass an array you just put the array name in the function when using it. and in the function definition you just use the type and then name of array. But it wont work for me..

I think you are asking for something like this?

Code:

#include <iostream>

using namespace std;


struct struc_name
{
	int num;
	char letter;
};


void funk(struc_name name_in_function[] );


int main()
{

	struc_name test[3] = { {1,'a'}, {2,'b'}, {3,'c'} } ;

	funk(test);

	cin.get();


	
return 0;
}

void funk(struc_name name_in_function[] )
{

	for (int i = 0; i< 3; i++)
	{
		cout << name_in_function[i].num << ',' << name_in_function[i].letter << endl;
	}

}

Using [] and * is the same thing, but I guess [] is more descriptive.

Works for me Hybird, but I wouldn't deep copy it.

Code:

#include <iostream>

struct stfirst
{
     int num;
     char let;
};

void show(stfirst me[3])
{
     for(int i = 0; i < 3; i++) {
             std::cout << me[i].num << std::endl;
             std::cout << me[i].let << std::endl;
     }
     
}

int main()
{
    stfirst me[3] = {
     {1,'a'},
     {2,'b'},
     {3,'c'}
     };
     
     show(me);

    std::cin.get();
}

Maybe its because you are missing a datatype? for the parameter in show(). Or mismatches because you have that global object Me, which isn't necessary.

stfirst Me[3]...

Is there a way to pass this struct of type first to a function called:

void show(stfirst);

No. 'stfirst' and 'stfirst[3]' are different types. The way parameter passing works is: when you call a function, the variable listed as the parameter of the function is first created, and then it is initialized with the value you send to the function. That is why the type of the argument you send to a function has to match the type of the function's parameter variable.

I also think you are misapplying Hungarian notation. I think the way you are supposed to use Hungarian notation is like this:

Code:

struct First
{
   ....
}

First stMyVarName;

In my opinion, the way you are using it now is confusing rather than informative.