How to find out what variable value is set for combining?

Question

I am working on project optimization. It contains a structure of options in which the user can select one parameter at a time. In addition to this option, we also use the flag variable to check what parameter value is set for this entry. To make memory efficient, I want to transform the structure into a union. But how do you know what variable value is set in union. Because there are no restrictions in the union to get the value of a variable, even if it is not set.

struct options{ int basicPay; int lumsumPay; int mothlyPay; int weeklyPay; int dailyPay; int anualPay; int payType; // Flag variable to check which option is selected }; union OptimizeOptions{ int basicPay; int lumsumPay; int mothlyPay; int weeklyPay; int dailyPay; int anualPay; int payType; // Confuse at here }; 

Answer 1

Have you tried union inside a struct ? Let's look at the following example:

 union options{ int basicPay; int lumsumPay; int mothlyPay; int weeklyPay; int dailyPay; int anualPay; }; struct record{ union options op; // Options union int payType; // Flag variable to check which option is selected } 

The union ( options ) will reserve memory for the largest variable, and you can set its value, and your structure (record) will track this block of the unification memory, and you can set the value of the payType flag, which will indicate to your program to get a percussion variable of union.

Answer 2

Unions suffer from the problem that there is no easy way to find out which member of the union was last changed. To track this information, you can embed union inside a structure that has one other member (called a "tag field" or "discriminant"). The purpose of the tag field is to remind which item has been changed / updated. You can try the following:

 typedef struct{ int payType; // Tag field union{ int basicPay; int lumsumPay; int mothlyPay; int weeklyPay; int dailyPay; int anualPay; }OptimizeOptions; }Options; 

But there is no need to write six separate members to combine in your case, since all are of type int . Therefore, it can be reduced to

 typedef struct{ enum{BASIC_PAY, LUMSUM_PAY, MONTHLU_PAY, WEEKLY_PAY, DAILY_PAY, ANNUAL_PAY} payType; int pay; }Options; 

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Let's look at using a tag field with a simple example. Suppose we need an array that can store data of type int and double . This is made possible through the use of union . So, first determine the type of union that will store either int or double .

 typedef union { int i; double d; } Num; 

Next we need to create an array whose elements are Num type

 Num num_arr[100]; 

Now suppose we want to assign element 0 to num_arr to store 25 , and element 1 stores 3.147 . It can be done as

 num_arr[0].i = 25; num_arr[1].d = 3.147; 

Now suppose we have to write a function that will print num_arr elements. The function will look like this:

 void print_num(Num n) { if(n contains integer) printf("%d", ni); else printf("%f", nd); } 

Wait! How can print_num decide if n an integer or double ?

This will be done using the tag field:

 typedef struct{ enum{INT, DOUBLE} kind; //Tag field union{ int i; double d; }u; }Num; 

So, every time a value is assigned to the u member, kind must ¹ be set to either int or double on Recall what type we actually saved. For example:

 nui = 100; n.kind = INT; 

The print_num function will look like this:

 void print_num(Num n) { if(n.kind == INT) printf("%d", ni); else printf("%f", nd); } 

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_{1 . The responsibility of the programmer is to update the tag field with each assignment to the union member. Forgetting to do this will result in an error, as noted in comment @ j_random_hacker .}