将值分配给结构的联合字段的问题

Question

I am new here, hope that I can get some help. After doing a tutorial about structs and unions , I practice what I learned, and want to combine two ideas. But I get an error that I've been trying to solve couple of hours, without any luck.

The code is:

void main(){

    typedef enum {INDIV, OUNCE, POUND } quantity;

    typedef union{

        short individual;
        float pound;
        float ounce;

    } amount;

    typedef struct{

        char *brand;
        amount theAmount;

    } orangeProduct;

    orangeProduct productOrdered;

    quantity quantityType = OUNCE;

    switch(quantityType)
    {
        case INDIV:
            productOrdered = {"Chiquita", .theAmount.individual = 13 };
            printf("You bought %d oranges\n\n", productOrdered.theAmount.individual)
            break;

    }
}

I have other cases as well, but this doesn't matter: if I solve the first one I will solve all of them. The goal is to have structure which can hold brands of oranges, and for each brand hold the amount bought, the correct weight type.

The problem is with the line:

productOrdered = {"Chiquita", productOrdered.theAmount.individual = 13 };

my gcc compiiler (using codeBlocks ) is shouting at me:

error: expected expression before '{' token|

and, no matter what I tried, it didn't help.

Please help.

Answer 1

Assigning To a Structure

The syntax of using braces ( {.... } ) with a list of initial values for a structure can be used only when defining a structure, as in:

orangeProduct productOrdered = { "Chiquita", { 13 } };

You cannot use it in an assignment; this statement does not have correct grammar:

productOrdered = { "Chiquita", { 13 } };

There are two ways to assign values to a structure, rather than initializing it when it is defined. One is to assign to each member individually:

productOrdered.brand = "Chiquita";
productOrdered.theAmount.individual = 13;

Another is to copy it from another structure:

productOrdered = SomeOtherProduct;

We can use the above with another C feature called a compound literal . A compound literal can be used in an expression to provide an object without giving it a name. The general form of a compound literal is:

(type) { initial values... }

Although (type) looks like a cast, it is not. This syntax defines an object, and it can use the same syntax for initial values that is used in declarations. That means we can initialize a compound literal and assign it to a structure:

productOrdered = (orangeProduct) { "Chiquita", { 13 } };

Other Notes About Assigning To a Structure

All of the following have the same effect:

productOrdered = (orangeProduct) { "Chiquita", { 13 } };
productOrdered = (orangeProduct) { "Chiquita", 13 };
productOrdered = (orangeProduct) { "Chiquita", { .individual = 13 } };
productOrdered = (orangeProduct) { "Chiquita", .theAmount.individual = 13 };

In the first of these, the inner braces indicate we are providing an initial value for a subobject of the main object—for the union amount inside the orangeProduct . As the second line shows, this is not required, because, if the braces are omitted, the compiler will take the ungrouped initial values and assign them to members of the subobjects anyway. However, it is good to include the braces explicitly, as it can avoid errors in more complicated aggregates and help convey intent to human readers.

The third and fourth forms show we can initial members using their names. This form is called designated initializers. They are most useful when you are skipping members in a structure or want to initialize a member of a union other than the first listed member.

However, another answer suggested this incorrect statement:

productOrdered = (orangeProduct){"Chiquita", productOrdered.theAmount.individual = 13 }

That is wrong because productOrdered.theAmount.individual = 13 is not just an initial value; it is an assignment expression. It assigns 13 to productOrdered.theAmount.individual . The problem with this is that the whole statement also assigns values to productOrdered , including to productOrdered.theAmount.individual . This violates the rule in C 2018 6.5 2, which says:

If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined…

The two assignments to productOrdered.theAmount.individual are unsequenced because no rule of the C standard says when the assignments occur relative to each other. (When an assignment expression updates the value of an object, that is called a side effect, and it is not part of the main evaluation operation of an expression.)

Other Notes

main should be declared with int main(void) or int main(int argc, char *argv) , not with void main() . It can also be declared with specific forms allowed by your C implementation.

Include #include <stdio.h> in your code to declare printf .

Append a semicolon to this line:

printf("You bought %d oranges\n\n", productOrdered.theAmount.individual)