Initializing an array of structs, why does using struct variables not work

问题

I am extremely new to C. Would appreciate if someone can help understand why code in lines 13,14 and 16 does not work, but lines 17-20 works.

With the first option (lines 13, 14 and 16) I get the error

error: initializer element is not constant

What does this mean? Also, does this mean one cannot use variables of certain type to generate new variables?

Thank you.

// Define structure for a good
  5 struct good {
  6     char goodname;
  7     double p; //starting proportion
  8     int theta; //average utility
  9     int sigma; //variance of error
 10 };
 11
 12 // The goods H and L
 13 struct good H = {.goodname = 'H', .p = 0.5, .theta = 100, .sigma = 20};
 14 struct good L = {.goodname = 'L', .p = 0.5, .theta = 75, .sigma = 20};
 15
 16 struct good goods[2] = {H, L}; // **Does not work**

 // ** Works**
 17 struct good goods[2] = {
 18     {.goodname = 'H', .p = 0.5, .theta = 100, .sigma = 20},
 19     {.goodname = 'L', .p = 0.5, .theta = 75, .sigma = 20}
 20 };

回答1:

H and L are storage locations that contain data. The statement:

 struct good goods[2] = {H, L}; // **Does not work**

implies that goods should point to H and L, or contain the same values as H and L.

Either, copy the data from H and L to goods[0] and [1] or modify goods to be an array of pointers, as:

 struct good *goods[2];

 goods[0] = &H;
 goods[1] = &L;

回答2:

The reason line 16 does not work is the same why this much simpler example would not work:

const int a = 5;
int b = a; // Does not work: "initializer element is not constant"

What this means is that you need to use only compile-time constant expressions in the initializers. C does not consider variables compile-time constants, even the const ones (C++, on the other hand, considers const variables compile-time constants).

回答3:

It means in C you can only use constant expressions to initialize an array or a structure with static storage duration.

In your example all objects have static storage duration but:

 struct good H = {.goodname = 'H', .p = 0.5, .theta = 100, .sigma = 20};

the initializers above are literals and literals are constant expressions in C, so it is OK.

 struct good goods[2] = {H, L};

but here H and L objects are not constant expressions so the compiler gives you an error.

C has a rather strict definition of what a constant expression is and the value of an object is not considered a constant expression (even if the object is const qualified or initialized with a literal).

回答4:

The value of a variable, even if the variable is already initialized globally¹ (and even if it is const-qualified), is not itself a constant-expression.

A pointer to a static-duration variable² is an "extended constant expression" and can be used to initialize pointer variables, so you could do:

struct good *goods[2] = {&H, &L};

for instance, if that suits the problem.

¹Yes, I know "initialized globally" is not really well-defined. :-) I mean, if it's a static-duration variable that has been initialized, and is visible at the point of the new initializer.

²I.e., any "global" (file scope) variable, or a static variable inside a block.

回答5:

If the object is static "goods[0]/goods[1]", you can only initialize with constants. So, the reason you did not get an error initializing "good H".

If you define your variable as non-static, (for e..g, in your main() or in any of your function()) you will not get this error, as the object will be treated as "auto".

来源：https://stackoverflow.com/questions/18495108/initializing-an-array-of-structs-why-does-using-struct-variables-not-work

标签

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