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

Question

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 };

Sergey Kalinichenko · Accepted Answer · 2013-08-28 18:08:58Z

4

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).

answered Aug 28, 2013 at 18:08

Sergey Kalinichenko

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JackCColeman · Accepted Answer · 2013-08-28 18:12:05Z

2

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;

answered Aug 28, 2013 at 18:12

JackCColeman

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ouah · Accepted Answer · 2013-08-28 18:08:50Z

2

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).

answered Aug 28, 2013 at 18:08

ouah

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torek · Accepted Answer · 2013-08-28 18:14:56Z

2

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.

edited Aug 28, 2013 at 18:14

answered Aug 28, 2013 at 18:09

torek

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parucra · Accepted Answer · 2013-08-28 19:17:31Z

1

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".

answered Aug 28, 2013 at 19:17

parucra

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