Initializer element is not constant in C [duplicate]

问题

Possible Duplicate:
Error “initializer element is not constant” when trying to initialize variable with const

I'm coming from javascript/php/python and probably I'm missing something, here is the code:

const int a = 50;
const int c = 100;
const int d = 100;
int endX = c + a;
int endY = d;
int startX, startY, b;

I get

ex1.4.c:6: error: initializer element is not constant
ex1.4.c:7: error: initializer element is not constant

Does someone have an explanation?

回答1:

If you are declaring endX as a global variable the error makes sense.

The reason is that global variables are initialized in compiling time, and you are trying to initialize endX as an operation that must be done in execution time.

回答2:

Unfortunately, in C const variables are not really const.

Below are the extracts from the c99 standard.

6.7.8 Initialization

All the expressions in an initializer for an object that has static storage duration shall be constant expressions or string literals.

The constants are defined as follows:

6.4.4 Constants

Syntax

constant:

integer-constant       (e.g. 4, 42L)
floating-constant      (e.g. 0.345, .7)
enumeration-constant   (stuff in enums)
character-constant     (e.g. 'c', '\0')

The standard defines constant expressions as follows:

6.6 Constant expressions

(7) More latitude is permitted for constant expressions in initializers. Such a constant expression shall be, or evaluate to, one of the following:

— an arithmetic constant expression,

— a null pointer constant,

— an address constant, or

— an address constant for an object type plus or minus an integer constant expression.

(8) An arithmetic constant expression shall have an arithmetic type and shall only have operands that are integer constants, floating constants, enumeration constants, character constants, and sizeof expressions. Cast operators in an arithmetic constant expression shall only convert arithmetic types to arithmetic types, except as part of an operand to a sizeof operator whose result is an integer constant.

Thus, c and a are not constant expressions and cannot be used as initializers in your case.

回答3:

const expressions must be a compile time constant in C unlike in C++ therefore c+a can't be used as a constant. The usual way to handle this problem in C is to use the preprocessor instead:

#define A 50
#define C 100
#define D 100
int endX = C + A;
int endY = D;
int startX, startY, b;

回答4:

Yeah, you can't initialize something to a variable. The compiler does the initialization and at compile time it doesn't know the value of c+a;

The int x = 1; type initialization is fine, the compiler just puts a 1 at the address of x in the object code.

To initialize something to c+a, you want to do it at runtime, in the startup code in c or constructor in C++.

回答5:

In C programming langages, objects with static storage duration must be initialized with constant expressions (or aggregate containing constant expressions). If endX has static storage duration, its initializer (c+a) is not a constant expression (i.e. the expression cannot be evaluated during translation phase).

来源：https://stackoverflow.com/questions/12750796/initializer-element-is-not-constant-in-c

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