C initializer is not a constant
Web16 rows · Others have provided the answer. The reason it works in C++ is that C++ can use non-constant initializers even for static objects, compiling into initialization code that … WebMay 11, 2024 · In C language objects with static storage duration have to be initialized with constant expressions or with aggregate initializers containing constant expressions. My guess is that you are trying to define several A as a global variable - Is that true? In that case, malloc() is not a a constant expression and cannot be used.
C initializer is not a constant
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WebJul 13, 2024 · Prior to C++20, we have the interesting situation that your constructor cannot be declared constexpr, but objects of the vector3 type can still be used in constant expressions, because a default constructor that is explicitly defaulted on its first declaration is not actually called during value-initialization unless it is nontrivial (C++17 ... WebThe first bullet (2.1) is satisfied because the variable b has an initializer. But the second bullet isn't satisfied because the full-expression of its initialization is not a constant expression because its initializer a is not a constant expression as aforementioned. Therefore, the variable b is not constant-initialized.
WebFrom the C Standard (6.7.9 Initialization) 4 All the expressions in an initializer for an object that has static or thread storage duration shall be constant expressions or string literals. And in this declaration. const char *text_entry = gtk_entry_buffer_get_text(p -> buffer); the initializer is not a constant expression. WebSep 23, 2016 · Thanks for contributing an answer to Stack Overflow! Please be sure to answer the question.Provide details and share your research! But avoid …. Asking for help, clarification, or responding to other answers.
WebFeb 21, 2024 · From the C Standard (6.6 Constant expressions) 6 An integer constant expression117) shall have integer type and shall only have operands that are integer constants, enumeration constants, character constants, sizeof expressions whose results are integer constants, and floating constants that are the immediate operands of casts. WebJan 28, 2015 · There is more detail though.As the standard quote says initializer should be constant expression, In C (unlike C++) const declarations do not produce constant expressions. So even with const int radius = 10; your code will not compile. In C the way to produce a constant expression is by using #define 's. – Alok Save.
WebUnfortunately, 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 …
WebDec 12, 2012 · 1 Answer. since val is declared at static file level, its value must be known to the compiler at compile time. This is not the case, since you initialize it by calling getenv (key). This because global values are placed in a special storage area of the binary file which is created at compile time. Nothing to do with the const qualifier; you ... redoute table basseredoute tapis rondWebThe problem with zero initializing variables automatically is that not all your variables should have zero as their initial value. If you forget to initialize a variable with its correct value, the compiler can easily warn you that you used an uninitialized variable.forget to initialize a variable with its correct value, the compiler can easily warn redoute tapis bainWebFeb 14, 2013 · I have to assign a static variable a value which I am getting from a function. I tried doing the following but I am getting initializer element is not constant. int countValue() { return 5; } void redoute traversinWebAug 21, 2013 · The trouble is that the 64-bit address constants don't fit into a 32-bit int, so the loader would have to generate code to truncate the addresses, which makes them insufficiently constant. With a 32-bit compilation, the code using int compiles cleanly too. richest comedians listWebJul 22, 2024 · Solution 1. It's not inside a function, which means it has to be an initializer - which is assigned only when the item is declared - which means it must be a constant … redout fatalWebMar 11, 2024 · The value of this variable can be altered every time the program is run. Moreover, dynamic initialization is of 3 kinds i.e. Unordered Dynamic Initialization; Partially-Ordered Dynamic Initialization; Ordered Dynamic Initialization; Different ways of Initializing a Variable in C++. There are 7 methods or ways to initialize a variable in C++: redoute\u0026apos s finest flowers in embroidery