Error with two pieces of code almost identical

**I am new to C so please keep that in mind when answering I'm currently working on a larger program where I use classes called integer.c (which defines an integer and functions used on it), real.c (which defines a double and functions used on it) an

C: About typedef and enum

I just started to learn C, and I think I didn't fully understand the use of typedef. Consider this code: #include <stdio.h> enum _season{spring=1, summer, autumn, winter} season; int main() { int val; printf(" enter the number of the season: &q

sizeof whole structure with member

Sorry for that dumb question but how can I know the actual size of a structure with severals members in it. For example here is my struct : typedef struct mastruct mastruct; struct mastruct{ int val; char *chaine; float decimal; }__attribute__((packe

Pointers in structures

I am new to c language and I tried to create a structure. so here is the my structure. typedef struct car{ int *transmission; int *year; char color[15]; }CAR; Then I tried to insert the values to the structure variables using below code, printf("Ente

typedef equivalent in C # to use C ++ DLL

I'm trying to call a C++ structure, exported to a DLL, from C#-Code This is the C++ interface to the Method I want to call: typedef void *Handle; typedef void (*Callback)(Info *info); typedef void (*Timeout)(Info *info); typedef struct { WORD port; W

typedef C ++ 11 template

I was trying to use C++11's "using =" syntax to typedef a templated comparison function. I then define a templated-function and want to assign it to comparisonFunc of type "ComparisonFunction". template<typename ValueType> using

What does this typedef statement do?

I was going through some code and am not able to understand the following piece of code. What does it do? What does it mean? typedef void*(*fun)[2]; fun new_array; OK, basically, this is how typedef works: first imagine that the typedef isn't there.

C ++ - using decltype with enum content in a template class

I have a template class that looks like this: template <typename T> class Foo { public: enum Mode { Mode1, Mode2, Mode3 }; // ... }; I instantiate this: Foo<float> foo; // ... auto m = Foo<float>::Mode1; But it seems to me an unnecessary

Type of table typedef in C

typedef int arr[10] I understand that the above statement defines a new name for int[10] so that arr intArray; is equivalent to int intArray[10]; However, I am confused by the convention for doing this. It seems to me that typedef int arr[10] is conf

error: contradictory statement for uint32_t

When I try to compile my program first I got this error: error: 'uint32_t' does not name a type Then I included #include <stdint.h> Now it turned out this error: /include/stdint.h:52: error: conflicting declaration 'typedef unsigned int uint32_t' /c

Forward declares typedef in the C ++ class

What's the best solution to forward declare a typedef within a class. Here's an example of what I need to solve: class A; class B; class A { typedef boost::shared_ptr<A> Ptr; B::Ptr foo(); }; class B { typedef boost::shared_ptr<B> Ptr; A::Ptr

Typedef enum declaration in C ++

My question is simple. Is it legal in C++ to postpone the typedef in a declaration like the following? gcc accepts the code without problems, but is it a deprecated way? I ask this because I found every time samples with typedef at the beginning of t

Typedeffed struct is undefined although it is defined

I'm receiving an error message from the compiler indicating (it's an embedded C compiler): Error[Pe020]: identifier "MPTR" is undefined in myflash.h (see code below) I'm using a typedeffed struct called MPTR. The code is: //datahandler.h #ifndef

Warning for const in typedef

When I was compiling a C++ program using icc 11, it gave this warning: warning #21: type qualifiers are meaningless in this declaration typedef const direction_vector_ref_t direction_vector_cref_t; It is saying const just meaningless. I am curious ab

Typedef templates in the global scope

using template classes I usually make some typedefs like: typedef super<puper<complex<template<type> > > > simple_name I usually do it in 2 ways: template <class A, ...> struct Types { typedef ... } template <class A, ...&

Do I need to convert variables using a typdef'd type?

If I have something like: typedef int MyType; is it good practice to cast the operands of an operation if I do something like this: int x = 5; int y = 6; MyType a = (MyType)(x + y); I know that I don't need to do that but wondering if it's better for

Enum & ldquo; copy & rdquo; problem

I have a class, let's call it A. It has a enum (E) and a method Foo(E e), with gets E in the arguments. I want to write a wrapper (decorator) W for A. So it'll have its own method Foo(A::E). But I want to have some kind of encapsulation, so this meth

Grammar and syntax of typedef in C language

I have a problem with the typedef keywords in C language. In my program, I use the following codes: typedef int* a[10]; int main(){ int a[10]; } they work well. But why there are no conflicts between a variable and a type sharing the same name a? Reg

Why should I use typedefs in C ++?

Lets say I have: set<int, less<int> > m_intset; That works, but now I change it to typedef, and I do end up with two lines of code: typedef set<int, less<int> > SetInt; SetInt m_intset; What's the advantage of such a typedef? Am I

Difference between 'struct' and 'typedef struct' in C ++?

In C++, is there any difference between: struct Foo { ... }; and typedef struct { ... } Foo; In C++, there is only a subtle difference. It's a holdover from C, in which it makes a difference. The C language standard (C89 §3.1.2.3, C99 §6.2.3, and C11