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#include <math.h> #include <windows.h> #include <stdio.h> void inputData (int *M11,int *M12,int *M13,int *M21,int *M22,int *M23,int *M31,int *M32,int *M33) { printf("введите оценки первого студента через пробелn"); scanf("%d%d%d",&M11,&M12,&M13); printf("введите оценки второго студента через пробелn"); scanf("%d%d%d",&M21,&M22,&M23); printf("введите оценки третьего студента через пробелn"); scanf("%d%d%d",&M31,&M32,&M33); } void calcuateScholarship(float *S1,float *S2, float *S3) { int M11,M12,M13,M21,M22,M23,M31,M32,M33; if (M11==2 || M12==2 || M13==2) S1=0; else { if (M11==3 || M12==3 || M13==3) S1=(M11*250)+(M12*250)+(M13*250); if (M11==5 && M12==5 && M13==5) S1=(M11*250)+(M12*250)+(M13*250)+(((M11*250)+(M12*250)+(M13*250))*0.5); else S1=(M11*250)+(M12*250)+(M13*250)+(((M11*250)+(M12*250)+(M13*250))*0.25);} if (M21==2 || M22==2 || M23==2) S2=0; else { if (M21==3 || M22==3 || M23==3) S2=(M21*250)+(M22*250)+(M23*250); if (M21==5 && M22==5 && M23==5) S2=(M21*250)+(M22*250)+(M23*250)+(((M21*250)+(M22*250)+(M23*250))*0.5); else S2=(M21*250)+(M22*250)+(M23*250)+(((M21*250)+(M22*250)+(M23*250))*0.25);} if (M31==2 || M32==2 || M33==2) S3=0; else { if (M31==3 || M32==3 || M33==3) S3=(M31*250)+(M32*250)+(M33*250); if (M31==5 && M32==5 && M33==5) S3=(M31*250)+(M32*250)+(M33*250)+(((M31*250)+(M32*250)+(M33*250))*0.5); else S3=(M31*250)+(M32*250)+(M33*250)+(((M31*250)+(M32*250)+(M33*250))*0.25);} } int main() { SetConsoleOutputCP(1251); int M11,M12,M13,M21,M22,M23,M31,M32,M33; float S1,S2,S3; inputData(&M11,&M12,&M13,&M21,&M22,&M23,&M31,&M32,&M33); calcuateScholarship(&S1,&S2,&S3); printf("S1=n%2.0f",S1); printf("S2=n%2.0f",S2); printf("S3=n%2.0f",S3); getchar();getchar(); } |
| description | title | ms.date | f1_keywords | helpviewer_keywords | no-loc | ms.assetid | ||
|---|---|---|---|---|---|---|---|---|
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Learn about type conversion errors that cause Compiler Error C2440. |
Compiler Error C2440 |
08/27/2021 |
C2440 |
C2440 |
|
36e6676c-f04f-4715-8ba1-f096c4bf3b44 |
Compiler Error C2440
‘initializing‘ : cannot convert from ‘type1‘ to ‘type2‘
‘conversion‘ : cannot convert from ‘type1‘ to ‘type2‘
The compiler can’t implicitly convert from *type1* to *type2*, or can’t use the specified cast or conversion operator.
Remarks
The compiler generates C2440 when it can’t convert from one type to another, either implicitly or by using the specified cast or conversion operator. There are many ways to generate this error. We’ve listed some common ones in the Examples section.
Examples
C++ string literals are const
C2440 can be caused if you attempt to initialize a non-const char* (or wchar_t*) by using a string literal in C++ code, when the compiler conformance option /Zc:strictStrings is set. In C, the type of a string literal is array of char, but in C++, it’s array of const char. This sample generates C2440:
// C2440s.cpp // Build: cl /Zc:strictStrings /W3 C2440s.cpp // When built, the compiler emits: // error C2440: 'initializing' : cannot convert from 'const char [5]' // to 'char *' // Conversion from string literal loses const qualifier (see // /Zc:strictStrings) int main() { char* s1 = "test"; // C2440 const char* s2 = "test"; // OK }
C++20 u8 literals are const char8_t
In C++20 or under /Zc:char8_t, a UTF-8 literal character or string (such as u8'a' or u8"String") is of type const char8_t or const char8_t[N], respectively. This sample shows how compiler behavior changes between C++17 and C++20:
// C2440u8.cpp // Build: cl /std:c++20 C2440u8.cpp // When built, the compiler emits: // error C2440: 'initializing' : cannot convert from 'const char8_t [5]' // to 'const char *' // note: Types pointed to are unrelated; conversion requires // reinterpret_cast, C-style cast or function-style cast) int main() { const char* s1 = u8"test"; // C2440 under /std:c++20 or /Zc:char8_t, OK in C++17 const char8_t* s2 = u8"test"; // OK under /std:c++20 or /Zc:char8_t, C4430 in C++17 const char* s3 = reinterpret_cast<const char*>(u8"test"); // OK }
Pointer to member
You may see C2440 if you attempt to convert a pointer to member to void*. The next sample generates C2440:
// C2440.cpp class B { public: void f(){;} typedef void (B::*pf)(); void f2(pf pf) { (this->*pf)(); void* pp = (void*)pf; // C2440 } void f3() { f2(f); } };
Cast of undefined type
The compiler emits C2440 if you attempt to cast from a type that’s only forward declared but not defined. This sample generates C2440:
// c2440a.cpp struct Base { }; // Defined struct Derived; // Forward declaration, not defined Base * func(Derived * d) { return static_cast<Base *>(d); // error C2440: 'static_cast' : cannot convert from 'Derived *' to 'Base *' }
Incompatible calling convention
The C2440 errors on lines 15 and 16 of the next sample are qualified with the Incompatible calling conventions for UDT return value message. A UDT is a user-defined type, such as a class, struct, or union. These kinds of incompatibility errors are caused when the calling convention of a UDT specified in the return type of a forward declaration conflicts with the actual calling convention of the UDT and when a function pointer is involved.
In the example, first there are forward declarations for a struct and for a function that returns the struct. The compiler assumes that the struct uses the C++ calling convention. Next is the struct definition, which uses the C calling convention by default. Because the compiler doesn’t know the calling convention of the struct until it finishes reading the entire struct, the calling convention for the struct in the return type of get_c2 is also assumed to be C++.
The struct is followed by another function declaration that returns the struct. At this point, the compiler knows that the struct’s calling convention is C++. Similarly, the function pointer, which returns the struct, is defined after the struct definition. The compiler now knows the struct uses the C++ calling convention.
To resolve C2440 errors caused by incompatible calling conventions, declare functions that return a UDT after the UDT definition.
// C2440b.cpp struct MyStruct; MyStruct get_c1(); struct MyStruct { int i; static MyStruct get_C2(); }; MyStruct get_C3(); typedef MyStruct (*FC)(); FC fc1 = &get_c1; // C2440, line 15 FC fc2 = &MyStruct::get_C2; // C2440, line 16 FC fc3 = &get_C3; class CMyClass { public: explicit CMyClass( int iBar) throw() { } static CMyClass get_c2(); }; int main() { CMyClass myclass = 2; // C2440 // try one of the following // CMyClass myclass{2}; // CMyClass myclass(2); int *i; float j; j = (float)i; // C2440, cannot cast from pointer to int to float }
Assign zero to interior pointer
C2440 can also occur if you assign zero to an interior pointer:
// C2440c.cpp // compile with: /clr int main() { array<int>^ arr = gcnew array<int>(100); interior_ptr<int> ipi = &arr[0]; ipi = 0; // C2440 ipi = nullptr; // OK }
User-defined conversions
C2440 can also occur for an incorrect use of a user-defined conversion. For example, when a conversion operator has been defined as explicit, the compiler can’t use it in an implicit conversion. For more information about user-defined conversions, see User-Defined Conversions (C++/CLI)). This sample generates C2440:
// C2440d.cpp // compile with: /clr value struct MyDouble { double d; // convert MyDouble to Int32 static explicit operator System::Int32 ( MyDouble val ) { return (int)val.d; } }; int main() { MyDouble d; int i; i = d; // C2440 // Uncomment the following line to resolve. // i = static_cast<int>(d); }
System::Array creation
C2440 can also occur if you try to create an instance of an array in C++/CLI whose type is a xref:System.Array. For more information, see Arrays. The next sample generates C2440:
// C2440e.cpp // compile with: /clr using namespace System; int main() { array<int>^ intArray = Array::CreateInstance(__typeof(int), 1); // C2440 // try the following line instead // array<int>^ intArray = safe_cast<array<int> ^>(Array::CreateInstance(__typeof(int), 1)); }
Attributes
C2440 can also occur because of changes in the attributes feature. The following sample generates C2440.
// c2440f.cpp // compile with: /LD [ module(name="PropDemoLib", version=1.0) ]; // C2440 // try the following line instead // [ module(name="PropDemoLib", version="1.0") ];
Component extensions down casts
The Microsoft C++ compiler no longer allows the const_cast operator to down cast when you compile source code under /clr.
To resolve this C2440, use the correct cast operator. For more information, see Casting operators.
This sample generates C2440:
// c2440g.cpp // compile with: /clr ref class Base {}; ref class Derived : public Base {}; int main() { Derived ^d = gcnew Derived; Base ^b = d; d = const_cast<Derived^>(b); // C2440 d = dynamic_cast<Derived^>(b); // OK }
Conforming template match changes
C2440 can occur because of conformance changes to the compiler in Visual Studio 2015 Update 3. Previously, the compiler incorrectly treated certain distinct expressions as the same type when identifying a template match for a static_cast operation. Now the compiler distinguishes the types correctly, and code that relied on the previous static_cast behavior is broken. To fix this issue, change the template argument to match the template parameter type, or use a reinterpret_cast or C-style cast.
This sample generates C2440:
// c2440h.cpp template<int *a> struct S1 {}; int g; struct S2 : S1<&g> { }; int main() { S2 s; static_cast<S1<&*&g>>(s); // C2440 in VS 2015 Update 3 // This compiles correctly: // static_cast<S1<&g>>(s); }
This error can appear in ATL code that uses the SINK_ENTRY_INFO macro defined in <atlcom.h>.
Copy-list-initialization
Visual Studio 2017 and later correctly raise compiler errors related to object creation using initializer lists. These errors weren’t caught in Visual Studio 2015 and could lead to crashes or undefined runtime behavior. In C++17 copy-list-initialization, the compiler is required to consider an explicit constructor for overload resolution, but must raise an error if that overload is actually chosen.
The following example compiles in Visual Studio 2015 but not in Visual Studio 2017.
// C2440j.cpp struct A { explicit A(int) {} A(double) {} }; int main() { const A& a2 = { 1 }; // error C2440: 'initializing': cannot // convert from 'int' to 'const A &' }
To correct the error, use direct initialization:
// C2440k.cpp struct A { explicit A(int) {} A(double) {} }; int main() { const A& a2{ 1 }; }
cv-qualifiers in class construction
In Visual Studio 2015, the compiler sometimes incorrectly ignores the cv-qualifier when generating a class object via a constructor call. This defect can potentially cause a crash or unexpected runtime behavior. The following example compiles in Visual Studio 2015 but raises a compiler error in Visual Studio 2017 and later:
struct S { S(int); operator int(); }; int i = (const S)0; // error C2440
To correct the error, declare operator int() as const.
Столкнулся с ошибкой при выделении памяти под структуру в Си. Вот строчка, на которую указывает ошибка компиляции:
RGBTRIPLE* StringPixels = malloc((abs(biCopy.biWidth) * sizeof(RGBTRIPLE)));
RGBTRIPLE это структура для пикселей:
typedef struct
{
BYTE rgbtBlue;
BYTE rgbtGreen;
BYTE rgbtRed;
} RGBTRIPLE;
Если кому не понятно, могу кинуть полный код. Сам код пробую скомпилировать на VS 2012.
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αλεχολυτ
28.2k10 золотых знаков52 серебряных знака114 бронзовых знаков
задан 31 июл 2016 в 17:36
Max_AdvancedMax_Advanced
3011 золотой знак3 серебряных знака12 бронзовых знаков
malloc возвращает указатель типа void * для последующего использования его необходимо явно привести в нужный тип, например так:
RGBTRIPLE* StringPixels = (RGBTRIPLE*) malloc((abs(biCopy.biWidth) * sizeof(RGBTRIPLE)));
ответ дан 31 июл 2016 в 17:37
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1
Ваша проблема в том, что Вы компилируете сишный код компилятором c++. В c допускается неявное преобразование void* (который возвращает malloc) в T*, а вот в c++ это уже запрещено.
Поэтому, если код именно сишный, нужно и исходник компилировать как сишный. Для Visual Studio для этого обычно достаточно изменить расширение файла на .c.
Если же требуется обеспечить сборку в режиме c++, то требуется явное приведение типов, о котором уже сказано в другом ответе.
ответ дан 31 июл 2016 в 17:52
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αλεχολυταλεχολυτ
28.2k10 золотых знаков52 серебряных знака114 бронзовых знаков
Это отличается от аналогичных вопросов, потому что я устанавливаю значение указателя на адрес, вместо того, чтобы пытаться назначить несовместимый тип … Я думаю.
template <class Type>
class ArrayStack
{
private:
int sz; // stack size
int asz; // array size (implementation)
Type* start; // address of first element
Type arr[]; // Might need to intialize each element to 0!?
public:
ArrayStack() { sz = 0; arr[0] = 0; asz = 0; start = &arr; }
/* other code... */
};
-1
Решение
start = arr; должен сделать свое дело.
- Вы можете назначить массив указателю, и указатель будет установлен на начало массива.
Кроме того, спецификация пустого массива:
Type arr[];
Не уверен, что это значит. Вероятно, так же, как:
Type arr[0];
Более нормально:
Type arr[asz];
Конечно, размер массива должен быть постоянным.
0
Другие решения
Предложить использование std::vector<Type> arr вместо Type arr[],
template <class Type>
class ArrayStack
{
private:
int sz; // stack size
int asz; // array size (implementation)
// Type* start; // address of first element
// Don't need this at all.
// You can use &arr[0] any time you need a pointer to the
// first element.
std::vector<Type> arr;
public:
// Simplified constructor.
ArrayStack() : sz(0), asz(0), arr(1, 0) {}
/* other code... */
};
0
I am converting my project from a 32 bit version to 64 bit version, when compiling the project in 64 bit, i get the below error.
Error 2 error C2440: 'static_cast' : cannot convert from 'long (__cdecl BrnDefDlgClass::* )(UINT,long)' to 'LRESULT (__cdecl CWnd::* )(WPARAM,LPARAM)' C:Program Files (x86)...CXXTEST.CXX 854 1 cfg
this error points me to below line of code.
ON_MESSAGE (WM_USER_AFTERCBNSELCHANGE, OnAfterCBSelectMsg)
advice please why would this error occur.
Thanks
asked Oct 15, 2012 at 7:23
ARVARV
1,1015 gold badges22 silver badges46 bronze badges
3
I suggest you look at the definition of OnAfterCBSelectMsg and modify it to use LRESULT as return type, and WPARAM and LPARAM as parameters (just as the second signature in your error message).
LRESULT is defined as LONG_PTR, and whenever PTR appears in a name, you should expect it to be large enough to hold a pointer. Pointers are 4 bytes on 32 Bit and 8 bytes on 64 bit, so an LRESULT won’t fit into a long (which is 4 bytes) if you compile for 64 bit.
Using WPARAM and LPARAM as parameter types is also important, as their size changes depending on the platform, too. See the answers to this question for more info: What are the definitions for LPARAM and WPARAM?
answered Oct 15, 2012 at 8:14
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Botz3000Botz3000
38.6k8 gold badges103 silver badges126 bronze badges
WPARAM is defined as unsigned __int64, LRESULT and LPARAM are __int64 in 64-bit Windows.
You have to change the signature of BrnDefDlgClass accordingly. If the WinAPI-provided typedefs were used instead of built-in types, it would be compilable on both 32 and 64-bit platforms.
answered Oct 15, 2012 at 8:15
AndreyAndrey
8,4263 gold badges26 silver badges51 bronze badges
I am converting my project from a 32 bit version to 64 bit version, when compiling the project in 64 bit, i get the below error.
Error 2 error C2440: 'static_cast' : cannot convert from 'long (__cdecl BrnDefDlgClass::* )(UINT,long)' to 'LRESULT (__cdecl CWnd::* )(WPARAM,LPARAM)' C:Program Files (x86)...CXXTEST.CXX 854 1 cfg
this error points me to below line of code.
ON_MESSAGE (WM_USER_AFTERCBNSELCHANGE, OnAfterCBSelectMsg)
advice please why would this error occur.
Thanks
asked Oct 15, 2012 at 7:23
ARVARV
1,1015 gold badges22 silver badges46 bronze badges
3
I suggest you look at the definition of OnAfterCBSelectMsg and modify it to use LRESULT as return type, and WPARAM and LPARAM as parameters (just as the second signature in your error message).
LRESULT is defined as LONG_PTR, and whenever PTR appears in a name, you should expect it to be large enough to hold a pointer. Pointers are 4 bytes on 32 Bit and 8 bytes on 64 bit, so an LRESULT won’t fit into a long (which is 4 bytes) if you compile for 64 bit.
Using WPARAM and LPARAM as parameter types is also important, as their size changes depending on the platform, too. See the answers to this question for more info: What are the definitions for LPARAM and WPARAM?
answered Oct 15, 2012 at 8:14
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Botz3000Botz3000
38.6k8 gold badges103 silver badges126 bronze badges
WPARAM is defined as unsigned __int64, LRESULT and LPARAM are __int64 in 64-bit Windows.
You have to change the signature of BrnDefDlgClass accordingly. If the WinAPI-provided typedefs were used instead of built-in types, it would be compilable on both 32 and 64-bit platforms.
answered Oct 15, 2012 at 8:15
AndreyAndrey
8,4263 gold badges26 silver badges51 bronze badges
87
28 апреля 2009 года
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Kogrom
2.7K / / 02.02.2008
Лучше все-таки изучить указатели. Особо не вникал — поправил чтоб только заработало:
Код:
#include <iostream>
#include <string>
using namespace std;
class String //user-defined string type
{
protected:
enum { SZ = 80 }; //size of all String objects
char str[SZ]; //holds a C-string
public:
String() //no-arg constructor
{ str[0] = »; }
String( char s[] ) //1-arg constructor
{ strcpy(str, s); } // convert C-string to String
void display() const //display the String
{ cout << str; }
operator char*() //conversion operator
{ return str; } //convert String to C-string
};
// Если длина ‘s’ больше 79, копируем только 80 символов
// Если меньше просто копируем s в str
class Pstring: public String
{
public:
Pstring():String()
{}
Pstring( char s[] )
{
if (strlen(s)>SZ-1)
{
for (int j=0; j<SZ-1; j++)
{
str[j]=s[j];
str[j+1]=»;
}
}
else
{
strcpy(str, s);
}
}
String& left(Pstring st, int n)
{
for (int i=0; i<=n; i++)
{
str = st.str;
str[i+1]=»;
}
return *this;
}
String& mid(Pstring st, int s, int n)
{
int u=0;
int start = s-1; // 9-1=8
int end = (n+start)-1; // 18+8=26
for (int i=start; i<=end; i++)
{
str = st.str;
str[i+1]=»;
u++;
}
return *this;
}
String& right(Pstring st, int n)
{
int u = 0;
//int len = strlen(st.str); // 54 — кол-во символов (54=»)
int start = strlen(st.str)-n; // 49 — первый копируемый символ ‘U’ser
int end = strlen(st.str)-1; // 53 — последный ‘.’
for (int i=start; i<=end; i++) // start=48, end=53;
{
str = st.str;
str[u+1]=»;
u++;
}
return *this;
}
};
////////////////////////////////////////////////////////////////
int main()
{
Pstring s1 = «It is a pleasure to welcome you as a new Premium User.»; // кол-во символов 54 (0…53)
Pstring s2,s3,s4;
s2.left(s1,7).display();
cout << endl;
s3.mid(s1,9,18).display();
cout << endl;
s4.right(s1,5).display();
cout << endl;
return 0;
}