I have this piece of code that works in C but not C++, is that any way to make it work on both C and C++ ?
void foo(void* b)
{
int *c = b;
printf("%d\n",*c);
}
int main ()
{
int a = 1000;
foo(&a);
return 0;
}
output:
C++:
1 In function 'void foo(void*)':
2 Line 3: error: invalid conversion from 'void*' to 'int*'
3 compilation terminated due to -Wfatal-errors.
C:
1000
Please help
invalid conversion from void* to int*
In order to make an conversion from void* to int* you will need to cast b as int* when assigning it to c. Do:
int *c = (int*) b;
This works in C++ and in C.
C allows implicit conversion between void* and any pointer to object type, C++ does not. To make your code compatible with both languages, you could type foo( (void*)&a );.
However, using void pointers is discouraged in both languages - they should only be used as a last resort. If you want the function to be type-generic in C, you'd use the _Generic keyword. In C++ you'd use templates.
Considering all the casting issues with both the languages, the correct way would be -
#ifdef __cplusplus
#define cast_to_intp(x) static_cast<int*>(x)
#else
#define cast_to_intp(x) (x)
#endif
And then use
int *c = cast_to_intp(b);
Upgrading some legacy code, I've started getting "warning: passing NULL to non-pointer argument 1 of ‘Identifier::Identifier(int)’ [-Wconversion-null]" messages from g++. Mostly this is good, but it doesn't seem to take classes with multiple constructors into account.
For example, take this test code:
#include <stdio.h>
class Identifier
{
private:
int m_iID;
public:
Identifier(const char* p_c) { m_iID = p_c ? p_c[0] : 0; }
Identifier(int i) { m_iID = i; }
};
int main(int argc, char* argv[])
{
Identifier* p_ID = new Identifier(NULL);
return 0;
}
(Please ignore what the constructors actually do, this is just an illustration of the issue. For context, the code in question is a class that stores an identifier in the form of a hash value. The constructors can either take a string to convert to a hash, or the hash value directly.)
The "new" statement on the third to last line throws this warning. The thing that is puzzling me is that g++ is apparently assuming that it needs to use the Identifier(int i) constructor, and ignores the Identifier(const char* p_c) constructor, which is a pointer.
Note that changing the ints to unsigned ints causes an ambiguity error instead, this being a 32-bit system.
I know specifying -Wno-conversion-null would fix the problem, as would passing 0 or explicitly casting NULL to const char*. But I'm curious as to why the seemingly valid constructor is being ignored. Plus, I'd like to avoid a massive search-and-replace job whilst keeping the warning active.
At least on this CentOS box, NULL is defined in linux/stddef.h:
#undef NULL
#if defined(__cplusplus)
#define NULL 0
#else
#define NULL ((void *)0)
#endif
As such, NULL in C++ is an integer; therefore the compiler chooses the int constructor by default and gives you the warning you're seeing.
I am interested in Judy Arrays and try to use it. But i had unable to do any useful thing using it. Every time it gives me casting errors.. Sample c++ code and the error given below.
#include "Judy.h"
#include <iostream>
using namespace std;
int main()
{
int Rc_int; // return code - integer
Word_t Rc_word; // return code - unsigned word
Word_t Index = 12, Index1 = 34, Index2 = 55, Nth;
Word_t PValue; // pointer to return value
//Pvoid_t PJLArray = NULL; // initialize JudyL array
Pvoid_t JudyArray = NULL;
char String[100];
PWord_t _PValue;
JSLI( JudyArray, _PValue, (uint8_t *) String);
return(0);
} // main()
This gives me the error
m.cpp: In function ‘int main()’:
m.cpp:19: error: invalid conversion from ‘long unsigned int**’ to ‘void**’
m.cpp:19: error: initializing argument 1 of ‘void** JudySLIns(void**, const uint8_t*, J_UDY_ERROR_STRUCT*)’
Please anyone help me to figure out what is the error what i'm doing..
Thanks
According to the documentation, you have the _PValue and JudyArray parameters reversed. Make your call look like this:
JSLI( _PValue, JudyArray, (uint8_t *) String);
Also, try not compiling it as C++ code. So far, your test uses no C++ features. I bet it will compile as C code. It looks like JudyArray relies on the fact that C will do certain kinds of implicit conversions between void * and other pointer types.
If this is the case, I'm not sure what to do about it. The error messages you're getting tell me that JSLI is a macro. In order to fix the error message you have in the comments on this answer, you'd have to reach inside the macro and add a typecast.
These kinds of implicit conversions are allowed in C because otherwise using malloc would always require ugly casts. C++ purposely disallows them because the semantics of new make the requirement that the result of malloc be cast to the correct type unimportant.
I don't think this library can be used effectively in C++ for this reason.
It seems that, you pass JudySLIns(void**, const uint8_t*, J_UDY_ERROR_STRUCT*) a wrong parameter, the first one, you'b better check it!
For integer keys there is a C++ wrapper at http://judyhash.sourceforge.net/
Let's say I have a macro called LengthOf(array):
sizeof array / sizeof array[0]
When I make a new array of size 23, shouldn't I get 23 back for LengthOf?
WCHAR* str = new WCHAR[23];
str[22] = '\0';
size_t len = LengthOf(str); // len == 4
Why does len == 4?
UPDATE: I made a typo, it's a WCHAR*, not a WCHAR**.
Because str here is a pointer to a pointer, not an array.
This is one of the fine differences between pointers and arrays: in this case, your pointer is on the stack, pointing to the array of 23 characters that has been allocated elsewhere (presumably the heap).
WCHAR** str = new WCHAR[23];
First of all, this shouldn't even compile -- it tries to assign a pointer to WCHAR to a pointer to pointer to WCHAR. The compiler should reject the code based on this mismatch.
Second, one of the known shortcomings of the sizeof(array)/sizeof(array[0]) macro is that it can and will fail completely when applied to a pointer instead of a real array. In C++, you can use a template to get code like this rejected:
#include <iostream>
template <class T, size_t N>
size_t size(T (&x)[N]) {
return N;
}
int main() {
int a[4];
int *b;
b = ::new int[20];
std::cout << size(a); // compiles and prints '4'
// std::cout << size(b); // uncomment this, and the code won't compile.
return 0;
}
As others have pointed out, the macro fails to work properly if a pointer is passed to it instead of an actual array. Unfortunately, because pointers and arrays evaluate similarly in most expressions, the compiler isn't able to let you know there's a problem unless you make you macro somewhat more complex.
For a C++ version of the macro that's typesafe (will generate an error if you pass a pointer rather than an array type), see:
Compile time sizeof_array without using a macro
It wouldn't exactly 'fix' your problem, but it would let you know that you're doing something wrong.
For a macro that works in C and is somewhat safer (many pointers will diagnose as an error, but some will pass through without error - including yours, unfortunately):
Is there a standard function in C that would return the length of an array?
Of course, using the power of #ifdef __cplusplus you can have both in a general purpose header and have the compiler select the safer one for C++ builds and the C-compatible one when C++ isn't in effect.
The problem is that the sizeof operator checks the size of it's argument. The argument passed in your sample code is WCHAR*. So, the sizeof(WCHAR*) is 4. If you had an array, such as WCHAR foo[23], and took sizeof(foo), the type passed is WCHAR[23], essentially, and would yield sizeof(WCHAR) * 23. Effectively at compile type WCHAR* and WCHAR[23] are different types, and while you and I can see that the result of new WCHAR[23] is functionally equivalent to WCHAR[23], in actuality, the return type is WCHAR*, with absolutely no size information.
As a corellary, since sizeof(new WCHAR[23]) equals 4 on your platform, you're obviously dealing with an architecture where a pointer is 4 bytes. If you built this on an x64 platform, you'd find that sizeof(new WCHAR[23]) will return 8.
You wrote:
WCHAR* str = new WCHAR[23];
if 23 is meant to be a static value, (not variable in the entire life of your program) it's better use #define or const than just hardcoding 23.
#define STR_LENGTH 23
WCHAR* str = new WCHAR[STR_LENGTH];
size_t len = (size_t) STR_LENGTH;
or C++ version
const int STR_LENGTH = 23;
WCHAR* str = new WCHAR[STR_LENGTH];
size_t len = static_cast<size_t>(STR_LENGTH);
C and C++ have many differences, and not all valid C code is valid C++ code.
(By "valid" I mean standard code with defined behavior, i.e. not implementation-specific/undefined/etc.)
Is there any scenario in which a piece of code valid in both C and C++ would produce different behavior when compiled with a standard compiler in each language?
To make it a reasonable/useful comparison (I'm trying to learn something practically useful, not to try to find obvious loopholes in the question), let's assume:
Nothing preprocessor-related (which means no hacks with #ifdef __cplusplus, pragmas, etc.)
Anything implementation-defined is the same in both languages (e.g. numeric limits, etc.)
We're comparing reasonably recent versions of each standard (e.g. say, C++98 and C90 or later)
If the versions matter, then please mention which versions of each produce different behavior.
Here is an example that takes advantage of the difference between function calls and object declarations in C and C++, as well as the fact that C90 allows the calling of undeclared functions:
#include <stdio.h>
struct f { int x; };
int main() {
f();
}
int f() {
return printf("hello");
}
In C++ this will print nothing because a temporary f is created and destroyed, but in C90 it will print hello because functions can be called without having been declared.
In case you were wondering about the name f being used twice, the C and C++ standards explicitly allow this, and to make an object you have to say struct f to disambiguate if you want the structure, or leave off struct if you want the function.
For C++ vs. C90, there's at least one way to get different behavior that's not implementation defined. C90 doesn't have single-line comments. With a little care, we can use that to create an expression with entirely different results in C90 and in C++.
int a = 10 //* comment */ 2
+ 3;
In C++, everything from the // to the end of the line is a comment, so this works out as:
int a = 10 + 3;
Since C90 doesn't have single-line comments, only the /* comment */ is a comment. The first / and the 2 are both parts of the initialization, so it comes out to:
int a = 10 / 2 + 3;
So, a correct C++ compiler will give 13, but a strictly correct C90 compiler 8. Of course, I just picked arbitrary numbers here -- you can use other numbers as you see fit.
The following, valid in C and C++, is going to (most likely) result in different values in i in C and C++:
int i = sizeof('a');
See Size of character ('a') in C/C++ for an explanation of the difference.
Another one from this article:
#include <stdio.h>
int sz = 80;
int main(void)
{
struct sz { char c; };
int val = sizeof(sz); // sizeof(int) in C,
// sizeof(struct sz) in C++
printf("%d\n", val);
return 0;
}
C90 vs. C++11 (int vs. double):
#include <stdio.h>
int main()
{
auto j = 1.5;
printf("%d", (int)sizeof(j));
return 0;
}
In C auto means local variable. In C90 it's ok to omit variable or function type. It defaults to int. In C++11 auto means something completely different, it tells the compiler to infer the type of the variable from the value used to initialize it.
Another example that I haven't seen mentioned yet, this one highlighting a preprocessor difference:
#include <stdio.h>
int main()
{
#if true
printf("true!\n");
#else
printf("false!\n");
#endif
return 0;
}
This prints "false" in C and "true" in C++ - In C, any undefined macro evaluates to 0. In C++, there's 1 exception: "true" evaluates to 1.
Per C++11 standard:
a. The comma operator performs lvalue-to-rvalue conversion in C but not C++:
char arr[100];
int s = sizeof(0, arr); // The comma operator is used.
In C++ the value of this expression will be 100 and in C this will be sizeof(char*).
b. In C++ the type of enumerator is its enum. In C the type of enumerator is int.
enum E { a, b, c };
sizeof(a) == sizeof(int); // In C
sizeof(a) == sizeof(E); // In C++
This means that sizeof(int) may not be equal to sizeof(E).
c. In C++ a function declared with empty params list takes no arguments. In C empty params list mean that the number and type of function params is unknown.
int f(); // int f(void) in C++
// int f(*unknown*) in C
This program prints 1 in C++ and 0 in C:
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int d = (int)(abs(0.6) + 0.5);
printf("%d", d);
return 0;
}
This happens because there is double abs(double) overload in C++, so abs(0.6) returns 0.6 while in C it returns 0 because of implicit double-to-int conversion before invoking int abs(int). In C, you have to use fabs to work with double.
#include <stdio.h>
int main(void)
{
printf("%d\n", (int)sizeof('a'));
return 0;
}
In C, this prints whatever the value of sizeof(int) is on the current system, which is typically 4 in most systems commonly in use today.
In C++, this must print 1.
Another sizeof trap: boolean expressions.
#include <stdio.h>
int main() {
printf("%d\n", (int)sizeof !0);
}
It equals to sizeof(int) in C, because the expression is of type int, but is typically 1 in C++ (though it's not required to be). In practice they are almost always different.
An old chestnut that depends on the C compiler, not recognizing C++ end-of-line comments...
...
int a = 4 //* */ 2
+2;
printf("%i\n",a);
...
The C++ Programming Language (3rd Edition) gives three examples:
sizeof('a'), as #Adam Rosenfield mentioned;
// comments being used to create hidden code:
int f(int a, int b)
{
return a //* blah */ b
;
}
Structures etc. hiding stuff in out scopes, as in your example.
Another one listed by the C++ Standard:
#include <stdio.h>
int x[1];
int main(void) {
struct x { int a[2]; };
/* size of the array in C */
/* size of the struct in C++ */
printf("%d\n", (int)sizeof(x));
}
Inline functions in C default to external scope where as those in C++ do not.
Compiling the following two files together would print the "I am inline" in case of GNU C but nothing for C++.
File 1
#include <stdio.h>
struct fun{};
int main()
{
fun(); // In C, this calls the inline function from file 2 where as in C++
// this would create a variable of struct fun
return 0;
}
File 2
#include <stdio.h>
inline void fun(void)
{
printf("I am inline\n");
}
Also, C++ implicitly treats any const global as static unless it is explicitly declared extern, unlike C in which extern is the default.
#include <stdio.h>
struct A {
double a[32];
};
int main() {
struct B {
struct A {
short a, b;
} a;
};
printf("%d\n", sizeof(struct A));
return 0;
}
This program prints 128 (32 * sizeof(double)) when compiled using a C++ compiler and 4 when compiled using a C compiler.
This is because C does not have the notion of scope resolution. In C structures contained in other structures get put into the scope of the outer structure.
struct abort
{
int x;
};
int main()
{
abort();
return 0;
}
Returns with exit code of 0 in C++, or 3 in C.
This trick could probably be used to do something more interesting, but I couldn't think of a good way of creating a constructor that would be palatable to C. I tried making a similarly boring example with the copy constructor, that would let an argument be passed, albeit in a rather non-portable fashion:
struct exit
{
int x;
};
int main()
{
struct exit code;
code.x=1;
exit(code);
return 0;
}
VC++ 2005 refused to compile that in C++ mode, though, complaining about how "exit code" was redefined. (I think this is a compiler bug, unless I've suddenly forgotten how to program.) It exited with a process exit code of 1 when compiled as C though.
Don't forget the distinction between the C and C++ global namespaces. Suppose you have a foo.cpp
#include <cstdio>
void foo(int r)
{
printf("I am C++\n");
}
and a foo2.c
#include <stdio.h>
void foo(int r)
{
printf("I am C\n");
}
Now suppose you have a main.c and main.cpp which both look like this:
extern void foo(int);
int main(void)
{
foo(1);
return 0;
}
When compiled as C++, it will use the symbol in the C++ global namespace; in C it will use the C one:
$ diff main.cpp main.c
$ gcc -o test main.cpp foo.cpp foo2.c
$ ./test
I am C++
$ gcc -o test main.c foo.cpp foo2.c
$ ./test
I am C
int main(void) {
const int dim = 5;
int array[dim];
}
This is rather peculiar in that it is valid in C++ and in C99, C11, and C17 (though optional in C11, C17); but not valid in C89.
In C99+ it creates a variable-length array, which has its own peculiarities over normal arrays, as it has a runtime type instead of compile-time type, and sizeof array is not an integer constant expression in C. In C++ the type is wholly static.
If you try to add an initializer here:
int main(void) {
const int dim = 5;
int array[dim] = {0};
}
is valid C++ but not C, because variable-length arrays cannot have an initializer.
Empty structures have size 0 in C and 1 in C++:
#include <stdio.h>
typedef struct {} Foo;
int main()
{
printf("%zd\n", sizeof(Foo));
return 0;
}
This concerns lvalues and rvalues in C and C++.
In the C programming language, both the pre-increment and the post-increment operators return rvalues, not lvalues. This means that they cannot be on the left side of the = assignment operator. Both these statements will give a compiler error in C:
int a = 5;
a++ = 2; /* error: lvalue required as left operand of assignment */
++a = 2; /* error: lvalue required as left operand of assignment */
In C++ however, the pre-increment operator returns an lvalue, while the post-increment operator returns an rvalue. It means that an expression with the pre-increment operator can be placed on the left side of the = assignment operator!
int a = 5;
a++ = 2; // error: lvalue required as left operand of assignment
++a = 2; // No error: a gets assigned to 2!
Now why is this so? The post-increment increments the variable, and it returns the variable as it was before the increment happened. This is actually just an rvalue. The former value of the variable a is copied into a register as a temporary, and then a is incremented. But the former value of a is returned by the expression, it is an rvalue. It no longer represents the current content of the variable.
The pre-increment first increments the variable, and then it returns the variable as it became after the increment happened. In this case, we do not need to store the old value of the variable into a temporary register. We just retrieve the new value of the variable after it has been incremented. So the pre-increment returns an lvalue, it returns the variable a itself. We can use assign this lvalue to something else, it is like the following statement. This is an implicit conversion of lvalue into rvalue.
int x = a;
int x = ++a;
Since the pre-increment returns an lvalue, we can also assign something to it. The following two statements are identical. In the second assignment, first a is incremented, then its new value is overwritten with 2.
int a;
a = 2;
++a = 2; // Valid in C++.