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++.
I was trying to explore the option of "solveInPlace()" function while using LLT in Eigen3.3.7 to speed up the matrix inverse computation in my application.
I used the following code to test it.
int main()
{
const int M=3;
Eigen::Matrix<MyType,Eigen::Dynamic,Eigen::Dynamic> R = Eigen::Matrix<MyType,Eigen::Dynamic,Eigen::Dynamic>::Zero(M,M);
// to make sure full rank
for(int i=0; i<M*2; i++)
{
const Eigen::Matrix<MyType, Eigen::Dynamic,1> tmp = Eigen::Matrix<MyType,Eigen::Dynamic,1>::Random(M);
R += tmp*tmp.transpose();
}
std::cout<<"R \n";
std::cout<<R<<std::endl;
decltype (R) R0 = R; // saving for later comparison
Eigen::LLT<Eigen::Ref<Eigen::Matrix<MyType,Eigen::Dynamic,Eigen::Dynamic> > > myllt(R);
const Eigen::Matrix<MyType,Eigen::Dynamic,Eigen::Dynamic> I = Eigen::Matrix<MyType,Eigen::Dynamic,Eigen::Dynamic>::Identity(R.rows(), R.cols());
myllt.solveInPlace(I);
std::cout<<"I: "<<I<<std::endl;
std::cout<<"Prod InPlace: \n"<<R0*I<<std::endl;
return 0;
}
After reading the Eigen documentation, I thought that the input matrix (here "R") will be modified while computing the transform. To my surprise, I found that the results is store in "I". This was not expected as I defined "I" as a constant. Please provide an explanation for this behaviour.
The simple non-compiler answer would be that you're asking for the LLT to solve in-place (i.e. in the passed parameter) so what would you expect the result to be? Apparently, you would expect it to be a compiler error, as the "in-place" means change the parameter, but you're passing a const object.
So, if we search the Eigen docs for solveInPlace, we find the only item that takes a const reference to have the following note:
"in-place" version of TriangularView::solve() where the result is written in other
Warning
The parameter is only marked 'const' to make the C++ compiler accept a temporary expression here. This function will const_cast it, so constness isn't honored here.
The non-in-place option would be:
R = myllt.solve(I);
but that won't really speed up the calculation. In any case, benchmark before you decide that you need the in-place option.
You're question is in place, as what const_cast is meant to do is strip references/pointers of their const-ness iff the underlying variable is not const qualified* (cppref). If you were to write some examples
const int i = 4;
int& iRef = const_cast<int&>(i); // UB, i is actually const
std::cout << i; // Prints "I want coffee", or it can as we like UB
int j = 4;
const int& jRef = j;
const_cast<int&>(jRef)++; // Legal. Underlying variable is not const.
std::cout << j; // Prints 5
The case with i may well work as expected or not, we're dependent on each implementation/compiler. It may work with gcc but not with clang or MSVC. There are no guarantees. As you are indirectly invoking UB in your example, the compiler can choose to do what you expect or something else entirely.
*Technically it's the modification that's UB, not the const_cast itself.
I am now studying C++ 11 and getting confused by value category of expressions in C++ 11. According to terminology the Lvalue is the top-left point of the W, that is iM (or i-m sometimes) meaning that "has identity but cannot be moved from". This really makes me confused. Please consider the example below:
#include <iostream>
int main()
{
int a = 0, b = 1, c = 2;
a = std::move(b = c);
std::cout << a << '\n';
}
This example compiled well.
We all know that the assignment b = c is an Lvalue then what do they mean by 'cannot be moved from'? Please give examples that can clearly illustrate this!
Thanks!
Roughly speaking:
an lvalue has identity because you can take its address
int x;
&x; // ok
an lvalue cannot be moved from because it cannot be used as an argument to move constructor/assignment
struct Foo
{
Foo(Foo&); // 0
Foo(Foo&&); // 1
};
Foo x;
Foo y{x}; // calls 0, not 1
in the example above, x is an lvalue: the copy constructor is invoked. If you want to move from x, you need to make it an rvalue: this is why you must use std::move(x) to cast it to an rvalue reference.
Foo y{std::move(x)}; // calls 1
In your example std::move(b = c) is an rvalue, as std::move is literally just a static_cast to an rvalue reference.
I'm developing an accelerated component in OpenCL, using Xcode 4.5.1 and Grand Central Dispatch, guided by this tutorial.
The full kernel kept failing on the GPU, giving signal SIGABRT. I couldn't make much progress interpreting the error beyond that.
But I broke out aspects of the kernel to test, and I found something very peculiar involving assigning certain values to positions in an array within a loop.
Test scenario: give each thread a fixed range of array indices to initialize.
kernel void zero(size_t num_buckets, size_t positions_per_bucket, global int* array) {
size_t bucket_index = get_global_id(0);
if (bucket_index >= num_buckets) return;
for (size_t i = 0; i < positions_per_bucket; i++)
array[bucket_index * positions_per_bucket + i] = 0;
}
The above kernel fails. However, when I assign 1 instead of 0, the kernel succeeds (and my host code prints out the array of 1's). Based on a handful of tests on various integer values, I've only had problems with 0 and -1.
I've tried to outsmart the compiler with 1-1, (int) 0, etc, with no success. Passing zero in as a kernel argument worked though.
The assignment to zero does work outside of the context of a for loop:
array[bucket_index * positions_per_bucket] = 0;
The findings above were confirmed on two machines with different configurations. (OSX 10.7 + GeForce, OSX 10.8 + Radeon.) Furthermore, the kernel had no trouble when running on CL_DEVICE_TYPE_CPU -- it's just on the GPU.
Clearly, something ridiculous is happening, and it must be on my end, because "zero" can't be broken. Hopefully it's something simple. Thank you for your help.
Host code:
#include <stdio.h>
#include <OpenCL/OpenCL.h>
#include "zero.cl.h"
int main(int argc, const char* argv[]) {
dispatch_queue_t queue = gcl_create_dispatch_queue(CL_DEVICE_TYPE_GPU, NULL);
size_t num_buckets = 64;
size_t positions_per_bucket = 4;
cl_int* h_array = malloc(sizeof(cl_int) * num_buckets * positions_per_bucket);
cl_int* d_array = gcl_malloc(sizeof(cl_int) * num_buckets * positions_per_bucket, NULL, CL_MEM_WRITE_ONLY);
dispatch_sync(queue, ^{
cl_ndrange range = { 1, { 0 }, { num_buckets }, { 0 } };
zero_kernel(&range, num_buckets, positions_per_bucket, d_array);
gcl_memcpy(h_array, d_array, sizeof(cl_int) * num_buckets * positions_per_bucket);
});
for (size_t i = 0; i < num_buckets * positions_per_bucket; i++)
printf("%d ", h_array[i]);
printf("\n");
}
Refer to the OpenCL standard, section 6, paragraph 8 "Restrictions", bullet point k (emphasis mine):
6.8 k. Arguments to kernel functions in a program cannot be declared with the built-in scalar types bool, half, size_t, ptrdiff_t, intptr_t, and uintptr_t. [...]
The fact that your compiler even let you build the kernel at all indicates it is somewhat broken.
So you might want to fix that... but if that doesn't fix it, then it looks like a compiler bug, plain and simple (of CLC, that is, the OpenCL compiler, not your host code). There is no reason this kernel should work with any constant other than 0, -1. Did you try updating your OpenCL driver, what about trying on a different operating system (though I suppose this code is OS X only)?
I am confused by the behavior of pointer arithmetics in C++. I have an array and I want to go N elements forward from the current one. Since in C++ pointer is memory address in BYTES, it seemed logical to me that the code would be newaddr = curaddr + N * sizeof(mytype). It produced errors though; later I found that with newaddr = curaddr + N everything works correctly. Why so? Should it really be address + N instead of address + N * sizeof?
Part of my code where I noticed it (2D array with all memory allocated as one chunk):
// creating pointers to the beginning of each line
if((Content = (int **)malloc(_Height * sizeof(int *))) != NULL)
{
// allocating a single memory chunk for the whole array
if((Content[0] = (int *)malloc(_Width * _Height * sizeof(int))) != NULL)
{
// setting up line pointers' values
int * LineAddress = Content[0];
int Step = _Width * sizeof(int); // <-- this gives errors, just "_Width" is ok
for(int i=0; i<_Height; ++i)
{
Content[i] = LineAddress; // faster than
LineAddress += Step; // Content[i] = Content[0] + i * Step;
}
// everything went ok, setting Width and Height values now
Width = _Width;
Height = _Height;
// success
return 1;
}
else
{
// insufficient memory available
// need to delete line pointers
free(Content);
return 0;
}
}
else
{
// insufficient memory available
return 0;
}
Your error in reasoning is right here: "Since in C++ pointer is memory address in BYTES, [...]".
A C/C++ pointer is not a memory address in bytes. Sure, it is represented by a memory address, but you have to differentiate between a pointer type and its representation. The operation "+" is defined for a type, not for its representation. Therefore, when it is called one the type int *, it respects the semantics of this type. Therefore, + 1 on an int * type will advance the pointer as much bytes as the underlying int type representation uses.
You can of course cast your pointer like this: (int)myPointer. Now you have a numeric type (instead of a pointer type), where + 1 will work as you would expect from a numeric type. Note that after this cast, the representation stays the same, but the type changes.
A "pointer" points to a location.
When you "increment", you want to go to the next, adjacent location.
Q: "Next" and "adjacent" depend on the size of the object you're pointing to, don't they?
Q: When you don't use "sizeof()", everything works, correct? Why? What do you think the compiler is doing for you, "behind your back"?
Q: What do you think should happen if you add your own "sizeof()"?
ON TOP OF the "everything works" scenario?
pointers point to addresses, so incrementing the pointer p by N will point to the Nth block of memory from p.
Now, if you were using addresses instead of pointers to addresses, then it would be appropriate to add N*sizeof(type).