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:
#ifdef __cplusplus, pragmas, etc.)转载于:https://stackoverflow.com/questions/12887700/can-code-that-is-valid-in-both-c-and-c-produce-different-behavior-when-compile
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;
}
#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.
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.
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.
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));
}
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 allows 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.
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
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.
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 correct C compiler 8. Of course, I just picked arbitrary numbers here -- you can use other numbers as you see fit.
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.
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.
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.
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
#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.