c++ variable size stack array in function that get const - c++

I'm learning now C++ I'm reading the book Effective C++ (Scott Meyers).
In the book, there is an item about const variables, and I try to work with them.
I notice something very interesting that I what to know if it bug in C++:
(I'm working with C++98 standard)
void Test(const int i)
{
int arr[i] = {0};
for (int j = 0; i > j; ++j)
{
arr[j] = i;
}
}
This function will compile and work exactly as I want (create int array on the stack with the size of 'i'. When I remove the 'const' from 'i' it won't compile.
I try this on gcc and clang.
Edit:
link to Compiler Explorer

To catch this kind of mistake in the future the compiler flag you want for both g++ and clang++ is -pedantic. And always remember to specify your language standard or you don't know what you'll get.
$ g++ -std=c++98 -pedantic c++-vla.cpp -o c++-vla
c++-vla.cpp: In function ‘void f(size_t)’:
c++-vla.cpp:3:30: warning: ISO C++ forbids variable length array ‘g’ [-Wvla]
3 | void f(const size_t x) { int g[x]; }
| ^
$ clang++ -std=c++98 -pedantic c++-vla.cpp -o c++-vla
c++-vla.cpp:3:31: warning: variable length arrays are a C99 feature [-Wvla-extension]
void f(const size_t x) { int g[x]; }
^
1 warning generated.

First of all, const in your function signature is ignored by the compiler. So the following two are equivalent:
Test(const int i) {}
Test(int i) {}
Secondly, this isn't valid C++ regardless of whether it compiles or not:
int arr[i] = {0};
It isn't valid because i is not a compile time constant i.e., the value of i has to be known at the time of compilation.
Try on Compiler Explorer

Related

Why am I able to assign a function reference to an anonymous function pointer variable?

The following code compiles just fine and I'm not sure why. Can someone please explain to me why this is legal?
I am using g++ (Debian 6.1.1-10) 6.1.1 20160724 to compile.
#include <iostream>
int sum(int x, int y) { return x + y; }
int main(int argc, char *argv[])
{
using std::cout;
int (*) (int, int) = ∑
cout << "what" << '\n';
}
Addendum
The following program compiles fine using g++ version 5.4.0 but fails to compile in gcc.
int main()
{
int (*) = 20;
}
It's very likely to be related to this bug reported by Zack Weinberg:
Bug 68265 - Arbitrary syntactic nonsense silently accepted after 'int (*){}' until the next close brace
(From Why does this invalid-looking code compile successfully on g++ 6.0? :)
The C++ compiler fails to diagnose ill-formed constructs such as
int main()
{
int (*) {}
any amount of syntactic nonsense
on multiple lines, with *punctuation* and ++operators++ even...
will be silently discarded
until the next close brace
}
With -pedantic -std=c++98 you do get "warning: extended initializer
lists only available with -std=c++11 or -std=gnu++11", but with
-std=c++11, not a peep.
If any one (or more) of the tokens 'int ( * ) { }' are removed, you do
get an error. Also, the C compiler does not have the same bug.
Of course, if you try int (*) (int, int) {} or other variants, it erroneously compiles. The interesting thing is that the difference between this and the previous duplicate/bug reports is that int (*) (int, int) = asdf requires asdf to be a name in scope. But I highly doubt that the bugs are different in nature, since the core issue is that GCC is allowing you to omit a declarator-id.
[n4567 §7/8]: "Each init-declarator in the init-declarator-list
contains exactly one declarator-id, which is the name declared by
that init-declarator and hence one of the names declared by the
declaration."
Here's an oddity:
int (*) (int, int) = main;
In this specific scenario, GCC doesn't complain about taking the address of main (like arrays, &main is equivalent to main).

Confusion about pointer values being compile-time constatns

In C++, it is possible for pointer values to be compile-time constants. This is true, otherwise, non-type template parameters and constexpr won't work with pointers. However, as far as I know, addresses of functions and objects of static storage are known (at least) at link-time rather than compile-time. Following is an illustration:
main.cpp
#include <iostream>
template <int* p>
void f() { std::cout << p << '\n'; }
extern int a;
int main() {
f<&a>();
}
a.cpp
int a = 0;
I'm just wondering how the address of a could possibly be known when compiling main.cpp. I hope somebody could explain this a little to me.
In particular, consider this
template <int* p, int* pp>
constexpr std::size_t f() {
return (p + 1) == (pp + 7) ? 5 : 10;
}
int main() {
int arr[f<&a, &b>()] = {};
}
How should the storage for arr be allocated?
PLUS: This mechanism seems to be rather robust. Even when I enabled Randomized Base Address, the correct output is obtained.
The compiler doesn't need to know the value of &a at compile time any more than it needs the value of function addresses.
Think of it like this: the compiler will instantiate your function template with &a as a parameter and generate "object code" (in whatever format it uses to pass to the linker). The object code will look like (well it won't, but you get the idea):
func f__<funky_mangled_name_to_say_this_is_f_for_&a>__:
reg0 <- /* linker, pls put &std::cout here */
reg1 <- /* hey linker, stuff &a in there ok? */
call std::basic_stream::operator<<(int*) /* linker, fun addr please? */
[...]
If you instantiate f<b&>, assuming b is another global static, compiler does the same thing:
func f__<funky_mangled_name_to_say_this_is_f_for_&b>__:
reg0 <- /* linker, pls put &std::cout here */
reg1 <- /* hey linker, stuff &b in there ok? */
call std::basic_stream::operator<<(int*) /* linker, fun addr please? */
[...]
And when your code calls for calling either of those:
fun foo:
call f__<funky_mangled_name_to_say_this_is_f_for_&a>__
call f__<funky_mangled_name_to_say_this_is_f_for_&b>__
Which exact function to call is encoded in the mangled function name.
The generated code doesn't depend on the runtime value of &a or &b.
The compiler knows there will be such things at runtime (you told it so), that's all it needs. It'll let the linker fill in the blanks (or yell at you if you failed to deliver on your promise).
For your addition I'm afraid I'm not familiar enough about the constexpr rules, but the two compilers I have tell me that this function will be evaluated at runtime, which, according to them, makes the code non-conforming. (If they're wrong, then the answer above is, at least, incomplete.)
template <int* p, int* pp>
constexpr std::size_t f() {
return (p + 1) == (pp + 7) ? 5 : 10;
}
int main() {
int arr[f<&a, &b>()] = {};
}
clang 3.5 in C++14 standards conforming mode:
$ clang++ -std=c++14 -stdlib=libc++ t.cpp -pedantic
t.cpp:10:10: warning: variable length arrays are a C99 feature [-Wvla-extension]
int arr[f<&a, &b>()];
^
1 warning generated.
GCC g++ 5.1, same mode:
$ g++ -std=c++14 t.cpp -O3 -pedantic
t.cpp: In function 'int main()':
t.cpp:10:22: warning: ISO C++ forbids variable length array 'arr' [-Wvla]
int arr[f<&a, &b>()];
As far as I know, the variables of static storage and functions are stored simply as symbols/place holders in the symbol table while compiling. It is in the linking phase when the place holders are resolved.
The compiler outputs machine code keeping the placeholders intact. Then the linker replaces the placeholders of the variables / functions with their respective memory locations. So in this case too, if you just compile main.cpp without compiling a.cpp and linking with it, you are bound to face linker error, as you can see here http://codepad.org/QTdJCgle (I compiled main.cpp only)

Why does g++ 4.8.1 issue a conversion warning

When I compile the code below with g++ 4.8.1 (64bit) in this way:
$ g++ -Wconversion -o main main.cpp
I get this result:
main.cpp: In function ‘int main()’:
main.cpp:12:20: warning: conversion to ‘int’ from ‘long unsigned int’ may alter its value [-Wconversion]
int i = sizeof(x)/sizeof(x[0]);
^
My expectation would be that the compiler should be able to evaluate the expression at compile time. If you make a similar program in plain c, gcc works like a charm.
Should this be considered a bug in g++ (e.g. clang++ does not have this problem)?
if you change the problematic line to something like:
char c = 0x10000000/0x1000000;
then the compiler does not complain. This suggest that some constant evaluation is done before warning generation.
main.cpp:
#include <iostream>
struct foo {
int a;
int b;
};
foo x[50];
int main()
{
int i = sizeof(x)/sizeof(x[0]);
std::cout << i << std::endl;
return 0;
}
int i = sizeof(x)/sizeof(x[0]);
//int <-- std::size_t <-- std::size_t / std::size_t
The type of the expression sizeof(x)/sizeof(x[0]) is std::size_t which on your machine is unsigned long int. So conversion from this type to int is data-loss, if the source is bigger in size than the target.
Though, I agree that in your case, there would not be actual data-loss if the compiler actually computes the value, but I guess it applies -Wconversion before the actual computation.
sizeof() returns you std::size_t not int! So cast it or declare i as std::size_t.
std::size_t i = sizeof(x)/sizeof(x[0]);

Arrays with a size determined at run time, is this valid in C++?

So I was talking to my friend, helping her with a piece of code, and I always thought that arrays needed to be compile-time constants, as they are on the stack. But she said that her friend did this using this code:
#include <iostream.h>
#include <stdlib.h>
int main()
{
int value = ' ' ;
int sum = 0;
int count = 0;
cout<<"Please enter the total number of employees" <<endl;;
cin>> value;
int numbers[value];
cout<<"Now enter the employees corresponding salaries" <<endl;;
for (int k = 0; k < value; k++)
{
cin >> numbers[k];
}
}
They are using Dev-C++.
Is this code suppose to work? I assume not.
Variable-length arrays are an extension in gcc and g++ ... so this won't work in every compiler.
For more information on gcc's support for variable length arrays, you can see the documentation here.
I believe that variable length arrays are officially unsupported in C++ but certain compilers and/or language extensions implement them.
If you want a variable length array I recommend using std::vector.
You can view its reference here:
http://www.cplusplus.com/reference/stl/vector/
#include <iostream.h>
^ is not a standard header. It used to be there in pre-standard times, i.e. before 1998. It's not there in e.g. modern Visual C++.
cin>> value;
int numbers[value];
Variable Length Arrays, or VLAs, were introduced in C99, a year after C++ was standardized. So they were not part of original standard C++, and happily they were not adopted in C++11 either. Instead of such beast, use e.g. std::vector from the vector header, or some other standard library container.
g++ supports variable length arrays as a language extension. You'd better turn off such extension. E.g.,
d:\dev\test> g++ foo.cpp
d:\dev\test> g++ -pedantic -std=c++0x -Wall -O foo.cpp
foo.cpp: In function 'int main()':
foo.cpp:11: warning: ISO C++ forbids variable length array 'numbers'
foo.cpp:7: warning: unused variable 'sum'
foo.cpp:8: warning: unused variable 'count'
d:\dev\test> _
I tried compiling it using GCC 4.6, and found that the code you've posted compiled successfully. I also tried running it and found that it worked, but I don't think the code is very good.

Piece of code that compiles with gcc but not g++ [duplicate]

This question already has answers here:
Closed 12 years ago.
Possible Duplicates:
Is this a legitimate C++ code?
“C subset of C++” -> Where not ? examples ?
Could anybody come up with a piece of code that compiles with gcc or any other C compiler, and doesn't compile g++ or any other C++ compiler?
UPDATE:
I don't mean just keywords
UPDATE2:
Thank you All for answers. Apparently moderators were less enthusiastic than I was about subtle differences between C and C++.
UPDATE3:
to moderators: could you merge it with my previous question on the topic, as you suggested? It makes perfect sense to keep these two question together.
#include <stdlib.h>
int main()
{
char* s = malloc(128);
return 0;
}
This will compile with gcc, but not with g++. C++ requires an explicit cast from void* here, whereas C does not.
int main(int argc, char **class)
{
return !argc;
}
Edit: another example
int foo();
int main(void) {
return foo(42);
}
int foo(int bar) {
return bar - 42;
}
Try
extern int getSize();
int main()
{
char x[getSize()];
x[0] = 0;
}
int getSize()
{
return 4;
}
Remember to compile with the strict flags.
> gcc -pedantic -std=c99 t.c
> g++ -pedantic t.c
t.c: In function `int main()':
t.c:6: error: ISO C++ forbids variable length array `x'
How about
/* Within a function */
{
enum {foo} bar;
bar++;
}
That seems a pretty big breaking change in the design of C++, but it is what it is.
What about character size:
Even worse is that it compiles but produces different output at runtime.
#include <stdio.h>
int main()
{
fprintf(stdout, "%s\n", (sizeof('\xFF') == sizeof(char))?"OK":"Fail");
}
> gcc -pedantic t.c
> ./a.exe
Fail
> g++ -pedantic t.c
> ./a.exe
OK
This actually makes we wonder why this works?
fprintf(stdout, "%c%c\n", 'A', 'B');
It works on both compilers even though the size of the objects are different.
pointer arithmetics on void*:
void* t;
t++; // compiles only in gcc