I have a struct like this
typedef struct _somestruct {
int a;
int b;
}SOMESTRUCT,*LPSOMESTRUCT;
I am creating an object for the struct and trying to print it's address like this
int main()
{
LPSOMESTRUCT val = (LPSOMESTRUCT)malloc(sizeof(SOMESTRUCT));
printf("0%x\n", val);
return 0;
}
..and I get this warning
warning C4313: 'printf' : '%x' in
format string conflicts with argument
1 of type 'LPSOMESTRUCT'
So, I tried to cast the address to int like this
printf("0%x\n", static_cast<int>(val));
But I get this error:
error C2440: 'static_cast' : cannot
convert from 'LPSOMESTRUCT' to 'int'
What am I missing here? How to avoid this warning?
Thanks.
%x expects an unsigned. What you're printing is a pointer. To do that correctly, you normally want to use %p. To be pedantically correct, that expects a pointer to void, so you'll need to cast it:
printf("%p\n", (void *)val);
In reality, most current implementations use the same format for all pointers, in which case the cast will be vacuous. Of course, given the C++ tag, most of the code you've included becomes questionable at best (other than the parts like LPSOMESTRUCT, which are questionable regardless). In C++, you normally want something more like:
struct somestruct {
int a;
int b;
};
somestruct *val = new somestruct; // even this is questionable.
std::cout << val;
Use the %p format specifier to print a pointer.
printf("%p\n", val);
If you want to cast then using reinterpret_cast instead of static_cast might do the trick here.
With printf try using the %zu instead of %x for printing out a pointer because the pointer is of unsigned integer type (ie %zu).
printf("%zu \n", val);
Just one other thing, being a c++ program is there a reason why you are using malloc instead of new?
As this is tagged C++, can I just point out that you do not need typedefs when creating structs in that language:
typedef struct _somestruct {
int a;
int b;
}SOMESTRUCT,*LPSOMESTRUCT;
should be:
struct SOMESTRUCT {
int a;
int b;
};
Also, it is considered by many to be bad practice to create typedefs like LPSOMESTRUCT which hide the fact that a type is a pointer.
Related
I'm playing a bit with the C++ syntax to figure out a generalized way to keep track of an offset within a class, sort of like offsetof, but in a type-safe way and without #defines
I know that a template class can be template-parametrized with fields, besides types and constants. So I came out with this prototype:
#include <iostream>
template <typename class_type, typename field_type>
struct offsetter
{
offsetter(const char* name, field_type class_type::*field)
: name(name)
{
fprintf(stderr, "%zu\n", field);
}
const char* const name;
};
struct some_struct
{
float avg;
int min;
int max;
struct internal
{
unsigned flag;
int x;
} test;
char* name;
};
int main()
{
offsetter<some_struct, float>("%h", &some_struct::avg);
offsetter<some_struct, int>("%h", &some_struct::min);
offsetter<some_struct, char*>("%h", &some_struct::name);
offsetter<some_struct, some_struct::internal>("x", &some_struct::test);
return 0;
}
This code is actually able to print the field offset, but I'm not really sure on what I'm doing here. Indeed it feels utterly wrong to reference field without referring to an instance (foo.*field).
But it does the job: it prints the offset. My guess is that I'm hitting on some loophole though, since for instance I can't assign size_t offset = field.
I figured out I probably want something like this:
size_t offset = (&(std::declval<class_type>().*field) - &(std::declval<class_type>()))
Which however wont' work as I can't take the address of an xvalue:
taking address of xvalue (rvalue reference)
Is there any way to do this?
AFAIK there isn't a standard way of doing this. Even the standard offsetof is defined only for standard layout types.
What you are doing is UB. You are using the wrong specifier zu. There isn't much you can do with a member pointer. You can't even do pointer arithmetics on them, you can't convert to char* nor to an integer type.
Also if your assumption is that a member pointer is just an integer representing the offset from the beginning of the structure that is false, not only in theory, but also in practice. Having multiple inheritance and virtual inheritance made sure of that.
I want to know whether we can memcpy a structure containing 2 float variables into another structure containing 2 int variable. This is what I have wriiten so far
struct stFloat
{
float a;
float b;
};
struct stInt
{
int a;
int b;
};
int main()
{
struct stFloat aa;
aa.a=12.234;
aa.b=673.797;
struct stInt bb;
memcpy(&bb,&aa,sizeof(stFloat));
printf("%d %d\n",bb.a,bb.b);
return 0;
}
But unfortunately I am not getting the desired result. The output that I was expecting is 12 673 but the output looks like some garbage. Can somebody help me resolving this issue.
Thanks
Integer and float have different internal representation, and memcpy is simply a bitwise copy so if you were expecting the numbers to be converted in some way it's not going to happen.
Instead you need to do it yourself, for example by declaring a constructor or function or assignment operator that allows you to assign stfloats to stint. This also allows you to explicitly indicate the conversion you want.
In general it is a bad idea to use memcpy (among other things, because it only works for PODs, and also because you get this sort of problem). It's a C thing that should be avoided in C++.
No, you can't, but you can write a function which will do it for you :
void cpy(struct stFloat *src, struct stInt *dest){
dest->a = (int)src->a;
dest->b = (int)src->b;
}
then, call it by passing your structures by references pointer (else it will only work on copies of the structures) :
struct stFloat f;
struct stFloat i;
cpy(&f,&i);
I have an array of unsigned integers that need to store pointers to data and functions as well as some data. In the device I am working with, the sizeof pointer is the same as sizeof unsigned int. How can I cast pointer to function into unsigned int? I know that this makes the code not portable, but it is micro controller specific. I tried this:
stackPtr[4] = reinterpret_cast<unsigned int>(task_ptr);
but it give me an error "invalid type conversion"
Casting it to void pointer and then to int is messy.
stackPtr[4] = reinterpret_cast<unsigned int>(static_cast<void *> (task_ptr));
Is there a clean way of doing it?
Edit - task_ptr is function pointer void task_ptr(void)
Love Barmar's answer, takes my portability shortcoming away. Also array of void pointer actually makes more sense then Unsigned Ints. Thank you Barmar and isaach1000.
EDIT 2: Got it, my compiler is thinking large memory model so it is using 32 bit pointers not 16 bit that I was expecting (small micros with 17K total memory).
A C-style cast can fit an octogonal peg into a trapezoidal hole, so I would say that given your extremely specific target hardware and requirements, I would use that cast, possibly wrapped into a template for greater clarity.
Alternately, the double cast to void* and then int does have the advantage of making the code stand out like a sore thumb so your future maintainers know something's going on and can pay special attention.
EDIT for comment:
It appears your compiler may have a bug. The following code compiles on g++ 4.5:
#include <iostream>
int f()
{
return 0;
}
int main()
{
int value = (int)&f;
std::cout << value << std::endl;
}
EDIT2:
You may also wish to consider using the intptr_t type instead of int. It's an integral type large enough to hold a pointer.
In C++ a pointer can be converted to a value of an integral type large enough to hold it. The conditionally-supported type std::intptr_t is defined such that you can convert a void* to intptr_t and back to get the original value. If void* has a size equal to or larger than function pointers on your platform then you can do the conversion in the following way.
#include <cstdint>
#include <cassert>
void foo() {}
int main() {
void (*a)() = &foo;
std::intptr_t b = reinterpret_cast<std::intptr_t>(a);
void (*c)() = reinterpret_cast<void(*)()>(b);
assert(a==c);
}
This is ansi compliant:
int MyFunc(void* p)
{
return 1;
}
int main()
{
int arr[2];
int (*foo)(int*);
arr[0] = (int)(MyFunc);
foo = (int (*)(int*))(arr[0]);
arr[1] = (*foo)(NULL);
}
I am a bit confused. There are two ways to return an array from a method. The first suggests the following:
typedef int arrT[10];
arrT *func(int i);
However, how do I capture the return which is an int (*)[]?
Another way is through a reference or pointer:
int (*func(int i)[10];
or
int (&func(int i)[10];
The return types are either int (*)[] or int (&)[].
The trouble I am having is how I can assign a variable to accept the point and I continue to get errors such as:
can't convert int* to int (*)[]
Any idea what I am doing wrong or what is lacking in my knowledge?
If you want to return an array by value, put it in a structure.
The Standard committee already did that, and thus you can use std::array<int,10>.
std::array<int,10> func(int i);
std::array<int,10> x = func(77);
This makes it very straightforward to return by reference also:
std::array<int,10>& func2(int i);
std::array<int,10>& y = func2(5);
First, the information you give is incorrect.
You write,
“There are two ways to return an array from a method”
and then you give as examples of the ways
typedef int arrT[10];
arrT *func(int i);
and
int (*func(int i))[10];
(I’ve added the missing right parenthesis), where you say that this latter way, in contrast to the first, is an example of
“through a reference or pointer”
Well, these two declarations mean exactly the same, to wit:
typedef int A[10];
A* fp1( int i ) { return 0; }
int (*fp2( int i ))[10] { return 0; }
int main()
{
int (*p1)[10] = fp1( 100 );
int (*p2)[10] = fp2( 200 );
}
In both cases a pointer to the array is returned, and this pointer is typed as "pointer to array". Dereferencing that pointer yields the array itself, which decays to a pointer to itself again, but now typed as "pointer to item". It’s a pointer to the first item of the array. At the machine code level these two pointers are, in practice, exactly the same. Coming from a Pascal background that confused me for a long time, but the upshot is, since it’s generally impractical to carry the array size along in the type (which precludes dealing with arrays of different runtime sizes), most array handling code deals with the pointer-to-first-item instead of the pointer-to-the-whole-array.
I.e., normally such a low level C language like function would be declared as just
int* func()
return a pointer to the first item of an array of size established at run time.
Now, if you want to return an array by value then you have two choices:
Returning a fixed size array by value: put it in a struct.
The standard already provides a templated class that does this, std::array.
Returning a variable size array by value: use a class that deals with copying.
The standard already provides a templated class that does this, std::vector.
For example,
#include <vector>
using namespace std;
vector<int> foo() { return vector<int>( 10 ); }
int main()
{
vector<int> const v = foo();
// ...
}
This is the most general. Using std::array is more of an optimization for special cases. As a beginner, keep in mind Donald Knuth’s advice: “Premature optimization is the root of all evil.” I.e., just use std::vector unless there is a really really good reason to use std::array.
using arrT10 = int[10]; // Or you can use typedef if you want
arrT10 * func(int i)
{
arrT10 a10;
return &a10;
// int a[10];
// return a; // ERROR: can't convert int* to int (*)[]
}
This will give you a warning because func returns an address of a local variable so we should NEVER code like this but I'm sure this code can help you.
In C/C++, how do I determine the size of the member variable to a structure without needing to define a dummy variable of that structure type? Here's an example of how to do it wrong, but shows the intent:
typedef struct myStruct {
int x[10];
int y;
} myStruct_t;
const size_t sizeof_MyStruct_x = sizeof(myStruct_t.x); // error
For reference, this should be how to find the size of 'x' if you first define a dummy variable:
myStruct_t dummyStructVar;
const size_t sizeof_MyStruct_x = sizeof(dummyStructVar.x);
However, I'm hoping to avoid having to create a dummy variable just to get the size of 'x'. I think there's a clever way to recast 0 as a myStruct_t to help find the size of member variable 'x', but it's been long enough that I've forgotten the details, and can't seem to get a good Google search on this. Do you know?
Thanks!
In C++ (which is what the tags say), your "dummy variable" code can be replaced with:
sizeof myStruct_t().x;
No myStruct_t object will be created: the compiler only works out the static type of sizeof's operand, it doesn't execute the expression.
This works in C, and in C++ is better because it also works for classes without an accessible no-args constructor:
sizeof ((myStruct_t *)0)->x
I'm using following macro:
#include <iostream>
#define DIM_FIELD(struct_type, field) (sizeof( ((struct_type*)0)->field ))
int main()
{
struct ABC
{
int a;
char b;
double c;
};
std::cout << "ABC::a=" << DIM_FIELD(ABC, a)
<< " ABC::c=" << DIM_FIELD(ABC, c) << std::endl;
return 0;
}
Trick is treating 0 as pointer to your struct. This is resolved at compile time so it safe.
You can easily do
sizeof(myStruct().x)
As sizeof parameter is never executed, you'll not really create that object.
Any of these should work:
sizeof(myStruct_t().x;);
or
myStruct_t *tempPtr = NULL;
sizeof(tempPtr->x)
or
sizeof(((myStruct_t *)NULL)->x);
Because sizeof is evaluated at compile-time, not run-time, you won't have a problem dereferencing a NULL pointer.
In C++11, this can be done with sizeof(myStruct_t::x). C++11 also adds std::declval, which can be used for this (among other things):
#include <utility>
typedef struct myStruct {
int x[10];
int y;
} myStruct_t;
const std::size_t sizeof_MyStruct_x_normal = sizeof(myStruct_t::x);
const std::size_t sizeof_MyStruct_x_declval = sizeof(std::declval<myStruct_t>().x);
From my utility macros header:
#define FIELD_SIZE(type, field) (sizeof(((type *)0)->field))
invoked like so:
FIELD_SIZE(myStruct_t, x);