Im trying to convert a program from C++/MEX to just C++ using MATIO and I am wondering if MATIO has an equivalent to mxGetPr(cal) and mxGetPi(cal)? I see in the struct typedef struct matvar_t it has void *data filed
Here is how to write a complex double into a file:
char* fieldname = "MyComplexDoubleVariable";
double real = 4.2;
double imag = 1.5;
mat_complex_split_t mycomplexdouble = {&real, &imag};
size_t dim[2]={ 1, 1 };
matvar_t *variable = Mat_VarCreate(fieldname, MAT_C_DOUBLE, MAT_T_DOUBLE, 2, dim, &mycomplexdouble, MAT_F_COMPLEX);
Mat_VarWrite(matfp,variable, MAT_COMPRESSION_NONE); //or MAT_COMPRESSION_ZLIB
Mat_VarFree(variable);
So working backwards I would think this would work
mat_complex_split_t cal_complex=cal->data; but I get this error when compiling
error: conversion from ‘void*’ to non-scalar type ‘mat_complex_split_t’ requested
mat_complex_split_t cal_complex=cal->data;
Any help would be greatly appreciated.
Here is my code to read complex data from mat-file.
I think you should use static_cast to convert void * to specify type.
const mat_complex_split_t* xData = static_cast<const mat_complex_split_t*>(matVar->data);
const double* real = static_cast<const double*>(xData->Re);
const double* imag = static_cast<const double*>(xData->Im);
// Read 10 complex num
for (int i = 0; i < 10; ++i) {
std::cout << real[i] << "+ 1j*"<< imag[i] << std::endl;
}
In the following code:
void benchmark(string desc, int(*pf)(int a[50])) {
printf("\n Benchmark for %s", desc.c_str());
double tStart = omp_get_wtime();
int result = pf(a[50]);
double tFinal = omp_get_wtime();
printf("\n\t Final result: %d", result);
printf("\n\t Duration: %f (s)", tFinal - tStart);
int main() {
int i, b;
int a[50];
for (i = 0; i < 50; i++)
{
b = rand() % 10000 + 1;
a[i] = b;
}
benchmark("Parallel solution without mutex for counting primes", Squential);
benchmark("Parallel solution with load balancing", Parallel1);
benchmark("Parallel solution with load balancing", Parallel2);
}
On line "int result = pf(a[50]);" the compiler says "identifier "a" is undefined" although I listed it in the parameters list. It may have something to do with the pointer, I'm not used to working with pointers.
This does not have a parameter a
void benchmark(string desc, int(*pf)(int a[50])) {
This has the parameters:
* desc: Type -> string
* pf: Type -> A function (pointer) that returns int and takes an array of int
This is a function pointer type:
int (*pf)(int a[50])
^^ This is the name of the paramter.
// This is the type
int (*)(int[50])
// This is a pointer to a function that returns an `int`
// and takes `int[50]` as a parameter.
Just to get this copiling do this:
int result = pf(a[50]);
// change into this:
int data[50]; // or get this from somewhere else.
int result = pf(data);
Let's say I have a .proto file with:
message Foo {
optional int32 x = 1;
enum y {
MOBILE = 0;
HOME = 1;
}
optional string z = 3;
}
Then I have this C++ code which prints all the types:
const Reflection *refl = Foo.GetReflection();
const Descriptor *desc = Foo.GetDescriptor();
int fieldCount = desc->field_count();
for(int i=0;i<fieldCount;i++){
const FieldDescriptor *field = desc->field(i);
cout << field->name().c_str() << " the type is "
<<field->type_name()<< ": Type Number "<< field->type() <<endl;
if(field->type()==FieldDescriptor::TYPE_ENUM){
//do something
}
The output is then:
x the type is int32: Type Number 5
z the type is string: Type Number 9
As seen in the output, the enum is skipped, how would I get the field descriptor to parse over the enum as well?
You do not have a field of your Enum type, you have only defined a type. Because of that, your iteration over fields doesn't yield anything related to the enum.
If you add a field of a given type, you will see your enum there.
The problem is as described above. When I try to read values from loaded *.so file (using libdl), whih are in struct I am getting wrong values
Code of application:
#include <dlfcn.h>
#include <iostream>
/* For face data type reproduction */
#define GET_FACE_XYZ_SIZE 1
/* For face_array reproduction */
#define GET_FACE_ARRAY_SIZE 2
#define GET_OBJECT_DATA 3
typedef struct face {
float x[1000];
float y[1000];
float z[1000];
int vertices;
} face;
int main()
{
void *hook;
int (*fn)(int request_type, void *ptr);
hook = dlopen("/root/osms/dlopen-test/lib.so", RTLD_LAZY);
if(!hook)
{
std::cout << "Couldn't find lib.so" << std::endl;
}
fn = dlsym(hook, "object_info");
int face_array_size = fn(GET_FACE_ARRAY_SIZE, NULL);
std::cout << "FACE_ARRAY_SIZE: " << face_array_size << std::endl;
face pointer[face_array_size];
fn(NULL, pointer);
dlclose(hook);
std::cout << "pointer[0].z[1]: " << pointer[0].z[1] << std::endl;
return 0;
}
and code of lib.so:
/* For face data type reproduction */
#define GET_FACE_XYZ_SIZE 1
/* For face array reproduction */
#define GET_FACE_ARRAY_SIZE 2
#define GET_OBJECT_DATA 3
typedef struct face {
float x[1000];
float y[1000];
float z[1000];
int vertices;
} face;
extern "C" int object_info(int request, void *ptr)
{
face face_array[2];
face_array[0].x[0] = 1.1;
face_array[0].y[0] = 0.5;
face_array[0].z[0] = 1.2;
face_array[0].x[1] = 1.6;
face_array[0].y[1] = -0.11;
face_array[0].z[1] = -12;
face_array[0].x[2] = -0.12;
face_array[0].y[2] = 0.24;
face_array[0].z[2] = -0.12;
face_array[0].vertices = 3;
face_array[1].x[0] = -1.1;
face_array[1].y[0] = 0.15;
face_array[1].z[0] = -1.2;
face_array[1].x[1] = -1.6;
face_array[1].y[1] = 0.11;
face_array[1].z[1] = 1.2;
face_array[1].x[2] = 0.12;
face_array[1].y[2] = -0.24;
face_array[1].z[2] = 0.12;
face_array[1].vertices = 3;
if(request == GET_FACE_ARRAY_SIZE)
{
return 2;
}
else
{
ptr = face_array;
}
}
The expected output is pointer[0].z[1]: -12 but I am getting pointer[0].z[1]: -0.12. What's wrong in my code ?
Thanks in advance
Accessing
pointer[0].z[1]
Has undefined behaviour, because it has an indeterminate value.
object_info never modifies the array pointed by ptr. It simply modifies a local array, and assigns the local ptr to point to that local array.
A solution: Don't declare a local array, and instead modify the array pointed by the argument. In other words, repace face face_array[2]; with:
face* face_array = (face*)ptr;
And get rid of the ptr = face_array; that does nothing meaningful.
object_info is declared to return int, but not all code paths return a value. When the function reaches the end of object_info without a return statement, the behaviour is undefined.
A solution: Always return a value if the function is not void.
face_array_size is not a compile time constant value, so face pointer[face_array_size]; will declare a variable length array. VLA are not allowed in C++.
Either use C (VLA are supported since C99, but only optionally supported since C11) instead or use a dynamic array: std::vector<face> or make peace with the fact that your program is not standard compliant.
The variable "face_array" in function object_info and the variable "pointer" in main are not the same variable.
The statement "ptr = face_array" does not change the content of "pointer".
extern "C" int object_info(int request, face *face_array)
{
if(request == GET_FACE_ARRAY_SIZE)
return 2;
face_array[0].x[0] = 1.1;
face_array[0].y[0] = 0.5;
face_array[0].z[0] = 1.2;
face_array[0].x[1] = 1.6;
face_array[0].y[1] = -0.11;
face_array[0].z[1] = -12;
face_array[0].x[2] = -0.12;
face_array[0].y[2] = 0.24;
face_array[0].z[2] = -0.12;
face_array[0].vertices = 3;
face_array[1].x[0] = -1.1;
face_array[1].y[0] = 0.15;
face_array[1].z[0] = -1.2;
face_array[1].x[1] = -1.6;
face_array[1].y[1] = 0.11;
face_array[1].z[1] = 1.2;
face_array[1].x[2] = 0.12;
face_array[1].y[2] = -0.24;
face_array[1].z[2] = 0.12;
face_array[1].vertices = 3;
}
I had an article, but I lost it. It showed and described a couple of C/C++ tricks that people should be careful. One of them interested me but now that I am trying to replicate it I'm not being able to put it to compile.
The concept was that it is possible to change by accident the value of a const in C/C++
It was something like this:
const int a = 3; // I promise I won't change a
const int *ptr_to_a = &a; // I still promise I won't change a
int *ptr;
ptr = ptr_to_a;
(*ptr) = 5; // I'm a liar; a is now 5
I wanted to show this to a friend but now I'm missing a step. Does anyone know what's missing for it to start compiling and working?
ATM I'm getting invalid conversion from 'const int*' to 'int*' but when I read the article I tried and it worked great.
you need to cast away the constness:
linux ~ $ cat constTest.c
#include <stdio.h>
void modA( int *x )
{
*x = 7;
}
int main( void )
{
const int a = 3; // I promisse i won't change a
int *ptr;
ptr = (int*)( &a );
printf( "A=%d\n", a );
*ptr = 5; // I'm a liar, a is now 5
printf( "A=%d\n", a );
*((int*)(&a)) = 6;
printf( "A=%d\n", a );
modA( (int*)( &a ));
printf( "A=%d\n", a );
return 0;
}
linux ~ $ gcc constTest.c -o constTest
linux ~ $ ./constTest
A=3
A=5
A=6
A=7
linux ~ $ g++ constTest.c -o constTest
linux ~ $ ./constTest
A=3
A=3
A=3
A=3
also the common answer doesn't work in g++ 4.1.2
linux ~ $ cat constTest2.cpp
#include <iostream>
using namespace std;
int main( void )
{
const int a = 3; // I promisse i won't change a
int *ptr;
ptr = const_cast<int*>( &a );
cout << "A=" << a << endl;
*ptr = 5; // I'm a liar, a is now 5
cout << "A=" << a << endl;
return 0;
}
linux ~ $ g++ constTest2.cpp -o constTest2
linux ~ $ ./constTest2
A=3
A=3
linux ~ $
btw.. this is never recommended... I found that g++ doesn't allow this to happen.. so that may be the issue you are experiencing.
Note any attempt to cast away constness is undefined by the standard. From 7.1.5.1 of the standard:
Except that any class member declared
mutable can be modified, any
attempt to modify a const object
during its lifetime
results in undefined behavior.
And right after this example is used:
const int* ciq = new const int (3); // initialized as required
int* iq = const_cast<int*>(ciq); // cast required
*iq = 4; // undefined: modifies a const object
So in short what you want to do isn't possible using standard C++.
Further when the compiler encounters a declaration like
const int a = 3; // I promisse i won't change a
it is free to replace any occurance of 'a' with 3 (effectively doing the same thing as #define a 3)
Just a guess, but a common question is why one can't convert an int** to a const int**, which at first appears to be reasonable (after all, you're just adding a const, which is normally ok). The reason is that if you could do this, you could accidentally modify a const object:
const int x = 3;
int *px;
const int **ppx = &px; // ERROR: conversion from 'int**' to 'const int**'
*ppx = &x; // ok, assigning 'const int*' to 'const int*'
*px = 4; // oops, just modified a const object
It's a very non-intuitive result, but the only way to make sure that you can't modify a const object in this case (note how there are no typecasts) is to make line 3 an error.
You're only allowed to add const without a cast at the FIRST level of indirection:
int * const *ppx = &px; // this is ok
*ppx = &x; // but now this is an error because *ppx is 'const'
In C++, it is impossible to modify a const object without using a typecast of some sort. You'll have to use either a C-style cast or a C++-style const_cast to remove the const-ness. Any other attempt to do so will result in a compiler error somewhere.
Back in the mists of time, we paleo-programmers used FORTRAN. FORTRAN passed all its parameters by reference, and didn't do any typechecking. This meant it was quite easy to accidentally change the value of even a literal constant. You could pass "3" to a SUBROUTINE, and it would come back changed, and so every time from then on where your code had a "3", it would actually act like a different value. Let me tell you, those were hard bugs to find and fix.
Did you try this?
ptr = const_cast<int *>(ptr_to_a);
That should help it compile but it's not really by accident due to the cast.
In C++, Using Microsoft Visual Studio-2008
const int a = 3; /* I promisse i won't change a */
int * ptr1 = const_cast<int*> (&a);
*ptr1 = 5; /* I'm a liar, a is now 5 . It's not okay. */
cout << "a = " << a << "\n"; /* prints 3 */
int arr1[a]; /* arr1 is an array of 3 ints */
int temp = 2;
/* or, const volatile int temp = 2; */
const int b = temp + 1; /* I promisse i won't change b */
int * ptr2 = const_cast<int*> (&b);
*ptr2 = 5; /* I'm a liar, b is now 5 . It's okay. */
cout << "b = " << b << "\n"; /* prints 5 */
//int arr2[b]; /* Compilation error */
In C, a const variable can be modified through its pointer; however it is undefined behavior. A const variable can be never used as length in an array declaration.
In C++, if a const variable is initialized with a pure constant expression, then its value cannot be modified through its pointer even after try to modify, otherwise a const variable can be modified through its pointer.
A pure integral const variable can be used as length in an array declaration, if its value is greater than 0.
A pure constant expression consists of the following operands.
A numeric literal (constant ) e.g. 2, 10.53
A symbolic constant defined by #define directive
An Enumeration constant
A pure const variable i.e. a const variable which is itself initialized with a pure constant expression.
Non-const variables or volatile variables are not allowed.
Some of these answers point out that the compiler can optimize away the variable 'a' since it is declared const. If you really want to be able to change the value of a then you need to mark it as volatile
const volatile int a = 3; // I promise i won't change a
int *ptr = (int *)&a;
(*ptr) = 5; // I'm a liar, a is now 5
Of course, declaring something as const volatile should really illustrate just how silly this is.
this will create a runtime fault. Because the int is static. Unhandled exception. Access violation writing location 0x00035834.
void main(void)
{
static const int x = 5;
int *p = (int *)x;
*p = 99; //here it will trigger the fault at run time
}
You probably want to use const_cast:
int *ptr = const_cast<int*>(ptr_to_a);
I'm not 100% certain this will work though, I'm a bit rusty at C/C++ :-)
Some readup for const_cast: http://msdn.microsoft.com/en-us/library/bz6at95h(VS.80).aspx
const int foo = 42;
const int *pfoo = &foo;
const void *t = pfoo;
void *s = &t; // pointer to pointer to int
int **z = (int **)s; // pointer to int
**z = 0;
The article you were looking at might have been talking about the difference between
const int *pciCantChangeTarget;
const int ci = 37;
pciCantChangeTarget = &ci; // works fine
*pciCantChangeTarget = 3; // compile error
and
int nFirst = 1;
int const *cpiCantChangePointerValue = &nFirst;
int nSecond = 968;
*pciCantChangePointerValue = 402; // works
cpiCantChangePointerValue = &ci; // compile error
Or so I recall-- I don't have anything but Java tools here, so can't test :)
#include<iostream>
int main( void )
{
int i = 3;
const int *pi = &i;
int *pj = (int*)&i;
*pj = 4;
getchar();
return 0;
}
I was looking on how to convert between consts and I found this one http://www.possibility.com/Cpp/const.html maybe it can be useful to someone. :)
I have tested the code below and it successfully changes the constant member variables.
#include <iostream>
class A
{
private:
int * pc1; // These must stay on the top of the constant member variables.
int * pc2; // Because, they must be initialized first
int * pc3; // in the constructor initialization list.
public:
A() : c1(0), c2(0), c3(0), v1(0), v2(0), v3(0) {}
A(const A & other)
: pc1 (const_cast<int*>(&other.c1)),
pc2 (const_cast<int*>(&other.c2)),
pc3 (const_cast<int*>(&other.c3)),
c1 (*pc1),
c2 (*pc2),
c3 (*pc3),
v1 (other.v1),
v2 (other.v2),
v3 (other.v3)
{
}
A(int c11, int c22, int c33, int v11, int v22, int v33) : c1(c11), c2(c22), c3(c33), v1(v11), v2(v22), v3(v33)
{
}
const A & operator=(const A & Rhs)
{
pc1 = const_cast<int*>(&c1);
pc2 = const_cast<int*>(&c2),
pc3 = const_cast<int*>(&c3),
*pc1 = *const_cast<int*>(&Rhs.c1);
*pc2 = *const_cast<int*>(&Rhs.c2);
*pc3 = *const_cast<int*>(&Rhs.c3);
v1 = Rhs.v1;
v2 = Rhs.v2;
v3 = Rhs.v3;
return *this;
}
const int c1;
const int c2;
const int c3;
int v1;
int v2;
int v3;
};
std::wostream & operator<<(std::wostream & os, const A & a)
{
os << a.c1 << '\t' << a.c2 << '\t' << a.c3 << '\t' << a.v1 << '\t' << a.v2 << '\t' << a.v3 << std::endl;
return os;
}
int wmain(int argc, wchar_t *argv[], wchar_t *envp[])
{
A ObjA(10, 20, 30, 11, 22, 33);
A ObjB(40, 50, 60, 44, 55, 66);
A ObjC(70, 80, 90, 77, 88, 99);
A ObjD(ObjA);
ObjB = ObjC;
std::wcout << ObjA << ObjB << ObjC << ObjD;
system("pause");
return 0;
}
The console output is:
10 20 30 11 22 33
70 80 90 77 88 99
70 80 90 77 88 99
10 20 30 11 22 33
Press any key to continue . . .
Here, the handicap is, you have to define as many pointers as number of constant member variables you have.
we can change the const variable value by the following code :
const int x=5;
printf("\nValue of x=%d",x);
*(int *)&x=7;
printf("\nNew value of x=%d",x);
#include<stdio.h>
#include<stdlib.h>
int main(void) {
const int a = 1; //a is constant
fprintf(stdout,"%d\n",a);//prints 1
int* a_ptr = &a;
*a_ptr = 4;//memory leak in c(value of a changed)
fprintf(stdout,"%d",a);//prints 4
return 0;
}
Final Solution: it will change const variable a value;
cont int a = 10;
*(int*)&a= 5; // now a prints 5
// works fine.
The step you're missing is that you don't need the int* pointer. The line:
const int *ptr_to_a = &a; // I still promiss i won't change a;
actually says you won't change ptr_to_a, not a. So if you changed your code to read like this:
const int a = 3; // I promise I won't change a
const int *ptr_to_a = &a; // I promise I won't change ptr_to_a, not a.
(*ptr_to_a) = 5; // a is now 5
a is now 5. You can change a through ptr_to_a without any warning.
EDIT:
The above is incorrect. It turns out I was confusing a similar trick with a shared_ptr, in which you can get access to the raw pointer and modify the internal data value without firing off any warnings. That is:
#include <iostream>
#include <boost/shared_ptr.hpp>
int main()
{
const boost::shared_ptr<int>* a = new boost::shared_ptr<int>(new int(3));
*(a->get()) = 5;
std::cout << "A is: " << *(a->get()) << std::endl;
return 0;
}
Will produce 5.