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Kindly tell me what's wrong with the code. It is showing many errors. The error statements and problem statement is given below:
First error is in the overloaded dereference operator. It says pmemory was not declared in this scope.
2nd error is in the main function where I use
ptrFloat ptr1 = &var1;
it shows error "conversion from 'Float' to non-scalar type 'ptrFloat' requested".
And in the next line of above mention statement save error..
in next line it show that (no match for operator <<) .
and in the very next line it show the same error as above.(no match for operator).
And at last it show again the same error that (no match for operator)
The question statement is given below:
Remember fmem_top is an index to fmemory which points to the next available place where a float value can be stored. And pmem_top is the similer index to pmemory.
Create a class called Float . We’ll use it to model numbers of type float that are stored in fmemory instead of real memory. The only instance data in Float is its own “address”; that is, the index where its float value is stored in fmemory. Call this instance variable addr. Class Float also needs two member functions. The first is a one-argument constructor to initialize the Float with a float value. This constructor stores the float value in the element of fmemory pointed to by fmem_top , and stores the value of fmem_top in addr . This is similar to how the compiler and linker arrange to store an ordinary variable in real memory. The second member function is the overloaded & operator. It simply returns the pointer (really the index, type int ) value in addr .
Create a second class called ptrFloat . The instance data in this class holds the address(index) in pmemory where some other address (index) is stored. A member function initializes this “pointer” with an int index value. The second member function is the overloaded * (dereference, or “contents of”) operator. Its operation is a tad more complicated.
It obtains the address from pmemory , where its data, which is also an address, is stored. It then uses this new address as an index into fmemory to obtain the float value pointed to by its address data.
float& ptrFloat::operator*()
{
return fmemory[pmemory[addr]];
}
#include<iostream>
#include<conio.h>
using namespace std;
class Float
{
protected:
float *fmem_top;
float **addr;
float fmemory[10];
public:
Float(float a)
{
fmem_top=fmemory;
addr=&fmem_top;
*fmem_top=a;
fmem_top++;
}
Float operator & ()
{
return *fmem_top;
}
};
class ptrFloat: public Float
{
private:
int *pmem_top;
int pmemory[10];
public:
ptrFloat(int abc) : Float(abc)
{
pmem_top=pmemory;
*pmem_top=abc;
pmem_top++;
}
ptrFloat operator * ()
{
return fmemory[pememory[addr]];
}
};
int main ()
{
Float var1 = 1.234;
Float var2 = 5.678;
ptrFloat ptr1 = &var1;
ptrFloat ptr2 = &var2;
cout<<"*ptr1 = "<< *ptr1;
cout<<"*ptr2 = "<< *ptr2;
*ptr1 = 7.123;
*ptr2 = 8.456;
cout<<"*ptr1 = "<< *ptr1;
cout<<"*ptr2 = "<< *ptr2;
getch();
return 0;
}
I'm not sure what's going on here, but this is what Clang has to say:
prog.cc:33:48: error: array subscript is not an integer
ptrFloat operator*() { return fmemory[pmemory[addr]]; }
^~~~~
There's a typo in the line here - you'd written pememory. But addr is a float**.
prog.cc:40:12: error: no viable conversion from 'Float' to 'ptrFloat'
ptrFloat ptr1 = &var1;
^ ~~~~~
ptrFloat is a class derived from Float, not a Float*.
What's your intent with this code?
As an aside, the clang compiler has great and helpful error messages. If you don't/can't install clang on your computer - I think it's a pfaff to set it up on Windows, I don't know - you can quickly test your code on an online compiler such as Wandbox and see if clang's output gives better clues. Here's your code running there - I had to strip out conio.h and getch as they're Windows-only: http://melpon.org/wandbox/permlink/4iCa61Rqysqc7Lpl
The first error is a typo you wrote pememory.
The second error appears because you didn't code any transformation between Float and ptrFloat.
You should add a copy constructor to Float:
Float(const Float& a) : Float(*a.fmem_top) {}
And then a conversion constructor to ptrFloat:
ptrFloat(Float a) : Float(a) {
pmem_top=pmemory;
*pmem_top=abc;
pmem_top++;
}
PS: your code has a lot of problems. I don't really get what you are trying to do, but, for instance, you can't have float typed variables as array subscripts.
Related
I'm trying to debug a small C++ program using gdb, but may be getting hung up on some pointer arithmetic:
A* get(int) returns a pointer to an instance of a class A I've defined. Internally, get(int) references an array of A, returning:
class A_list {
private:
A* A_array;
int count;
public:
A_list(int c): count(c) { A = new A[c]; }
void insertAt(A a, int idx) {
A_array[idx] = a;
}
A* get(int);
};
A* A_list::get(int idx) {
...
A* result = A_array + idx;
return result;
}
presumably, when dealing with an array of A, I can simply add the index (times the size of a single A) to get the address of the idx'th.
This seems to work as expected. However, when calling get(int) from within another member function of A_list, I watch the value assignment in gdb and see two different values:
void A_list::foo() {
A* a = nullptr; // I declare my pointer, and initialize to 0x0
...
a = get(0); // I store the address of `A_array[0]`
The gdb watchpoint outputs:
Old value = (Number *) 0x0
New value = (Number *) 0x55555556b2c0
However, when I print the address stored in a, I get a completely different value, with an unrecognized message attached.
(gdb) p a
(Number *) 0x7ffff7b4e5c0 <_IO_file_overflow+256>
attempting to dereference any of the member values gives unexpected results
I can't find <_IO_file_overflow+256> defined anywhere in the gdb sources. What does it mean?
Why might the value stored in a appear to be different from the value returned when get() is called from inside a member function of A_list? From outside (eg - in main()) I get the expected value.
Edit 9-08:
Changed assignment in get() based on feedback. Still getting the same arbitrary address when I return from the get() function.
When doing pointer arithmetic, it's done in elements and not in units of bytes.
Therefore the multiplication with sizeof(A) is invalid and wrong: The expression A_array + (idx * sizeof(A)) should be plain A_array + idx.
Or you could be explicit and return &A_array[idx].
All this means that for any pointer or array a and (valid) index i, the expression *(a + i) is exactly the same as a[i]. And from that follows that &a[i] will be exactly the same as a + i.
To answer your last question -- <_IO_file_overflow+256> is telling you gdb's best guess as to what that address (0x7ffff7b4e5c0) refers to -- in this case, the address is pointing into some shared libaray, and the symbol _IO_file_overflow is the closest symbol defined in that library (and specifically, this address is 256 bytes past that symbol). This looks to be part of libc.
You can get more detail about what addresses correspond to what in your program by examining the file /proc/<pid>/maps -- you just need to know the pid of the process you are debugging and you can look at that file in another window.
As to why you're getting this odd value when it looks like you've just assigned a different value, it may be that gdb is getting confused and you have another a defined somewhere and gdb is printing that. Or you may have incomplete/incorrect debugging info in your program -- make sure that you compile with -O0 -g if you want accurate debug info.
I am reading a book called "Teach Yourself C in 21 Days" (I have already learned Java and C# so I am moving at a much faster pace). I was reading the chapter on pointers and the -> (arrow) operator came up without explanation. I think that it is used to call members and functions (like the equivalent of the . (dot) operator, but for pointers instead of members). But I am not entirely sure.
Could I please get an explanation and a code sample?
foo->bar is equivalent to (*foo).bar, i.e. it gets the member called bar from the struct that foo points to.
Yes, that's it.
It's just the dot version when you want to access elements of a struct/class that is a pointer instead of a reference.
struct foo
{
int x;
float y;
};
struct foo var;
struct foo* pvar;
pvar = malloc(sizeof(struct foo));
var.x = 5;
(&var)->y = 14.3;
pvar->y = 22.4;
(*pvar).x = 6;
That's it!
I'd just add to the answers the "why?".
. is standard member access operator that has a higher precedence than * pointer operator.
When you are trying to access a struct's internals and you wrote it as *foo.bar then the compiler would think to want a 'bar' element of 'foo' (which is an address in memory) and obviously that mere address does not have any members.
Thus you need to ask the compiler to first dereference whith (*foo) and then access the member element: (*foo).bar, which is a bit clumsy to write so the good folks have come up with a shorthand version: foo->bar which is sort of member access by pointer operator.
a->b is just short for (*a).b in every way (same for functions: a->b() is short for (*a).b()).
foo->bar is only shorthand for (*foo).bar. That's all there is to it.
Well I have to add something as well. Structure is a bit different than array because array is a pointer and structure is not. So be careful!
Lets say I write this useless piece of code:
#include <stdio.h>
typedef struct{
int km;
int kph;
int kg;
} car;
int main(void){
car audi = {12000, 230, 760};
car *ptr = &audi;
}
Here pointer ptr points to the address (!) of the structure variable audi but beside address structure also has a chunk of data (!)! The first member of the chunk of data has the same address than structure itself and you can get it's data by only dereferencing a pointer like this *ptr (no braces).
But If you want to acess any other member than the first one, you have to add a designator like .km, .kph, .kg which are nothing more than offsets to the base address of the chunk of data...
But because of the preceedence you can't write *ptr.kg as access operator . is evaluated before dereference operator * and you would get *(ptr.kg) which is not possible as pointer has no members! And compiler knows this and will therefore issue an error e.g.:
error: ‘ptr’ is a pointer; did you mean to use ‘->’?
printf("%d\n", *ptr.km);
Instead you use this (*ptr).kg and you force compiler to 1st dereference the pointer and enable acess to the chunk of data and 2nd you add an offset (designator) to choose the member.
Check this image I made:
But if you would have nested members this syntax would become unreadable and therefore -> was introduced. I think readability is the only justifiable reason for using it as this ptr->kg is much easier to write than (*ptr).kg.
Now let us write this differently so that you see the connection more clearly. (*ptr).kg ⟹ (*&audi).kg ⟹ audi.kg. Here I first used the fact that ptr is an "address of audi" i.e. &audi and fact that "reference" & and "dereference" * operators cancel eachother out.
struct Node {
int i;
int j;
};
struct Node a, *p = &a;
Here the to access the values of i and j we can use the variable a and the pointer p as follows: a.i, (*p).i and p->i are all the same.
Here . is a "Direct Selector" and -> is an "Indirect Selector".
I had to make a small change to Jack's program to get it to run. After declaring the struct pointer pvar, point it to the address of var. I found this solution on page 242 of Stephen Kochan's Programming in C.
#include <stdio.h>
int main()
{
struct foo
{
int x;
float y;
};
struct foo var;
struct foo* pvar;
pvar = &var;
var.x = 5;
(&var)->y = 14.3;
printf("%i - %.02f\n", var.x, (&var)->y);
pvar->x = 6;
pvar->y = 22.4;
printf("%i - %.02f\n", pvar->x, pvar->y);
return 0;
}
Run this in vim with the following command:
:!gcc -o var var.c && ./var
Will output:
5 - 14.30
6 - 22.40
#include<stdio.h>
int main()
{
struct foo
{
int x;
float y;
} var1;
struct foo var;
struct foo* pvar;
pvar = &var1;
/* if pvar = &var; it directly
takes values stored in var, and if give
new > values like pvar->x = 6; pvar->y = 22.4;
it modifies the values of var
object..so better to give new reference. */
var.x = 5;
(&var)->y = 14.3;
printf("%i - %.02f\n", var.x, (&var)->y);
pvar->x = 6;
pvar->y = 22.4;
printf("%i - %.02f\n", pvar->x, pvar->y);
return 0;
}
The -> operator makes the code more readable than the * operator in some situations.
Such as: (quoted from the EDK II project)
typedef
EFI_STATUS
(EFIAPI *EFI_BLOCK_READ)(
IN EFI_BLOCK_IO_PROTOCOL *This,
IN UINT32 MediaId,
IN EFI_LBA Lba,
IN UINTN BufferSize,
OUT VOID *Buffer
);
struct _EFI_BLOCK_IO_PROTOCOL {
///
/// The revision to which the block IO interface adheres. All future
/// revisions must be backwards compatible. If a future version is not
/// back wards compatible, it is not the same GUID.
///
UINT64 Revision;
///
/// Pointer to the EFI_BLOCK_IO_MEDIA data for this device.
///
EFI_BLOCK_IO_MEDIA *Media;
EFI_BLOCK_RESET Reset;
EFI_BLOCK_READ ReadBlocks;
EFI_BLOCK_WRITE WriteBlocks;
EFI_BLOCK_FLUSH FlushBlocks;
};
The _EFI_BLOCK_IO_PROTOCOL struct contains 4 function pointer members.
Suppose you have a variable struct _EFI_BLOCK_IO_PROTOCOL * pStruct, and you want to use the good old * operator to call it's member function pointer. You will end up with code like this:
(*pStruct).ReadBlocks(...arguments...)
But with the -> operator, you can write like this:
pStruct->ReadBlocks(...arguments...).
Which looks better?
#include<stdio.h>
struct examp{
int number;
};
struct examp a,*b=&a;`enter code here`
main()
{
a.number=5;
/* a.number,b->number,(*b).number produces same output. b->number is mostly used in linked list*/
printf("%d \n %d \n %d",a.number,b->number,(*b).number);
}
output is 5
5 5
Dot is a dereference operator and used to connect the structure variable for a particular record of structure.
Eg :
struct student
{
int s.no;
Char name [];
int age;
} s1,s2;
main()
{
s1.name;
s2.name;
}
In such way we can use a dot operator to access the structure variable
I am trying to overwrite a char and a function pointer on the stack. Based on what I found on this question (How can I store a value at a specific location in the memory?) I was able to figure out how to overwrite the character. My problem now is that I get a compile error saying I am casting it wrong.
void foo(char letter);
void bar(char letter);
void function1()
{
void (*pointer)(char);
pointer = foo;
letter = 'B';
function2();
(*pointer)(letter);
}
void function2()
{
int number; // Used in omitted code
*(char *)(&number + 75) = 'A';
*(void (*)(char)) (&number + 42) = &bar; // This is the line with the error
}
The first injection works but the second one gives me a compile error.
I am running Redhat Linux using a g++ compiler. The error I get from the compiler is:
"cannot convert ‘void (*)(char)’ to ‘void(char)’ in assignment"
If I change that line to *(void(char)) then the compiler says:
"invalid cast to function type ‘void(char)’"
What is the proper syntax for this?
(This is modified code from a school security assignment, I'm not writing malware)
Your goal is to write the address of pass to memory, so why are you casting (&number + 13) to a function pointer? Just do what you did before:
*(long *)(&number + 13) = (long)&pass;
And you won't get a compiler error. As to what will happen when this undefined behavior is invoked, you'll just have to see.
Edit: As #DavidGrayson pointed out, if we deference the right side of the equation, we'd get the contents of the function, not its pointer. So we have to cast it to a POD type, not a pointer.
Im a little bit confused about passing structs into functions. I understand pointers and everything.
But for instance:
struct stuff
{
int one
int two
};
int main{
stuff fnc;
fnc.two = 2;
fnc.one = 1;
multiply(&fnc);
}
void multiply(const stuff * pm){
cout << pm->one * pm->two;
}
First of all....am i even doing this right.
And second of all, why do we use the address operator when we pass the function, but use the * pointer operator in the actual function call?
Im confused?
Yes, your code is compilable other than the missing semicolons in the defintion of struct stuff. I'm not quite sure exactly what you're asking about passing the function and the actual function call, but I think you're wondering why the function call uses &fnc, but the parameter is stuff *pm? In that case, the fnc variable declared is a plain stuff. It is not a pointer, it refers to the actual instance of that struct.
Now the multiply function is declared as taking a stuff* -- a pointer to a stuff. This means that you can't pass fnc directly -- it's a stuff and multiply expects a *stuff. However, you can pass fnc as a stuff* by using the & operator to take the address, and &fnc is a valid stuff* that can be passed to multiply.
Once you're in the multiply function, you now have a stuff* called pm. To get the one and two variables from this stuff*, you use the pointer to member operator (->) since they are pointers to a stuff and not a plain stuff. After obtaining those values (pm->one and pm->two), the code then multiples them together before printing them out (pm->one * pm->two).
The * and & operands mean different things depending on whether they describe the type or describe the variable:
int x; // x is an integer
int* y = &x; // y is a pointer that stores the address of x
int& z = x; // z is a reference to x
int a = *y; // a in an integer whose value is the deference of y
Your pm variable is declared as a pointer, so the stuff type is modified with *. Your fnc variable is being used (namely for its address), and thus the variable itself is marked with &.
You can imagine the above examples as the following (C++ doesn't actually have these, so don't go looking for them):
int x;
pointer<int> y = addressof(x);
reference<int> z = x;
int a = dereference(y);
It the difference between describing a type and performing an operation.
In
void multiply(const stuff * pm){
cout << pm->one * pm->two;
}
The stuff * pm says that pm is an address of a stuff struct.
The
&fnc
says "the address of fnc".
When a variable is declared like:
stuff *pm;
it tells us that pm should be treated like an address whose underlying type is stuff.
And if we want to get the address of a variable stuff fnc, we must use
&fnc
You need the address of operator so you can get the address of the object, creating a pointer, which the function expects. The '*' in the parameter list is not a pointer operator, it simply says that the variable is a pointer.
Your code is correct. In the main, you successfully create a 'stuff' object and set its values. Then, you pass a constant address to the object into the function multiply. The multiply function then uses that address to access the two variables of the structure to output the multiplication of the variables.
The * in "const stuff * pm" means that it takes a constant pointer to a stuff object.
Here is a working example of what you would like to see.
#include <iostream>
using namespace std;
struct stuff
{
int one;
int two;
};
void multiply(stuff* pm)
{
cout << pm->one * pm->two;
}
int main()
{
stuff* fnc = new stuff;
fnc->two = 1;
fnc->one = 2;
multiply(fnc);
delete fnc;
cin.ignore(1000, 10);
return 0;
}
Sure, this would work, aside from your erroneous main function definition.
The reason why this works is because when you use the unary & operator, it essentially returns a pointer to the operand, so in your case, fnc, which is of type stuff, if you did &fnc, that would return a stuff *. This is why the multiply function must take in a stuff *.
struct stuff
{
int one, two;
};
int main(int argc, const char* argv[]) {
stuff fnc;
fnc.two = 2;
fnc.one = 1;
multiply(&fnc); //passes a pointer to fnc
}
void multiply(const stuff * pm){
//the "*" operator is the multiplication operator, not a pointer dereference
cout << pm->one * pm->two; //->one is equivalent to (*pm).one
}
You have a couple of syntactic errors in your program, but other than that, the basic idea is fine. Here are the syntax problems I had to fix before your program would compile:
#include <iostream>
using namespace std;
struct stuff
{
int one;
int two;
};
void multiply(const stuff * pm) {
cout << pm->one * pm->two;
}
int main() {
stuff fnc;
fnc.two = 2;
fnc.one = 1;
multiply(&fnc);
}
To answer your questions about difference between the '&' (address of) operator and the '*' (pointer dereference) operator though, we just need to think about the types you're passing in to the function.
Take the function multiply:
void multiply(stuff *fnc) {
...
}
In the definition of this function, you are describing something that takes a pointer to a stuff struct. In that first line, you aren't saying you are dereferencing that object, just that you are expecting a pointer to a stuff object.
Now when you call multiply:
stuff fnc;
multiply(&fnc);
You are using the '&' (address of) operator to get a pointer to the object. Since the multiply function expects a pointer, and you have the plain old object, you need to use the & operator to get a pointer to give to the multiply function.
Perhaps it is clearer to think call it like this:
stuff fnc; //The actual object
stuff* fnc_ptr = &fnc; //A pointer to a stuff object, initialized to point at fnc created above
multiply(fnc_ptr); //Call the function with the pointer directly
Following code will tell you about the pointer illustration
A struct address is passed into the function named multiply and this
function perform some operations with the element of the passed
structure and store the result in the result variable.
you can see here clearly that the result variable is previously zero then after passing the address of the structure to the function multiply the result variable value gets updated to value 6. this is how pointer works.
#include <iostream.h>
struct stuff
{
int one;
int two ;
int result;
};
void multiply(stuff *pm);
int main(){
stuff fnc;
fnc.two = 2;
fnc.one = 3;
fnc.result = 0;
multiply(&fnc);
cout<<fnc.result;
return 0;
}
void multiply(stuff *pm)
{
pm->result = pm->one * pm->two;
}
Any ideas for this typecasting problem?
Here's what I am trying to do. This is not the actual code:
LinkedList* angles;
double dblangle;
dblangle = (some function for angle calculation returning double value);
(double*)LinkedListCurrent(angles) = &double;
I hope you get the idea. The last line is causing the problem. Initially angles is void* type so I have to first convert it to double*.
You use the unary * operator to dereference a pointer. Dereferencing a pointer means extracting the value pointed to, to get a value of the original type.
// Create a double value
double foo = 1.0;
// Create a pointer to that value
double *bar = &foo;
// Extract the value from that pointer
double baz = *bar;
edit 2: (deleted edit 1 as it was not relevant to your actual question, but was based on a miscommunication)
From your clarification, it looks like you are wondering how to set a value pointed to by a pointer that has been cast from a void * to a double *. In order to do this, we need to use the unary * on the left hand side of the assignment, in order to indicate that we want to write to the location pointed to by the pointer.
// Get a void * pointing to our double value.
void *vp = &foo;
// Now, set foo by manipulating vp. We need to cast it to a double *, and
// then dereference it using the * operator.
*(double *)vp = 2.0;
// And get the value. Again, we need to cast it, and dereference it.
printf("%F\n", *(double *)vp);
So, I'm assuming that your LinkedListCurrent is returning a void * pointing to some element in the linked list that you would like to update. In that case, you would need to do the following:
*(double*)LinkedListCurrent(angles) = dbangle;
This updated the value pointed to by the pointer returned from LinkedListCurrent to be equal to dbangle. I believe that is what you are trying to do.
If you are trying to update the pointer returned by LinkedListCurrent, you can't do that. The pointer has been copied into a temporary location for returning from the function. If you need to return a pointer that you can update, you would have to return a pointer to a pointer, and update the inner one.
My explanation above is based on what I believe you are trying to do, based on the example snippet you posted and some guesses I've made about the interface. If you want a better explanation, or if one of my assumptions was bad, you might want to try posting some actual code, any error messages you are getting, and a more detailed explanation of what you are trying to do. Showing the interface to the linked list data type would help to provide some context for what your question is about.
edit 3: As pointed out in the comments, you probably shouldn't be casting here anyhow; casts should be used as little as possible. You generally should use templated collection types, so your compiler can actually do typechecking for you. If you need to store heterogenous types in the same structure, they should generally share a superclass and have virtual methods to perform operations on them, and use dynamic_cast if you really need to cast a value to a particular type (as dynamic_cast can check at runtime that the type is correct).
why on earth would you want to use a memory address as a floating point number?
if you meant dereference:
double d = 1.0; // create a double variable with value 1.0
double *dp = &d; // store its address in a pointer
double e = *dp; // copy the value pointed at to another variable
Note this line:
(double*)LinkedListCurrent(angles) = &double;
where you've written &double, it i think should be &dbangle. To improve readability, I would write:
((double*)LinkedListCurrent(angles)) = &dbangle;
However, you should not do this type of conversion as others mentioned.
Use a union. If you want the store two variables in one memory location (but not at the same time), you don't have to pretend that one is the other.
union double_and_ptr {
double d;
double *p;
};
double_and_ptr x, y;
x.d = 0.1;
y.p = &x.d; // store a pointer in the same place as a double
y.d = x.d * 1.2; // store a double in the same place as a ptr
Use reinterpret_cast.
double foo = 3.0;
double* pfoo = &foo;
double bar = reinterpret_cast<double>(pfoo);
In answer to this:
I want to do the opposite of what you have done here. I want to copy the value from the pointer to the d floating point number. How can I do that?
You would do something like this:
// declare a pointer to a double
double *pointer_to_double;
// declare a double
double my_double = 0.0;
// use the indirection operator (*) to dereference the pointer and get the value
// that it's pointing to.
my_double = *pointer_to_double;
This might be done like so in a real program:
void print_chance_of_snow(double *chance)
{
double d = *chance;
d = d * 100; // convert to a percentage
printf("Chance of snow is: %.2f%%\n", d);
}
int main(int argc, char *argv[])
{
double chance_of_snow = 0.45;
print_chance_of_snow(&chance_of_snow);
}