How to check a function pointer exists - c++

In C++, I'm trying to write a function with function pointers. I want to be able to throw an exception if a function pointer is passed for a function that does not exist. I tried to handle the function pointer like a normal pointer and check if it is null
#include <cstddef>
#include <iostream>
using namespace std;
int add_1(const int& x) {
return x + 1;
}
int foo(const int& x, int (*funcPtr)(const int& x)) {
if (funcPtr != NULL) {
return funcPtr(x);
} else {
throw "not a valid function pointer";
}
}
int main(int argc, char** argv) {
try {
int x = 5;
cout << "add_1 result is " << add_1(x) << endl;
cout << "foo add_1 result is " << foo(x, add_1) << endl;
cout << "foo add_2 result is " << foo(x, add_2) << endl; //should produce an error
}
catch (const char* strException) {
cerr << "Error: " << strException << endl;
}
catch (...) {
cerr << "We caught an exception of an undetermined type" << endl;
}
return 0;
}
but that doesn't seem to work. What is the best way to do this?

Checking for NULL is ok. But it is not possible to pass a pointer to a function that does not exist in the first place. So you don't have to worry about this. Although it is possible to just declare a function without defining it and pass the address of it. In that case you will get linker error.

It will automatically throw an error if you are passing pointer which does not exist, if you are declaring a pointer then you have to initialize it with null to avoid garbage value, so comparing with null will not serve any purpose.
you still you want to check then try to assign some function(like add, sub etc.), if it takes then ok , if not then it will show again error as previously mentioned.

#include<cstddef>
#include <iostream>
using namespace std;
int foo(const int& x, int (*funcPtr)(const int& x)) {
if (*funcPtr != NULL) {
return funcPtr(x);
}
else
{
cout << "not a valid function pointer";
}
}
If you want to 'throw' exception then you need to 'catch' it as well.
Your code is failing because of two reasons in short,
1) You are not checking value of function pointer.
2) You are not properly catching the thrown exception.

Related

Can any primitive type be passed into a function expecting a pointer?

Basically I'm wondering what the rules are for passing in pointers vs references to functions in C++. I couldn't find them stated anywhere. Can you pass a primitive type integer, for example, into a function expecting a pointer? Can't you only pass in pointers to methods expecting pointers?
A pointer is just a memory address and in c++ you can get the address of a variable by using the &. Here is an example
#include <iostream>
void increment(int& x)
{
++x;
}
void increment2(int* x)
{
++(*x);
}
int main()
{
int i = 1;
int * p = new int(1);
increment2(&i);
increment2(p);
std::cout << i << std::endl;
std::cout << *p << std::endl;
increment(i);
increment(*p);
std::cout << i << std::endl;
std::cout << *p << std::endl;
}
output
2
2
3
3
try it:
https://godbolt.org/z/br9APq

No output while doing try catch in c++

I am trying to catch bad allocation error. When input length will be in order of 10000000000000000000000 or something, then bad allocation error should come. I don't know why its not being caught.
Any help will be appreciated!
# include <vector>
# include <iostream>
using namespace std;
void length(int m)
{
vector<int> x;
try
{
x.resize(m);
}
catch(std::bad_alloc&)
{
cout << "caught bad alloc exception" << std::endl;
}
}
int main()
{
int l;
cout << "Length" ;
cin >> l ;
length(l);
return 0;
}
UPDATED:
When I am hard coding the value for input, then it is throwing an exception. I don't know why its working this way.
# include <vector>
# include <iostream>
using namespace std;
void length(int m)
{
vector<int> x;
try
{
x.resize(m);
}
catch(std::bad_alloc&)
{
cout << "caught bad alloc exception" << std::endl;
}
}
int main()
{
int m= 100000000000000000000;
length(m);
return 0;
}
You ought to write
if (!(cin >> l)){
// I could not read that into `l`
}
The lack of an exception being caught could be down to
Your int value being smaller than you think (perhaps some undefined wrap-around behaviour), and an exception is not thrown since the allocation is successful.
The allocation being lazy in the sense that the memory is not allocated until you actually use it.
If std::bad_alloc is thrown as an anonymous temporary then it will not be caught at your catch site. (Unless your naughty compiler allows non-const references to bind to anonymous temporaries, which some do as an extension). Write catch (const std::bad_alloc&) instead, and it will be caught there.
The maximum length of an integer type int is 2.147.483.647 . Are you sure you have actually used an higher number to test it?
You're passing Integer variable which has the limit.
Minimum value for a variable of type short: –32768
Maximum value for a variable of type short: 32767
The error which you will get from your code is std::length_error
To raise the bad allocation error dynamically you can try malloc() with incorrect size OR try below code.
#include <iostream>
#include <new>
int main()
{
try {
while (true) {
new int[100000000ul];
}
} catch (const std::bad_alloc& e) {
std::cout << "Allocation failed: " << e.what() << '\n';
}
}
</i>
No exception is thrown because the int that makes it through to your function void length(int m) is capped at its max value that is much less than vector::max_size(). Consider:
void length(int m)
{
cout << "m is: " << m << " which has a max value: " << numeric_limits<int>::max() << endl;
// ...
}
with the output:
Length10000000000000000000000
m is: 2147483647 and has a max value: 2147483647

calling function depending on index of array array of functions' names

making random actions in a game makes it really look like real...
so if a character has many capabilities like move, work, study... so in programming a function of those is called depending on some conditions. what we want is a more random and real-looking like action where no condition is there but depending on a random condition the character takes a random actions..
I thought to make actions (functions) in an array then declare a pointer to function and the program can randomly generate an index which on which the pointer to function will be assigned the corresponding function name from the array:
#include <iostream>
void Foo() { std::cout << "Foo" << std::endl; }
void Bar() { std::cout << "Bar" << std::endl; }
void FooBar(){ std::cout << "FooBar" << std::endl; }
void Baz() { std::cout << "Baz" << std::endl; }
void FooBaz(){ std::cout << "FooBaz" << std::endl; }
int main()
{
void (*pFunc)();
void* pvArray[5] = {(void*)Foo, (void*)Bar, (void*)FooBar, (void*)Baz, (void*)FooBaz};
int choice;
std::cout << "Which function: ";
std::cin >> choice;
std::cout << std::endl;
// or random index: choice = rand() % 5;
pFunc = (void(*)())pvArray[choice];
(*pFunc)();
// or iteratley call them all:
std::cout << "calling functions iteraely:" << std::endl;
for(int i(0); i < 5; i++)
{
pFunc = (void(*)())pvArray[i];
(*pFunc)();
}
std::cout << std::endl;
return 0;
}
the program works fine but I only is it good or there's an alternative. every comment is welcome
There is absolutely no point in converting function pointers to void* and back. Define an array of function pointers, and use it as a normal array. The syntax for the declaration is described in this Q&A (it is for C, but the syntax remains the same in C++). The syntax for the call is a straightforward () application after the indexer [].
void (*pFunc[])() = {Foo, Bar, FooBar, Baz, FooBaz};
...
pFunc[choice]();
Demo.
Note: Although function pointers work in C++, a more flexible approach is to use std::function objects instead.

How to create exceptions?

So I have an upcoming assignment dealing with exceptions and using them in my current address book program that most of the homework is centered around. I decided to play around with exceptions and the whole try catch thing, and using a class design, which is what I will eventually have to do for my assignment in a couple of weeks. I have working code that check the exception just fine, but what I want to know, is if there is a way to standardize my error message function, (i.e my what() call):
Here s my code:
#include <iostream>
#include <exception>
using namespace std;
class testException: public exception
{
public:
virtual const char* what() const throw() // my call to the std exception class function (doesn't nessasarily have to be virtual).
{
return "You can't divide by zero! Error code number 0, restarting the calculator..."; // my error message
}
void noZero();
}myex; //<-this is just a lazy way to create an object
int main()
{
void noZero();
int a, b;
cout << endl;
cout << "Enter a number to be divided " << endl;
cout << endl;
cin >> a;
cout << endl;
cout << "You entered " << a << " , Now give me a number to divide by " << endl;
cin >> b;
try
{
myex.noZero(b); // trys my exception from my class to see if there is an issue
}
catch(testException &te) // if the error is true, then this calls up the eror message and restarts the progrm from the start.
{
cout << te.what() << endl;
return main();
}
cout <<endl;
cout << "The two numbers divided are " << (a / b) << endl; // if no errors are found, then the calculation is performed and the program exits.
return 0;
}
void testException::noZero(int &b) //my function that tests what I want to check
{
if(b == 0 ) throw myex; // only need to see if the problem exists, if it does, I throw my exception object, if it doesn't I just move onto the regular code.
}
What I would like to be able to do is make it so my what() function can return a value dependent on what type of error is being called on. So for instance, if I were calling up an error that looked a the top number,(a), to see if it was a zero, and if it was, it would then set the message to say that "you can't have a numerator of zero", but still be inside the what() function. Here's an example:
virtual const char* what() const throw()
if(myex == 1)
{
return "You can't have a 0 for the numerator! Error code # 1 "
}
else
return "You can't divide by zero! Error code number 0, restarting the calculator..."; // my error message
}
This obviously wouldn't work, but is there a way to make it so I'm not writing a different function for each error message?
Your code contains a lot of misconceptions. The short answer is yes, you can change what() in order to return whatever you want. But let's go step by step.
#include <iostream>
#include <exception>
#include <stdexcept>
#include <sstream>
using namespace std;
class DivideByZeroException: public runtime_error {
public:
DivideByZeroException(int x, int y)
: runtime_error( "division by zero" ), numerator( x ), denominator( y )
{}
virtual const char* what() const throw()
{
cnvt.str( "" );
cnvt << runtime_error::what() << ": " << getNumerator()
<< " / " << getDenominator();
return cnvt.str().c_str();
}
int getNumerator() const
{ return numerator; }
int getDenominator() const
{ return denominator; }
template<typename T>
static T divide(const T& n1, const T& n2)
{
if ( n2 == T( 0 ) ) {
throw DivideByZeroException( n1, n2 );
}
return ( n1 / n2 );
}
private:
int numerator;
int denominator;
static ostringstream cnvt;
};
ostringstream DivideByZeroException::cnvt;
In the first place, runtime_error, derived from exception, is the adviced exception class to derive from. This is declared in the stdexcept header. You only have to initialize its constructor with the message you are going to return in the what() method.
Secondly, you should appropriately name your classes. I understand this is just a test, but a descriptive name will always help to read and understand your code.
As you can see, I've changed the constructor in order to accept the numbers to divide that provoked the exception. You did the test in the exception... well, I've respected this, but as a static function which can be invoked from the outside.
And finally, the what() method. Since we are dividing two numbers, it would be nice to show that two numbers that provoked the exception. The only way to achieve that is the use of ostringstream. Here we make it static so there is no problem of returning a pointer to a stack object (i.e., having cnvt a local variable would introduce undefined behaviour).
The rest of the program is more or less as you listed it in your question:
int main()
{
int a, b, result;
cout << endl;
cout << "Enter a number to be divided " << endl;
cout << endl;
cin >> a;
cout << endl;
cout << "You entered " << a << " , Now give me a number to divide by " << endl;
cin >> b;
try
{
result = DivideByZeroException::divide( a, b );
cout << "\nThe two numbers divided are " << result << endl;
}
catch(const DivideByZeroException &e)
{
cout << e.what() << endl;
}
return 0;
}
As you can see, I've removed your return main() instruction. It does not make sense, since you cannot call main() recursively. Also, the objective of that is a mistake: you'd expect to retry the operation that provoked the exception, but this is not possible, since exceptions are not reentrant. You can, however, change the source code a little bit, to achieve the same effect:
int main()
{
int a, b, result;
bool error;
do {
error = false;
cout << endl;
cout << "Enter a number to be divided " << endl;
cout << endl;
cin >> a;
cout << endl;
cout << "You entered " << a << " , Now give me a number to divide by " << endl;
cin >> b;
try
{
result = DivideByZeroException::divide( a, b ); // trys my exception from my class to see if there is an issue
cout << "\nThe two numbers divided are " << result << endl;
}
catch(const DivideByZeroException &e) // if the error is true, then this calls up the eror message and restarts the progrm from the start.
{
cout << e.what() << endl;
error = true;
}
} while( error );
return 0;
}
As you can see, in case of an error the execution follows until a "proper" division is entered.
Hope this helps.
You can create your own exception class for length error like this
class MyException : public std::length_error{
public:
MyException(const int &n):std::length_error(to_string(n)){}
};
class zeroNumerator: public std::exception
{
const char* what() const throw() { return "Numerator can't be 0.\n"; }
};
//...
try
{
myex.noZero(b); // trys my exception from my class to see if there is an issue
if(myex==1)
{
throw zeroNumerator(); // This would be a class that you create saying that you can't have 0 on the numerator
}
}
catch(testException &te)
{
cout << te.what() << endl;
return main();
}
You should always use std::exception&e. so do
catch(std::exception & e)
{
cout<<e.what();
}
You should consider a hierarchy of classes.
The reason for it might not be obvious when trying to use exceptions just for transferring a string, but actual intent of using exceptions should be a mechanism for advanced handling of exceptional situations. A lot of things are being done under the hood of C++ runtime environment while call stack is unwound when traveling from 'throw' to corresponded 'catch'.
An example of the classes could be:
class CalculationError : public std::runtime_error {
public:
CalculationError(const char * message)
:runtime_error(message)
{
}
};
class ZeroDeviderError : public CalculationError {
public:
ZeroDeviderError(int numerator, const char * message)
: CalculationError(message)
, numerator (numerator)
{
}
int GetNumerator() const { return numerator; }
private:
const int numerator;
};
Providing different classes for the errors, you give developers a chance to handle different errors in particular ways (not just display an error message)
Providing a base class for the types of error, allows developers to be more flexible - be as specific as they need.
In some cases, they might want to be specific
} catch (const ZeroDividerError & ex) {
// ...
}
in others, not
} catch (const CalculationError & ex) {
// ...
}
Some additional details:
You should not create objects of your exceptions before throwing in the manner you did. Regardless your intention, it is just useless - anyway, you are working with a copy of the object in the catch section (don't be confused by access via reference - another instance of the exception object is created when throwing)
Using a const reference would be a good style catch (const testException &te) unless you really need a non-constant object.
Also, please note that the type (classes) used for exceptions are not permitted to throw exceptions out of their copy constructors since, if the initial exception is attempted to be caught by value, a call of copy constructor is possible (in case is not elided by the compiler) and this additional exception will interrupt the initial exception handling before the initial exception is caught, which causes calling std::terminate.
Since C++11 compilers are permitted to eliminate the copying in some cases when catching, but both the elision is not always sensible and, if sensible, it is only permission but not obligation (see https://en.cppreference.com/w/cpp/language/copy_elision for details; before C++11 the standards of the language didn’t regulate the matter).
Also, you should avoid exceptions (will call them the additional) to be thrown out of constructors and move constructors of your types (classes) used for exceptions (will call them initial) since the constructors and move constructors could be called when throwing objects of the types as initial exceptions, then throwing out an additional exception would prevent creation of an initial exception object, and the initial would just be lost. As well as an additional exception from a copy constructor, when throwing an initial one, would cause the same.

Printing a pointer-to-member-field

I was debugging some code involving pointers to member fields, and i decided to print them out to see their values. I had a function returning a pointer to member:
#include <stdio.h>
struct test {int x, y, z;};
typedef int test::*ptr_to_member;
ptr_to_member select(int what)
{
switch (what) {
case 0: return &test::x;
case 1: return &test::y;
case 2: return &test::z;
default: return NULL;
}
}
I tried using cout:
#include <iostream>
int main()
{
std::cout << select(0) << " and " << select(3) << '\n';
}
I got 1 and 0. I thought the numbers indicated the position of the field inside the struct (that is, 1 is y and 0 is x), but no, the printed value is actually 1 for non-null pointer and 0 for null pointer. I guess this is a standard-compliant behavior (even though it's not helpful) - am i right? In addition, is it possible for a compliant c++ implementation to print always 0 for pointers-to-members? Or even an empty string?
And, finally, how can i print a pointer-to-member in a meaningful manner? I came up with two ugly ways:
printf("%d and %d\n", select(0), select(3)); // not 64-bit-compatible, i guess?
ptr_to_member temp1 = select(0); // have to declare temporary variables
ptr_to_member temp2 = select(3);
std::cout << *(int*)&temp1 << " and " << *(int*)&temp2 << '\n'; // UGLY!
Any better ways?
Pointers to members are not as simple as you may think. Their size changes from compiler to compiler and from class to class depending on whether the class has virtual methods or not and whether it has multiple inheritance or not. Assuming they are int sized is not the right way to go. What you can do is print them in hexadecimal:
void dumpByte(char i_byte)
{
std::cout << std::hex << static_cast<int>((i_byte & 0xf0) >> 4);
std::cout << std::hex << static_cast<int>(i_byte & 0x0f));
} // ()
template <typename T>
void dumpStuff(T* i_pStuff)
{
const char* pStuff = reinterpret_cast<const char*>(i_pStuff);
size_t size = sizeof(T);
while (size)
{
dumpByte(*pStuff);
++pStuff;
--size;
} // while
} // ()
However, I'm not sure how useful that information will be to you since you don't know what is the structure of the pointers and what each byte (or several bytes) mean.
Member pointers aren't ordinary pointers. The overloads you expect for << aren't in fact there.
If you don't mind some type punning, you can hack something up to print the actual values:
int main()
{
ptr_to_member a = select(0), b = select(1);
std::cout << *reinterpret_cast<uint32_t*>(&a) << " and "
<< *reinterpret_cast<uint32_t*>(&b) << " and "
<< sizeof(ptr_to_member) << '\n';
}
You can display the raw values of these pointer-to-members as follows:
#include <iostream>
struct test {int x, y, z;};
typedef int test::*ptr_to_member;
ptr_to_member select(int what)
{
switch (what) {
case 0: return &test::x;
case 1: return &test::y;
case 2: return &test::z;
default: return NULL;
}
}
int main()
{
ptr_to_member x = select(0) ;
ptr_to_member y = select(1) ;
ptr_to_member z = select(2) ;
std::cout << *(void**)&x << ", " << *(void**)&y << ", " << *(void**)&z << std::endl ;
}
You get warnings about breaking strict anti-aliasing rules (see this link), but the result is what you might expect:
0, 0x4, 0x8
Nevertheless, the compiler is free to implement pointer-to-member functionality however it likes, so you can't rely on these values being meaningful.
I think you should use printf to solve this problen
#include <stdio.h>
struct test{int x,y,z;}
int main(int argc, char* argv[])
{
printf("&test::x=%p\n", &test::x);
printf("&test::y=%p\n", &test::y);
printf("&test::z=%p\n", &test::z);
return 0;
}