CODEWITHSUNDEEP
c++function
user2953119
user2953119

Reputation:

Assignment function to function

Please explain me why the code below doesn't work

#include <stdio.h>

int foo() { return 1; }
int bar() { return 2; }

void ass()
{
    foo=bar;
}

int main()
{
    ass()
}

The following error

test.cpp: In function ‘void ass()’:
test.cpp:8:8: error: assignment of function ‘int foo()’
test.cpp:8:8: error: cannot convert ‘int()’ to ‘int()’ in assignment

caused.

Upvotes: 0

Views: 1022

Answers (3)

phyrrus9
phyrrus9

Reputation: 1467

try this:

typedef int (*int_funcptr_void)(void);

then, you can simply:

int foo() { return 1; }
int bar() { return 2; }

int_funcptr_void func;

void ass()
{
    func = (int_funcptr_void)foo;
}

int main()
{
    ass(); //you also forgot a semicolon here, but nice naming
    //then, we can call it:
    printf("%d\n", func());
}

and get this:

hydrogen:tmp phyrrus9$ ./a.out
1

Hope that helps.

Upvotes: 1

Rakib
Rakib

Reputation: 7625

You can not assign function to function, as you can not assign int to int. Think naturally, you assign an int variable to another int variable, which means you are assigning rvalue of the second variable to the lvalue(address) of the first one. Same rule for function, you can assign a function object to another function object. The difference is, function is code, not data, but it has an address ( the starting point). So assigning function to one another generally means assigning the address of the function to a variable which can hold an address, i.e. a function pointer.

void f(){}

typedef void(*pF)(); //typedef for easy use

pf foo;  //create a function pointer object
foo = &f; //assign it the address of the function

Upvotes: 0

CuriousGeorge
CuriousGeorge

Reputation: 7400

You must use a function pointer. You cannot assign to the function itself.

int(*baz)() = &foo;
baz();

Upvotes: 3

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