Error C2664 Int
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Error C2664 Cannot Convert Argument
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Error C2664 Cannot Convert Parameter From Const Char To Lpcwstr
developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question mfc error c2664 x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; error c2664 cannot convert argument 1 from it only takes a minute: Sign up error C2664: cannot convert parameter 1 from 'int' to 'int (__cdecl *)(int)' up vote 1 down vote favorite 2 having some trouble passing a function as a parameter of another function... ERROR: http://stackoverflow.com/questions/22150353/error-c2664-cannot-convert-parameter-1-from-int-to-int Error 1 error C2664: 'wrapper' : cannot convert parameter 1 from 'int' to 'int (__cdecl *)(int)' int inc( int n ) { return n + 1 ; } int dec( int n ) { return n - 1 ; } int wrapper( int i, int func(int) ) { return func( i ) ; } int main(){ int a = 0 ; a = wrapper( 3, inc( 3 ) ) ; return 0 ; } c++ function parameters c2664 share|improve this http://stackoverflow.com/questions/6332078/error-c2664-cannot-convert-parameter-1-from-int-to-int-cdecl-int question edited Jun 22 '11 at 18:07 user195488 asked Jun 13 '11 at 14:52 tuk 4718 add a comment| 5 Answers 5 active oldest votes up vote 5 down vote accepted You're passing the result of a function call inc(3) to wrapper, NOT a function pointer as it expects. a = wrapper(3, &inc) ; share|improve this answer answered Jun 13 '11 at 14:54 Mark B 76.4k465138 add a comment| up vote 1 down vote Your call is passing an integer, the return value from calling inc(3), i.e. 4. That is not a function pointer. Perhaps you meant: a = wrapper(3, inc); This would work, and assign a to the value of calling int with the parameter 3. share|improve this answer answered Jun 13 '11 at 14:54 unwind 254k38331460 add a comment| up vote 1 down vote The line: a = wrapper( 3, inc( 3 ) ) ; is effectively: a = wrapper(3, 4); I think you mean: a = wrapper(3, inc); This passes a pointer to the inc() function as the second argument to wrapper(). share|improve this answer answered Jun 13 '11 at 14:55 janm 12.6k12646 add a comment| up vote 1 down vote As it is now, wrapper takes an int and a pointer to a function that takes one int and returns an int. You are trying to pass it an int and an int, because instead of passing the a pointer to
here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company Business Learn more http://stackoverflow.com/questions/28757722/c-error-c2664-int-scanfconst-char-cannot-convert-argument-1-f about hiring developers or posting ads with us Stack Overflow Questions Jobs Documentation Tags Users Badges Ask Question x Dismiss Join the Stack Overflow Community Stack Overflow is a community of 4.7 million programmers, just like you, helping each other. Join them; it only takes a minute: Sign up C++ — error C2664: 'int scanf(const char *,…)' : cannot convert argument 1 from 'int' to 'const char *' up vote 4 down vote favorite 1 I'm error c2664 very new to C++ and I'm trying to build this very simple code, but I don't understand why I get this error: Error 1 error C2664: 'int scanf(const char *,...)' : cannot convert argument 1 from 'int' to 'const char *' Here is the code: // lab.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include