Try this video series: https://youtu.be/F3ntGDm6hOs
This is a "pre-introduction" series of 5 videos, to the Handmade Hero video series. HH is an ongoing series and may/may not be valuable to you, but the prelim is a good, fast intro that explains a bunch of concepts on Windows.
You got caught by the spam filter that put your message in moderation. Next time be a bit patient please instead of posting again.
I have let both messages through and deleted the other one.
If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about?
Help me to understand what I'm saying, and I'll explain it better to you
Rating helpful answers is nice, but saying thanks can be even nicer.
C:\Program Files (x86)\Windows Kits\8.1\include\\shared\secext.h(117): error C2146: syntax error: missing ';' before identifier 'GetUserNameExA'
C:\Program Files (x86)\Windows Kits\8.1\include\\shared\secext.h(121): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
C:\Program Files (x86)\Windows Kits\8.1\include\\shared\secext.h(126): error C2086: 'BOOLEAN SEC_ENTRY': redefinition
C:\Program Files (x86)\Windows Kits\8.1\include\\shared\secext.h(117): note: see declaration of 'SEC_ENTRY'
C:\Program Files (x86)\Windows Kits\8.1\include\\shared\secext.h(126): error C2146: syntax error: missing ';' before identifier 'GetUserNameExW'
C:\Program Files (x86)\Windows Kits\8.1\include\\shared\secext.h(130): error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
I have modified program according to your suggestion but I am not getting expected result. I want to use NameFullyQualifiedDN enum value and get the fully qualified distinguished name (for example, CN=Jeff Smith,OU=Users,DC=Engineering,DC=Microsoft,DC=Com).
You are putting holes in your arrays of even and odd elements. In order to avoid that, you need to maintain two separate variables to keep track of how many even and odd items have been added so far. Try
char str = "Hello";
char *const p=str;
As per my knowledge when we use const then we can't change a char in t then how output is Mello instead of Hello.
const applies to what's on its immediate left: a pointer (*) or a type (char and many others, even ones that you've defined). If there's nothing on its left--that is, when it appears first--const applies to what's on its immediate right.
I am learning callbacks in c.I found an example in callback but could not understand .Can some one please explain me how the flow of execution happens and brief explaination of program.
PFA code what I found.
<pre>* callback.c */
/* callback function definition goes here */
/* initialize function pointer to
printf("This is a program demonstrating function callback\n");
/* register our callback function */
printf("back inside main program\n");
void my_callback(void)... /* This is the callback function that will be invoked. */
callback ptr_my_callback = my_callback; /* ptr_my_callback points to the function my_callback */
register_callback(ptr_my_callback); /* registers the callback */
/* This defines the type "callback". The '*' before "callback", which is in
parentheses, means that it's a pointer to a function. The function returns
void and takes (void) as its parameter. */
typedef void (*callback)(void);
/* This function is called to register the callback. It actually doesn't do
that, because registering would mean saving a pointer to it. All that it does
is invoke the function immediately, which is what it could do later if it
saved its parameter somewhere and used it to call the function when it was
time to do so. To call the function, it dereferences the pointer to it and
invokes it with no arguments. */void register_callback(callback ptr_reg_callback)
The printfs are just there so that you can observe the flow of execution.