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Likewise, array names can be implicitly cast to be pointers to their first element.

Char g video xxx-53

see also ma set mark a at current cursor location 'a jump to line of mark a (first non-blank character in line) `a jump to position (line and column) of mark a d'a delete from current line to line of mark a d`a delete from current cursor position to position of mark a c'a change text from current line to line of mark a y`a yank text to unnamed buffer from cursor to position of mark a :marks list all the current marks :marks a B list marks a, B x Delete char UNDER cursor X Delete char BEFORE cursor #x Delete the next # chars.

starting from char under cursor dw Delete next word d W Delete UP TO the next word d^ Delete up unto the beginning of the line d$ Delete until end of the line D See d$, delete until end of the line dd Delete whole line dib Delete contents in parenthesis '(' ')' block (e.g.

There are a lot of similarities between arrays and pointers.

Particularly, pointers are designed in a way which allows them to be used as if they pointed to array data.

The thing that might be tripping you up is that arrays can be implicitly cast to pointers to their first element.

So in most cases you can treat array names as if they were pointers -- even though they're not.

Gli e Book vengono realizzati e pubblicati nei più svariati formati, molti dei quali però non sono stati originariamente concepiti per essere dei veri e propri formati di e Book.

The GN-XXX Gundam Rasiel (aka Gundam Rasiel, Rasiel), is the experimental 3rd generation Gundam in Mobile Suit Gundam 00P.

So this is NOT a pointer, but is just a single character.

, è un libro in formato digitale a cui si può avere accesso mediante computer e dispositivi mobili, come smartphone, tablet PC e dispositivi appositamente ideati per la lettura di testi lunghi in digitale, detti e Reader (ebook reader).

like sizeof: /* a = char* (as we defined it above) a[0] = char (this takes 1 single character (the 'H') from that pointer &a[0] = char* (this give us a pointer to the 'H') So really: &a[0] == a They're both pointers, they both point to the exact same address, and they both are of the exact same type (char*). Note that &a[1] == a 1 This is still of type char*, but it no longer points to the 'H'. So when we send this to cout, it will print string data starting from 'e' until it reaches the null terminator at the end.

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