what is wrong here

Question

   unsigned __int8 result[]= new unsigned __int8[sizeof(username) * 4];

IntelliSense: initialization with '{...}' expected for aggregate object

Do you have a [C++ book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) you're learning from? — GManNickG, Feb 09 '11 at 11:15
@GMan nope, plus what if i want to return result how can i do that!? — TQCopier, Feb 09 '11 at 11:25
Well before you can program in C++ you need to learn C++, so do pick up one of those books. — GManNickG, Feb 09 '11 at 11:33

score 1 · Answer 1 · answered Feb 09 '11 at 11:16

1

The types are not the same; you cannot initialize an array with a pointer.

new unsigned __int8[sizeof(username) * 4]; returns a unsigned __int8*, not unsigned __int8[]

change your code to

unsigned __int8* result = new unsigned __int8[sizeof(username) * 4];

answered Feb 09 '11 at 11:16

tenfour

what if i want to return result how can i do that!? – TQCopier Feb 09 '11 at 11:22
@Kevin i mean if result is = 5; i want to return 5 , coz i cant return reslut while it's pointing for something else , i need to get it's value from the memory or something – TQCopier Feb 09 '11 at 11:35
@TQCopier Do you mean the value pointed by `result`? In that case you cold do this: `return *result;` which returns the value pointed by `result`. – Kevin Feb 09 '11 at 14:08

score 0 · Answer 2 · answered Feb 09 '11 at 11:16

0

unsigned __int8 *result = new unsigned __int8[sizeof(username) * 4];

answered Feb 09 '11 at 11:16

Thomas

score 0 · Answer 3 · answered Feb 09 '11 at 11:16

0

new returns a pointer, not an array. You should declare

unsigned __int8* result = ....

answered Feb 09 '11 at 11:16

Francesco

score 0 · Answer 4 · answered Feb 09 '11 at 11:16

0

here, result is an array of __int8, so you can't assign one value into the entire array. You actually want:

unsigned __int8* p_result = new unsigned __int8[sizeof username * 4];

answered Feb 09 '11 at 11:16

Tony Delroy

4 Answers4