Hi,
I want to fill up an array of int with 1. I have such problem:
and when i run it:Code:#include <cstring> #include <cstdio> int main() { int y[4]={9,9,9,9}, i; memset(y,1,4 * sizeof(int)); for(i=0;i<4;i++) printf("%d",y[i]); return 0; }
$ ./a.out
16843009168430091684300916843009$
I have no clue what's wrong, I expect something with third argument of memset() but 4 * sizeof(int) is a size of y. Can anyone help me ?
Regards.
Because the 1 fills each byte, not each int.
Each int is now 0x01010101 (which is some large number, probably the one you're seeing)
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Hm... so how shouid i call memset?
Regards.
This will fix your problem:
Code:void memset32(void *Buffer,DWORD value,DWORD size) { asm { les edi,[Buffer] mov ecx,[value] mov eax,[size] rep stosd and ecx,03h rep stosb } }
tx Bubba but is there no way to use standard memset? Btw. is this asm code portable, I mean if it depends on arch?
Regards
If you want to set something other than bytes, then use a for loop
Code:for(i=0;i<4;i++) y[i] = 1;
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Code:#include <algorithm> int y[4]; std::fill(y,y+4,1); // Set all 4 elements of array "y" to 1
"Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
-Christopher Hitchens
Ok, now i have no doubt. Thank you.
Regards.
