I have initialised the entire array with value 1 but the output is showing some garbage value. But this program works correctly if i use 0 or -1 in place of 1. So are there some restrictions on what type of values can be initialised using memset.
int main(){
int a[100];
memset(a,1,sizeof(a));
cout<<a[5]<<endl;
return 0;
}
memset, as the other say, sets every byte of the array at the specified value.
The reason this works with 0 and -1 is because both use the same repeating pattern on arbitrary sizes:
(int) -1 is 0xffffffff
(char) -1 is 0xff
so filling a memory region with 0xff will effectively fill the array with -1.
However, if you're filling it with 1, you are setting every byte to 0x01; hence, it would be the same as setting every integer to the value 0x01010101, which is very unlikely what you want.
Memset fills bytes, from cppreference:
Converts the value ch to unsigned char and copies it into each of the first count characters of the object pointed to by dest.
Your int takes several bytes, e.g. a 32bit int will be filled with 1,1,1,1 (in base 256, endianess doesn't matter in this case), which you then falsly interpreted as a "garbage" value.
0x01010101 which is what you'll be forming in each a is 16843009 decimal, which I predict is the "garbage" value you're seeing. -1 "works" (for the wrong definition of "works") because it fills the array's memory with 0xff and 0xffffffff is the 32-bit representation of -1.The other answers have explained std::memset already. But it's best to avoid such low level features and program at a higher level. So just use the Standard Library and its C++11 std::array
#include <array>
std::array<int, 100> a;
a.fill(1);
Or if you prefer C-style arrays, still use the Standard Library with the std::fill algorithm as indicated by @BoPersson
#include <algorithm>
#include <iterator>
int a[100];
std::fill(std::begin(a), std::end(a), 1);
In most implementations, both versions will call std::memset if it is safe to do so.
memset is an operation that sets bits.
If you want to set a value use a for-loop.
Consider a 4-bit-integer:
Its value is 1 when the bits are 0001 but memset sets it to 1111
sizeof(a)does, and then you should figure out what is the type of the third argument ofmemset(). This should clear it up. PS. Don't do this in C++, it is unnecessary, and as you see, wrong.std::fill(std::begin(a), std::end(a), 1);