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.
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
