Why is memset() incorrectly initializing int?
c, c++, memset
Solution
`memset` sets each byte of the destination buffer to the specified value. On your system, an `int` is four bytes, each of which is 5 after the call to `memset`. Thus, `grid[0]` has the value `0x05050505` (hexadecimal), which is `84215045` in decimal.
Some platforms provide alternative APIs to `memset` that write wider patterns to the destination buffer; for example, on OS X or iOS, you could use:
int pattern = 5;
memset_pattern4(grid, &pattern, sizeof grid);
to get the behavior that you seem to expect. What platform are you targeting?
In C++, you should just use `std::fill_n`:
std::fill_n(grid, 100, 5);
Problem
Why is the output of the following program `84215045`? ``` int grid[110]; int main() { memset(grid, 5, 100 * sizeof(int)); printf("%d", grid[0]); return 0; } ```