Why is memset() incorrectly initializing int?

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);

memset(grid, 5, 100 * sizeof(int));

You are setting 400 bytes, starting at (char*)grid and ending at (char*)grid + (100 * sizeof(int)), to the value 5 (the casts are necessary here because memset deals in bytes, whereas pointer arithmetic deals in objects.

84215045 in hex is 0x05050505; since int (on your platform/compiler/etc.) is represented by four bytes, when you print it, you get "four fives."

memset is about setting bytes, not values. One of the many ways to set array values in C++ is std::fill_n:

std::fill_n(grid, 100, 5);

Don't use memset.

You set each byte [] of the memory to the value of 5. Each int is 4 bytes long [5][5][5][5], which the compiler correctly interprets as 5*256*256*256 + 5*256*256 + 5*256 + 5 = 84215045. Instead, use a for loop, which also doesn't require sizeof(). In general, sizeof() means you're doing something the hard way.

for(int i=0; i<110; ++i)
    grid[i] = 5;