• Deewiant (27/27) Sep 18 2005 std.string.atoi() claims to return a long. However, it's just a wrapper,

Deewiant <deewiant.doesnotlike.spam gmail.com> writes:

std.string.atoi() claims to return a long. However, it's just a wrapper,
consisting of no more than:

long atoi(char[] s)
{
    return std.c.stdlib.atoi(toStringz(s));
}

Which is basically fine, but std.c.stdlib.atoi() returns an int. Thus, atoi()
anything greater than int.max or lesser than int.min and it wraps around. The
only thing wrong with that is that to match the function signature, it should be
long.max and long.min. Some example code, which fails:

import std.string;
void main() {
	assert (atoi( "12345678900") ==  12345678900);
	assert (atoi("-12345678900") == -12345678900);
}

So, either change the function to return an int (and correct the docs at
http://www.digitalmars.com/d/phobos/std_string.html to match) or make the
function work throughout [long.min, long.max].

I've put such a function up at http://per.us.to/~deewiant/extra/atoi.d - like it
says in the comment, it's basically just a non-error-checking version of
std.conv.toLong(). But that's what I expected std.string.atoi() to be in the
first place, after reading the docs...

I suggest that atoi() is simply changed to return an int, as that's what its
name implies - "alpha to integer". Plus, it wouldn't impact any existing code at
all, unless something relies on the bug, which is doubtful.

But an atol() would be handy nevertheless, and I've just supplied a ready-made
one... ;)