I'm converting some 32-bit compatible code into 64-bit - and I've hit a snag. I'm compiling a VS2008 x64 project, and I receive this warning:
warning C4334: '<<' : result of 32-bit shift implicitly converted to 64 bits
(was 64-bit shift intended?)
Here's the original line of code:
if ((j & (1 << k)) != 0) {
And here's what it looks like if I follow Microsoft's advice:
if ((j & (1i64 << k)) != 0) {
Is this safe to do, when the code will be compiled on both 32-bit and 64-bit systems? If so, please explain why I must add "i64" to the end, and why this will not affect a 32-bit compilation. Otherwise, a work-around would be much appreciated.
Beyond this, I have what looks like an even trickier bit of code.
if (id[j] != id[j ^ (1u << k)]) {
I understand that the "u" means that the digit is unsigned, but what's the point in specifying that on a value that doesn't exceed the signed maximum value... I'm guessing this has something to do with the bit shift?