Does C++ round floats to the nearest int when casted into an int?

[johnhoffman]

I performed the following casting in C++.

int n = (int) sqrt(3);

The square root of 3 is about 1.732.

When I checked the value of n in gdb, the value I got was 2. Does this mean that casting into an int rounds a float to the nearest integer?

I find this outlandish because c rounds down.

Wait... perhaps I erred. Let me check.

[Athar]

It's the same as in C, so n=1.

[The Palm Tree Magician]

C++ always truncates, aka rounds down. If you want it to round to the nearest intager, add 0.5 or 0.5f before casting.

int n = (int)(sqrt(3)+0.5f);

[Peter87]

The Palm Tree Magician wrote:

C++ always truncates, aka rounds down.

That's not true for negative numbers. It rounds towards zero.

[MrHutch]

It truncates, the rounding is irrelevant.

Basically, anything after the decimal point is chopped off.

1.7 would become 1.

-3.9 would become -3.

[z05DSL3A]

The Palm Tree Magician is correct in that it truncates. However truncation is just the fractional part is discarded. ... So I guess technically it is rounding towards zero...I'l go back to sleep. ;0)