Use in C

int C = var_in_float;

They will convert implicit

If you want to round it to lower, just cast it.

float my_float = 42.8f;
int my_int;
my_int = (int)my_float;          // => my_int=42

For other purpose, if you want to round it to nearest, you can make a little function or a define like this:

#define FLOAT_TO_INT(x) ((x)>=0?(int)((x)+0.5):(int)((x)-0.5))

float my_float = 42.8f;
int my_int;
my_int = FLOAT_TO_INT(my_float); // => my_int=43

Be careful, ideally you should verify float is between INT_MIN and INT_MAX before casting it.

double a = 100.3;
printf("%f %d\n", a, (int)(a* 10.0));

Output Cygwin 100.3 1003
Output MinGW: 100.3 1002

Using (int) to convert double to int seems not to be fail-safe

You can find more about that here: Convert double to int?

Good guestion! -- where I have not yet found a satisfying answer for my case, the answer I provide here works for me, but may not be future proof...

If one uses gcc (clang?) and have -Werror and -Wbad-function-cast defined,

int val = (int)pow(10,9);

will result:

error: cast from function call of type 'double' to non-matching type 'int' [-Werror=bad-function-cast]

(for a good reason, overflow and where values are rounded needs to be thought out)

EDIT: 2020-08-30: So, my use case casting the value from function returning double to int, and chose pow() to represent that in place of a private function somewhere. Then I sidestepped thinking pow() more. (See comments more why pow() used below could be problematic...).

After properly thought out (that parameters to pow() are good), int val = pow(10,9); seems to work with gcc 9.2 x86-64 ...

but note:

printf("%d\n", pow(10,4));

may output e.g.

-1121380856

(did for me) where

int i = pow(10,4); printf("%d\n", i);

printed

10000

in one particular case I tried.