Can someone explains me this line of code?

The function returns the value of the loc bit of the integer variable num. Note you could rewrite it this way

int float_bit_return(int num, int loc)
{
    return ((num >> loc) & 1);
}

(the inner braces are not needed due to C operator precedence rules, but I find the code more readable with them).

You may actually use use such a function to inspect the bits of a float value, as the following code shows (it assumes sizeof(float)==sizeof(int))

#include <stdio.h>
int float_bit_return(int num, int loc)
{
    return ((num >> loc) & 1);
}

int main()
{
  int i = 5;
  float f = 3.25f;

  printf("bits of the integer value %d\n", i);
  for (int n=0; n<sizeof(int)*8; ++n)
    printf("%02d %d\n", n, float_bit_return(i, n));


  printf("bits of the float value  %f\n", f);
  for (int n=0; n<sizeof(int)*8; ++n)
    printf("%02d %d\n", n, float_bit_return(*(int*)&f, n));

  return 0;
}

Output:

[...]
bits of the float value  3.250000
00 0
01 0
02 0
03 0
04 0
05 0
06 0
07 0
08 0
09 0
10 0
11 0
12 0
13 0
14 0
15 0
16 0
17 0
18 0
19 0
20 1
21 0
22 1
23 0
24 0
25 0
26 0
27 0
28 0
29 0
30 1
31 0

with
sign: bit 31 equal to zero -> positive number
exponent bits 30-23 = 128 -> 2^1
mantissa bits 22-0 = .101b -> 1.101b

so,
f = 1.101b * 2^1 = 11.01b = 3.25

[update]
The same result could be obtained using a (possibly more intuite) union:

#include <stdio.h>
int float_bit_return(int num, int loc)
{
    return ((num >> loc) & 1);
}

int main()
{
  union // the int i and the float f share the same memory area (hence the same bit pattern)
  {
    int i;
    float f;
  } u;

  u.f = 3.25;

  printf("bits of the float value  %f\n", u.f);
  for (int n=0; n<sizeof(int)*8; ++n)
    printf("%02d %d\n", n, float_bit_return( u.i, n ));

  return 0;
}

[/update]