function
<cmath> <ctgmath>

# scalbln

```     double scalbln  (double x     , long int n);
float scalblnf (float x      , long int n);
long double scalblnl (long double x, long int n);
```
```     double scalbln (double x     , long int n);
float scalbln (float x      , long int n);
long double scalbln (long double x, long int n);
double scalbln (T x          , long int n); // additional overloads for integral types```
Scale significand using floating-point base exponent (long)
Scales x by FLT_RADIX raised to the power of n, returning the result of computing:

`scalbn(x,n)` = x * FLT_RADIXn

Presumably, x and n are the components of a floating-point number in the system; In such a case, this function may be optimized to be more efficient than the theoretical operations to compute the value explicitly.

There also exists another version of this function: scalbn, which is identical, except that it takes an `int` as second argument.

Header <tgmath.h> provides a type-generic macro version of this function.
Additional overloads are provided in this header (`<cmath>`) for the integral types: These overloads effectively cast x to a `double` before calculations (defined for T being any integral type).

### Parameters

Value representing the significand.</dd>
exp
Value of the exponent.

### Return Value

If the magnitude of the result is too large to be represented by a value of the return type, the function returns HUGE_VAL (or HUGE_VALF or HUGE_VALL) with the proper sign, and an overflow range error may occur (if too small, the function returns zero, and an underflow range error may occur).

If a range error occurs:
- And math_errhandling has MATH_ERRNO set: the global variable errno is set to ERANGE.
- And math_errhandling has MATH_ERREXCEPT set: either FE_OVERFLOW or FE_UNDERFLOW is raised.

### Example

 ``12345678910111213141516`` ``````/* scalbln example */ #include /* printf */ #include /* FLT_RADIX */ #include /* scalbn */ int main () { double param, result; long n; param = 1.50; n = 4L; result = scalbln (param , n); printf ("%f * %d^%d = %f\n", param, FLT_RADIX, n, result); return 0; }``````

Output:
 ``` 1.500000 * 2^4 = 24.000000 ```