feat(math): add math/base/special/round10f

Author: vivekmaurya001

lib/node_modules/@stdlib/math/base/special/round10f/src/main.c

@@ -26,11 +26,11 @@
 #include "stdlib/constants/float32/max_base10_exponent.h"
 #include "stdlib/constants/float32/min_base10_exponent_subnormal.h"
 
-// 10^308:
-static const double HUGE_VALUE = 1.0e308;
+// 10^38:
+static const float HUGE_VALUE = 1.0e38;
 
-// 10^-323
-static const double TINY_VALUE = 1.0e-323;
+// 10^-45
+static const float TINY_VALUE = 1.0e-45;
 
 /**
 * Rounds a numeric value to the nearest power of `10` on a linear scale.
@@ -39,18 +39,18 @@ static const double TINY_VALUE = 1.0e-323;
 * @return     rounded value
 *
 * @example
-* double out = stdlib_base_round10( 3.141592653589793 );
+* float out = stdlib_base_round10f( 3.141592653589793 );
 * // returns 1.0
 */
-double stdlib_base_round10( const double x ) {
-	double sign;
-	double half;
-	double p1;
-	double p2;
-	double y1;
-	double y2;
-	double xc;
-	double p;
+float stdlib_base_round10f( const float x ) {
+	float sign;
+	float half;
+	float p1;
+	float p2;
+	float y1;
+	float y2;
+	float xc;
+	float p;
 
 	if ( stdlib_base_is_nan( x ) || stdlib_base_is_infinite( x ) || x == 0.0 ) {
 		return x;
@@ -64,25 +64,25 @@ double stdlib_base_round10( const double x ) {
 	}
 
 	// Solve the equation `10^p = x` for `p`:
-	p = stdlib_base_log10( xc );
+	p = stdlib_base_logf( xc, 10.0 );
 
 	// Find the previous and next integer powers:
-	p1 = stdlib_base_floor( p );
-	p2 = stdlib_base_ceil( p );
+	p1 = stdlib_base_floorf( p );
+	p2 = stdlib_base_ceilf( p );
 
 	// Handle tiny:
-	if ( p1 == STDLIB_CONSTANT_FLOAT64_MIN_BASE10_EXPONENT_SUBNORMAL ) {
+	if ( p1 == STDLIB_CONSTANT_FLOAT32_MIN_BASE10_EXPONENT_SUBNORMAL ) {
 		return sign * TINY_VALUE;
 	}
 
 	// Handle huge:
-	if ( p1 == STDLIB_CONSTANT_FLOAT64_MAX_BASE10_EXPONENT ) {
+	if ( p1 == STDLIB_CONSTANT_FLOAT32_MAX_BASE10_EXPONENT ) {
 		return sign * HUGE_VALUE;
 	}
 
 	// Compute previous and next powers of 10:
-	y1 = stdlib_base_pow( 10.0, p1 );
-	y2 = stdlib_base_pow( 10.0, p2 );
+	y1 = stdlib_base_pow( 10.0, (float)p1 );
+	y2 = stdlib_base_pow( 10.0, (float)p2 );
 
 	// Find the closest power of 10:
 	half = ( y2 - y1 ) / 2.0;