Fixed to point out that that frexp() returns a number whose

*absolute* value is >= 0.5 and < 1.  Amended the example
program to demonstrate this.

man3/frexp.3

@@ -48,9 +48,10 @@ Link with \-lm.
 The \fBfrexp\fP() function is used to split the number \fIx\fP into a
 normalized fraction and an exponent which is stored in \fIexp\fP.
 .SH "RETURN VALUE"
-The \fBfrexp\fP() function returns the normalized fraction.  If the
-argument \fIx\fP is not zero, the normalized fraction is \fIx\fP
-times a power of two, and is always in the range 1/2 (inclusive) to
+The \fBfrexp\fP() function returns the normalized fraction.  
+If the argument \fIx\fP is not zero, 
+the normalized fraction is \fIx\fP times a power of two, 
+and its absolute value is always in the range 1/2 (inclusive) to
 1 (exclusive).  If \fIx\fP is zero, then the normalized fraction is
 zero and zero is stored in \fIexp\fP.
 .SH "CONFORMING TO"
@@ -58,23 +59,33 @@ SVID 3, POSIX, 4.3BSD, ISO 9899.
 The float and the long double variants are C99 requirements.
 .SH EXAMPLE
 .nf
-#include <stdio.h>
 #include <math.h>
 #include <float.h>
-int main () {
-        double d = 2560;
-        int e;
-        double f = frexp(d, &e);
-        printf("frexp(%g, &e) = %g: %g * %d^%d = %g\en",
-               d, f, f, FLT_RADIX, e, d);
-        return 0;
-}
+#include <stdio.h>
+#include <stdlib.h>
+
+int
+main(int argc, char *argv[])
+{
+    double x, r;
+    int exp;
+
+    x = strtod(argv[1], NULL);
+    r = frexp(x, &exp);
+
+    printf("frexp(%g, &e) = %g: %g * %d^%d = %g\n",
+		x, r, r, FLT_RADIX, exp, x);
+    exit(EXIT_SUCCESS);
+} /* main */
 .fi
 .sp
-This program prints
+This program produces results such as the following:
 .sp
 .in +5
+$ ./a.out 2560
 frexp(2560, &e) = 0.625: 0.625 * 2^12 = 2560
+$ ./a.out -4
+frexp(-4, &e) = -0.5: -0.5 * 2^3 = -4
 .in
 .SH "SEE ALSO"
 .BR ldexp (3),