A call to type suffixed with one parameter checks whether a symbol or string x has the same type as y, and x begins with the characters of y.

With two parameters, x is checked to see if it has the same type of and begins with the characters of y and that the balance, when parsed, is of type stype.

If y is a set or  list, then x is checked against each operand of y.

Note that x and "x" are considered to be of type suffixed(x) and suffixed("x"), respectively.  Also note that parse will automatically simplify the balance, so "x6/2" and "x4!" is an "x" suffixed by the integers 3 and 24, respectively.

$\mathrm{type}\left(\mathrm{\_Z1}\,'\mathrm{suffixed}\left(\mathrm{\_Z}\,\mathrm{integer}\right)'\right)$

$\mathrm{true}$

$\mathrm{type}\left(''x1.32''\,'\mathrm{suffixed}\left(''x''\,\left\{\mathrm{float}\,\mathrm{integer}\right\}\right)'\right)$

$\mathrm{true}$

$\mathrm{discont}\left(\tan \left(x\right)\,x\right)$

$\left\{\frac{1}{2}\mathrm{\pi }+\mathrm{\pi }\mathrm{\_Z1~}\right\}$

$\mathrm{hastype}\left(\,'\mathrm{suffixed}\left(\left\{\mathrm{\_B}\,\mathrm{\_N}\,\mathrm{\_NN}\,\mathrm{\_Z}\right\}\,\mathrm{integer}\right)'\right)$

$\mathrm{true}$

$\mathrm{dsolve}\left(\left\{\mathrm{diff}\left(y\left(t\right)\,t\right)=y\left(t\right)\right\}\,y\left(t\right)\right)$

$\left\{y\left(t\right)=\mathrm{c\_\_1}{\ⅇ}^t\right\}$

$\mathrm{hastype}\left(\,'\mathrm{suffixed}\left(\mathrm{\_C}\,\mathrm{integer}\right)'\right)$

$\mathrm{true}$

$\mathrm{type}\left(''x4/2''\,'\mathrm{suffixed}\left(''x''\,\mathrm{integer}\right)'\right)$

$\mathrm{true}$