Infer generic type argument from parameter type of function in TypeScript -

i want create toplainobject() function in typescript , have come working example:

function toplainobject<s extends d, d>(source: s) {     return json.parse(json.stringify(source)) d; } 

now can invoke function this:

interface isample {} class sample implements isample {}  let plain: isample = toplainobject<sample, isample>(new sample()); 

now question: there way declare toplainobjectwithout need of first generic type argument s extends d using first parameter type (which s) invoke function doing:

let plain: isample = toplainobject<isample>(new sample()); 

the signature function toplainobject<d>(source: s extends d) { ... } not work , results in syntax error.

maybe misunderstand after, don't see why can't do:

interface isample {} class sample implements isample {}  function toplainobject<tinterface>(source: tinterface) : tinterface {     return json.parse(json.stringify(source)) tinterface; }  let plain: isample = toplainobject(new sample()); 

also sample works me no problem (typescript 1.8.10)

interface isample {} class sample implements isample {}  function toplainobject<s extends d, d>(source: s) {     return json.parse(json.stringify(source)) d; }  let plain: isample = toplainobject(new sample());