type conversion - Typed integer constant in OCaml -


आप कहते हैं कि ऐसा कुछ कोड था 

  type foo_t = int64 do_something_with_foo (f : Foo_t) = (* आपकी कल्पना को छोड़ दिया *)   
 और आप इसे "निरंतर" फू के साथ कॉल करना चाहते हैं, जैसे:  
  do_something_with_foo 99   
 आप संकलक / दुभाषिया को कैसे समझते हैं कि आपका निरंतर 99 वास्तव में एक  foo_t ?   
 
 
  do_something_with_foo 99L  का उपयोग करें। निरंतर  99  प्रकार  int ,  99L  प्रकार  int64  है।  
 यह कुछ नहीं है टाइप उपनाम  foo_t = int64  के साथ करने के लिए जब तक  foo_t  की परिभाषा दिखाई देती है (अर्थात किसी युक्त मॉड्यूल के हस्ताक्षर द्वारा नकाबपोश नहीं),  int64  के किसी भी मान में भी  foo_t  और उपाध्यक्ष।

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