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suppose I declare:

struct [[eosio::table]] tst {
   asset component;
   uint64_t primary_key() const {
      return component.symbol.code().raw();
   }   
};  

where I mean to keep a list of assets. when it's time to find a record in the table I want to do something like:

void token::find(char *symbol) {
   tst t(_self, _self.value);
   auto o = t.find(symbol);

but this results in an error:

t.cpp:127:20: error: cannot initialize a parameter of type 'uint64_t' (aka 'unsigned long long') with an lvalue of type 'char *' auto o = t.find(symbol);

changing the parameter definition to:

void token::find(std::string symbol) {

produces a similar message:

t.cpp:127:20: error: no viable conversion from 'std::string' (aka 'basic_string, allocator >') to 'uint64_t' (aka 'unsigned long long') auto o = t.find(symbol);

in other code I've seen these finds done given an asset. something like this works:

auto o = t.find(someAsset.symbol.code().raw());

which ultimately resolves to a uint64_t

so I need somehow to convert my string to an integer. can anyone provide a hint as to how I might do this?

1

When you are lost with EOSIO native types, you can look it up at eosio.cdt's eosiolib library (github)

A explicit constructor is defined as following inside symbol.hpp

constexpr explicit symbol_code( std::string_view str )
      :value(0)
      {
         if( str.size() > 7 ) {
            eosio::check( false, "string is too long to be a valid symbol_code" );
         }
         for( auto itr = str.rbegin(); itr != str.rend(); ++itr ) {
            if( *itr < 'A' || *itr > 'Z') {
               eosio::check( false, "only uppercase letters allowed in symbol_code string" );
            }
            value <<= 8;
            value |= *itr;
         }
      }

symbol.hpp is included in asset.hpp so it should be fine to include just theeosiolib/asset.hpp as usual.

  • thanks for the reply. I guess the answer is I can do something like symbol::symbol_code("USD"). I've marked your reply as the answer even though it doesn't actually answer the question – ekkis Jan 25 '19 at 23:05
  • actually, the above doesn't work. so I'll reopen this question – ekkis Jan 2 at 5:43

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