c++ - How to avoid scientific notation and show full number instead? -

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i have below code in 1 of library causing numbers shown in form of scientific notation.

t value = 0;  template<typename u> void process(u& buf, dataoption holder) const {     if (holder == dataoption::types) {         switch (type_) {         case teck::proc_int:             buf << "{\"int\":" << value << "}";             break;         case teck::proc_long:             buf << "{\"long\":" << value << "}";             break;         case teck::proc_float:             buf << "{\"float\":" << value << "}";             break;         case teck::proc_double:             buf << "{\"double\":" << value << "}";             break;         default:             buf << "{\"" << type_ << "\":" << value << "}";         }     } }

for of different cases above, "value" coming in scientific notation. how can avoid showing scientific notation , instead show full number? did research , can use "std::fixed" should using this?

std::fixed works within same stream, won't work in case you're working stateless stream

case teck::proc_double:     buf << std::fixed;     buf << "{\"double\":" << value << "}";

instead should this

case teck::proc_double:     buf << "{\"double\":" << std::fixed << value << "}";

so function can simplified for better readability too

template<typename u> void process(u& buf, dataoption holder) const {     if (holder == dataoption::types)     {         buf << "{\"";          switch (type_)         {             case teck::proc_int:    buf << "int";    break;             case teck::proc_long:   buf << "long";   break;             case teck::proc_float:  buf << "float";  break;             case teck::proc_double: buf << "double"; break;             default:                buf <<  type_;         }          buf << "\":" << std::fixed << value << "}";     } }

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