C++ (Cpp) npuParseBiggestFloat示例

示例#1

文件： parseutils.c 项目： ddlsmurf/Nimrod

N_NIMCALL(NI, npuParseFloat)(NimStringDesc* S_23425, NF* Number_23427, NI Start_23428) {
NI Result_23429;
NF64 Bf_23430;
Result_23429 = 0;
Bf_23430 = 0;
Result_23429 = npuParseBiggestFloat(S_23425, &Bf_23430, Start_23428);
(*Number_23427) = ((NF) (Bf_23430));
return Result_23429;
}

示例#2

文件： pure_parseutils.c 项目： reactormonk/csources

N_NIMCALL(NI, npuParseFloat)(NimStringDesc* s, NF* number, NI start) {
	NI result;
	NF bf;
	nimfr("parseFloat", "parseutils.nim")
	result = 0;
	bf = 0;
	nimln(323, "parseutils.nim");
	result = npuParseBiggestFloat(s, &bf, start);
	nimln(324, "parseutils.nim");
	(*number) = bf;
	popFrame();
	return result;
}N_NOINLINE(void, pureparseutilsInit)(void) {