In my C++ program, I need to know if a Lua variable is an integer number or a floating-point number. The C API provides lua_isnumber()
but this function does not distinguish between int/float/double
.
So far I have worked around this by using modf()
:
if (lua_isnumber(luaCtx, -1)) // int / unsigned int / float:
{
luaVarName = lua_tostring(luaCtx, -2);
double n = static_cast<double>(lua_tonumber(luaCtx, -1));
// Figure out if int or float:
double fractPart, intPart;
fractPart = modf(n, &intPart);
if (fractPart != 0.0)
{
luaVarType = ScriptVar::TypeTag::Float;
luaVarData.asFloat = static_cast<float>(n);
}
else
{
luaVarType = ScriptVar::TypeTag::Integer;
luaVarData.asInteger = static_cast<int>(n);
}
}
Does the Lua API provide a way to infer the variable type more precisely?
double n = lua_tonumber(L, -1);
if (n == (int)n) {
// n is an int
} else {
// n is a double
}
What this code does is just checking if n has any decimals or not. If n is 1.5, then casting it to int ((int)n) will floor the value to 1, so:
1.5 == 1 is false, n is a double
But if n is lets say 4:
4 == 4 is true, n is a int
This works because to lua, the only numeric number that exist is double. So when converting a number from lua to C, we can choose to use int if the number is a integer(whole number).
来源:https://stackoverflow.com/questions/25196042/test-if-lua-number-is-integer-or-float