测试Lua号是整数还是浮点数 [英] Test if Lua number is integer or float

问题描述

在我的C ++程序中，我需要知道Lua变量是整数还是浮点数. C API提供了lua_isnumber()，但是此功能不能区分int/float/double.

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.

到目前为止，我已经通过使用modf()来解决此问题:

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);
    }
}

Lua API是否提供一种更精确地推断变量类型的方法?

Does the Lua API provide a way to infer the variable type more precisely?

推荐答案

double n = lua_tonumber(L, -1);
if (n == (int)n) {
    // n is an int
} else {
    // n is a double
}

此代码的作用只是检查n是否有小数. 如果n为1.5，则将其强制转换为int((int)n)会将值底限为1，因此:

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为假，n为双精度

1.5 == 1 is false, n is a double

但是，如果n为4:

4 == 4为真，n为整数

4 == 4 is true, n is a int

之所以起作用，是因为lua唯一存在的数字是double.因此，当将数字从lua转换为C时，如果数字是整数(整数)，我们可以选择使用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).

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