未定义的函数引用（uint8_t）

Question

在LiquidCrystal.h文件中声明了这样的构造函数：

class LiquidCrystal {    
    public :
    LiquidCrystal(uint8_t rs, 
            uint8_t enable,
            uint8_t d0, 
            uint8_t d1, 
            uint8_t d2, 
            uint8_t d3);
}

LiquidCrystal.cpp:

#include "LiquidCrystal.h"

LiquidCrystal::LiquidCrystal(uint8_t rs,  uint8_t enable,
                 uint8_t d0, uint8_t d1, uint8_t d2, uint8_t d3)
{
  init(1, rs, 255, enable, d0, d1, d2, d3, 0, 0, 0, 0);
}

void LiquidCrystal::init(uint8_t fourbitmode, uint8_t rs, uint8_t rw, uint8_t enable,
             uint8_t d0, uint8_t d1, uint8_t d2, uint8_t d3,
             uint8_t d4, uint8_t d5, uint8_t d6, uint8_t d7)
{
  // Some code 
}

我试图像这样打电话给那个人：

#include <LiquidCrystal.h>

int main(void) {

    LiquidCrystal* lcd = new LiquidCrystal(6, 7, 5, 4, 3, 8);
    // Rest of the code
}

它出现以下错误：undefined reference to 'LiquidCrystal::LiquidCrystal(unsigned char, unsigned char, unsigned char, unsigned char, unsigned char, unsigned char)'

正确包含头文件，因为如果我喜欢这样，它会正确编译：

#include <LiquidCrystal.h>

int main(void) {

    uint8_t number6 = 6;
    uint8_t number7 = 7;
    uint8_t number5 = 5;
    uint8_t number4 = 4;
    uint8_t number3 = 3;
    uint8_t number8 = 8;

    LiquidCrystal* lcd = new LiquidCrystal(number6, number7, number5, number4, number3, number8);
    // Rest of the code
}

那么，有没有办法避免逐个声明uint8_t变量并直接传递整数（比如将它转换为uint_8）？

EDIT1：头文件可以在这里找到：https://github.com/arduino-libraries/LiquidCrystal/blob/master/src/LiquidCrystal.h

EDIT2：

LiquidCrystal库添加如下： （在Eclipse中）右键单击项目 - ＆gt;项目属性 - ＆gt; C / C ++ Build - ＆gt;编译器 - ＆gt;一般 - ＆gt;包含路径 - ＆gt;添加 - ＆gt; （LiquidCrystal.cpp和LiquidCrystal.h存在的文件夹）