c - 某些脚本语言的算术运算

Question

我正在研究某种脚本语言。包含结构的值是

struct myvar
{
 char  name[NAMELEN];
 int   type;
 void* value;
}
type = 0  --> int* value
type = 1  --> char* value
type = 2  --> float* value

我在算术运算方面遇到了一些问题。似乎我需要对每个操作进行各种类型转换，这发展成为为每个操作编写一大堆代码，如:

case 0:  // "="
 if(factor1.name)
 {
    if((factor1.type == 1) && (factor2.type==1))
    {
        free(factor1.value);
        int len = (strlen((STRING)factor2.value)+1)*sizeof(char);
        factor1.value = malloc(len);
        memcpy(factor1.value,factor2.value,len);
    }
    else if((factor1.type == 2) && (factor2.type==2))
    *(FLOAT*)factor1.value = *(FLOAT*)factor2.value;
    else if((factor1.type == 0) && (factor2.type==0))
    *(INTEGER*)factor1.value = *(INTEGER*)factor2.value;
    else if((factor1.type == 0) && (factor2.type==2))
    *(INTEGER*)factor1.value = *(FLOAT*)factor2.value;
    else if((factor1.type == 2) && (factor2.type==0))
    *(FLOAT*)factor1.value = *(INTEGER*)factor2.value;
    else
     GetNextWord("error");
 }
 break;

有什么方法可以避免这个烦人的过程吗？否则我别无选择，只能为每个 "=","~","+","-","*","/","%",">","复制粘贴这段代码<",">=","<=","==","~=","AND","OR"

Answer 1

对值使用union 而不是struct:

struct myvar {
  enum {
    STRING, INT, FLOAT,
  } type;

  union {
    char  strval[NAMELEN];
    int   intval;
    float fltval;
  } val;
};

然后在您的脚本语言中执行赋值运算符时，您只需执行以下操作:

factor1 = factor2;

要根据您要执行的类型获取正确的值:

switch (operand.type) {
  case STRING:
    printf("%s", operand.val.strval);
    break;

  case INT:
    printf("%d", operand.val.intval);
    break;

  case FLOAT:
    printf("%f", operand.val.fltval);
    break;
}