我正在制作一个语言解释器,我已经到了需要评估 if 语句的地步。起初我认为这很简单,我能够让我的解释器评估简单的 if 条件,10 == 10
但是当我试图让它评估更复杂的条件时, 10 == 10 和 9 > 2
例如,它搞砸了。
例如,我编写了一些 C++ 代码,可以单独评估条件的每个部分。
"Hello World" == "Hello World" or "Test" == "Test"
它目前是这样工作的,上面给出了实际结果,下面给出了预期结果。这个结果将是:
TRUE or TRUE <- Actual
-----------------------
TRUE or TRUE <- Expected
-----------------------
Mã này như sau:
#include
#include
#include
#include "cond.h"
sử dụng không gian tên std;
vector cond_holder;
vector res;
vector expects;
bool eval_cond(string cond) {
int tôi;
int i2;
bool state = 0;
bool final_return = false;
string c = "";
for (i = 0; i < cond.length();i++) {
c += cond[i];
if (cond[i] == '"') {
if (state == 0)
state = 1;
khác
state = 0;
} else if (cond[i] == ' ' && state == 0) {
c = c.substr(0,c.length()-1);
cond_holder.push_back(c);
c = "";
}
if (i == cond.length()-1) {
cout << c << endl;
}
}
for (i = 0; i < cond_holder.size();i++) {
if (cond_holder[i+1] == "eqeq") {
expects.push_back("TRUE");
if (cond_holder[i] == cond_holder[i+2]) {
res.push_back("TRUE");
} khác {
res.push_back("FALSE");
}
i+=3;
}
if (cond_holder[i] == "and") {
res.push_back("and");
expects.push_back("and");
} else if (cond_holder[i] == "or") {
res.push_back("or");
expects.push_back("or");
}
}
for (i = 0; i < res.size();i++) {
cout << res[i] << " ";
}
cout << endl << "-----------------------" << endl;
for (i = 0; i < expects.size();i++) {
cout << expects[i] << " ";
}
cout << endl << "-----------------------" << endl;
return final_return;
}
int chính() {
cout << eval_cond("string:\"Hello World\" eqeq string:\"Hello World\" or string:\"H\" eqeq string:\"H\" ") << endl;
trả về 0;
}
但老实说,我只是边做边编代码,所以如果有人能告诉我更好的方法,我将不胜感激。我什至不确定接下来要如何处理这段代码。
Variable is an expression
Number is an expression
String is an expression
Expression == Expression is an expression
Expression AND Expression is an expression
Expression ; is a statement
IF Expression { statement } is a statement
用这样的比特构建你的语言,然后让它一起崩溃。 Flex 和 Yacc 的旧 unix 手册很好地介绍了该主题。
Tôi là một lập trình viên xuất sắc, rất giỏi!