调度场算法

C++程序實現

#include <cstring>#include <cstdio>// 操作符// 优先级符号运算顺序// 1!从右至左// 2* / %从左至右// 3+ -从左至右// 4=从右至左int op_preced(const char c){
    switch(c)    {
        case '!':
            return 4;
        case '*':  case '/': case '%':
            return 3;
        case '+': case '-':
            return 2;
        case '=':
            return 1;
    }
    return 0;}unsigned int op_arg_count(const char c){
    switch(c)  {
        case '*': case '/': case '%': case '+': case '-': case '=':
            return 2;
        case '!':
            return 1;
        default:
            return c - 'A';
    }
    return 0;}#define op_left_assoc(c) (c == '+' || c == '-' || c == '/' || c == '*' || c == '%')#define is_operator(c)   (c == '+' || c == '-' || c == '/' || c == '*' || c == '!' || c == '%' || c == '=')#define is_function(c)   (c >= 'A' && c <= 'Z')#define is_ident(c)      ((c >= '0' && c <= '9') || (c >= 'a' && c <= 'z'))bool shunting_yard(const char *input, char *output){
    const char *strpos = input, *strend = input + strlen(input);
    char c, stack[32], sc, *outpos = output;
    unsigned int sl = 0;
    while(strpos < strend)   {
        c = *strpos;
        if(c != ' ')    {
            // 如果输入为一个数字，则直接入输出队列
            if(is_ident(c))  {
                *outpos = c; ++outpos;
            }
            // 如果输入为一个函数记号，则压入堆栈
            else if(is_function(c))   {
                stack[sl] = c;
                ++sl;
            }
            // 如果输入为函数分割符（如:逗号）
            else if(c == ',')   {
                bool pe = false;
                while(sl > 0)   {
                    sc = stack[sl - 1];
                    if(sc == '(')  {
                        pe = true;
                        break;
                    }
                    else  {
                        // 直到栈顶元素是一个左括号
                        // 从堆栈中弹出元素入输出队列
                        *outpos = sc; ++outpos;
                        sl--;
                    }
                }
                // 如果没有遇到左括号，则有可能是符号放错或者不匹配
                if(!pe)   {
                    printf("Error: separator or parentheses mismatched\n");
                    return false;
                }
            }
            // 如果输入符号为操作符，op1，然后:
            else if(is_operator(c))  {
                while(sl > 0)    {
                    sc = stack[sl - 1];
                    // While there is an operator token, o2, at the top of the stack
                    // op1 is left-associative and its precedence is less than or equal to that of op2,
                    // or op1 is right-associative and its precedence is less than that of op2,
                    if(is_operator(sc) &&
                        ((op_left_assoc(c) && (op_preced(c) <= op_preced(sc))) ||
                           (!op_left_assoc(c) && (op_preced(c) < op_preced(sc)))))   {
                        // Pop o2 off the stack, onto the output queue;
                        *outpos = sc; ++outpos;
                        sl--;
                    }
                    else   {
                        break;
                    }
                }
                // push op1 onto the stack.
                stack[sl] = c;
                ++sl;
            }
            // If the token is a left parenthesis, then push it onto the stack.
            else if(c == '(')   {
                stack[sl] = c;
                ++sl;
            }
            // If the token is a right parenthesis:
            else if(c == ')')    {
                bool pe = false;
                // Until the token at the top of the stack is a left parenthesis,
                // pop operators off the stack onto the output queue
                while(sl > 0)     {
                    sc = stack[sl - 1];
                    if(sc == '(')    {
                        pe = true;
                        break;
                    }
                    else  {
                        *outpos = sc; ++outpos;
                        sl--;
                    }
                }
                // If the stack runs out without finding a left parenthesis, then there are mismatched parentheses.
                if(!pe)  {
                    printf("Error: parentheses mismatched\n");
                    return false;
                }
                // Pop the left parenthesis from the stack, but not onto the output queue.
                sl--;
                // If the token at the top of the stack is a function token, pop it onto the output queue.
                if(sl > 0)   {
                    sc = stack[sl - 1];
                    if(is_function(sc))   {
                        *outpos = sc; ++outpos;
                        sl--;
                    }
                }
            }
            else  {
                printf("Unknown token %c\n", c);
                return false; // Unknown token
            }
        }
        ++strpos;
    }
    // When there are no more tokens to read:
    // While there are still operator tokens in the stack:
    while(sl > 0)  {
        sc = stack[sl - 1];
        if(sc == '(' || sc == ')')   {
            printf("Error: parentheses mismatched\n");
            return false;
        }
        *outpos = sc; ++outpos;
        --sl;
    }
    *outpos = 0; // Null terminator
    return true;}bool execution_order(const char *input) {
    printf("order: (arguments in reverse order)\n");
    const char *strpos = input, *strend = input + strlen(input);
    char c, res[4];
    unsigned int sl = 0, sc, stack[32], rn = 0;
// While there are input tokens left
    while(strpos < strend)  {
// Read the next token from input.
        c = *strpos;
// If the token is a value or identifier
        if(is_ident(c))    {
// Push it onto the stack.
            stack[sl] = c;
            ++sl;
        }
// Otherwise, the token is an operator  (operator here includes both operators, and functions).
        else if(is_operator(c) || is_function(c))    {
sprintf(res, "_%02d", rn);
printf("%s = ", res);
++rn;
// It is known a priori that the operator takes n arguments.
unsigned int nargs = op_arg_count(c);
// If there are fewer than n values on the stack
if(sl < nargs) {
// (Error) The user has not input sufficient values in the expression.
return false;
}
// Else, Pop the top n values from the stack.
// Evaluate the operator, with the values as arguments.
if(is_function(c)) {
printf("%c(", c);
while(nargs > 0){
sc = stack[sl - 1];
sl--;
if(nargs > 1){
printf("%s, ", &sc);
}
else {
printf("%s)\n", &sc);
}
--nargs;
}
}
else{
if(nargs == 1) {
sc = stack[sl - 1];
sl--;
printf("%c %s;\n", c, &sc);
}
else{
sc = stack[sl - 1];
sl--;
printf("%s %c ", &sc, c);
sc = stack[sl - 1];
sl--;
printf("%s;\n",&sc);
}
}
// Push the returned results, if any, back onto the stack.
            stack[sl] = *(unsigned int*)res;
            ++sl;
        }
        ++strpos;
    }
// If there is only one value in the stack
// That value is the result of the calculation.
if(sl == 1) {
sc = stack[sl - 1];
sl--;
printf("%s is a result\n", &sc);
return true;
}
// If there are more values in the stack
// (Error) The user input has too many values.
return false;}int main() {
    // functions: A() B(a) C(a, b), D(a, b, c) ...
    // identifiers: 0 1 2 3 ... and a b c d e ...
    // operators: = - + / * % !
    const char *input = "a = D(f - b * c + d, !e, g)";
    char output[128];
    printf("input: %s\n", input);
    if(shunting_yard(input, output))    {
        printf("output: %s\n", output);
        if(!execution_order(output))
            printf("\nInvalid input\n");
    }
    return 0;}