本文整理汇总了C++中SS::top方法的典型用法代码示例。如果您正苦于以下问题：C++ SS::top方法的具体用法？C++ SS::top怎么用？C++ SS::top使用的例子？那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类SS的用法示例。
在下文中一共展示了SS::top方法的3个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C++代码示例。
示例1: play
void play(MSI& rank, VS& toks) {
SS oprs;
SD nums;
string out = "";
char str[100];
cout << "# from infix to postfix!!!" << endl;
for (VS::iterator i = toks.begin(); i != toks.end(); i++) {
string tok = *i;
cout << "There is a " << tok << ": ";
if (is_num(tok)) {
cout << "push to output" << endl;
sprintf(str, "%.6lf ", stod(tok));
out += str;
nums.push(stod(tok));
}
else if (tok == ",") {
cout << "flush the stack until '('" << endl;
while (oprs.top() != "(")
calc(oprs, nums, out);
}
else if (tok == ")") {
cout << "flush the stack until '(' and check if there's a func" << endl;
while (oprs.top() != "(")
calc(oprs, nums, out);
oprs.pop();
if (!oprs.empty() && is_func(oprs.top()))
calc(oprs, nums, out);
}
else {
cout << "push to stack" << endl;
if ( is_unary(tok) && (i == toks.begin() || rank[*(i - 1)] > 0) )
oprs.push("u" + tok);
else {
while (!is_func(tok) && tok != "(" && !oprs.empty() && rank[oprs.top()] <= rank[tok]) {
cout << "\t*** stack.top() has higher precedence" << endl;
calc(oprs, nums, out);
}
oprs.push(tok);
}
}
cout << "\tcurrent output: " << out << endl
<< "\tcurrent stack: " << oprs << endl;
}
while (!oprs.empty())
calc(oprs, nums, out);
cout << "There is nothing left, so flush the stack to output" << endl
<< "\tcurrent output: " << out << endl
<< "#transformation completed!!!" << endl
<< "Postfix Exp: " << out << endl
<< "RESULT: " << nums.top() << endl << endl;
}示例2: calc
void calc(SS& oprs, SD& nums, string& s) {
string opr = oprs.top(); oprs.pop();
if (opr[0] == 'u') s += opr[1];
else s += opr;
s += ' ';
double b = nums.top(); nums.pop();
if (opr[0] == 'u' || (is_func(opr) && opr != "pow")) {
if (opr == "u+") nums.push(b);
else if (opr == "u-") nums.push(-b);
else if (opr == "sin") nums.push(sin(b));
else if (opr == "cos") nums.push(cos(b));
else if (opr == "exp") nums.push(exp(b));
else if (opr == "log") nums.push(log(b));
else if (opr == "sqrt") nums.push(sqrt(b));
else if (opr == "fabs") nums.push(fabs(b));
return;
}
double a = nums.top(); nums.pop();
if (opr == "*") nums.push(a * b);
else if (opr == "+") nums.push(a + b);
else if (opr == "-") nums.push(a - b);
else if (opr == "pow") nums.push(pow(a, b));
}示例3:
ostream& operator<<(ostream& out, const SS& s) {
VS v;
SS ss = s;
while (!ss.empty()) {
v.PB(ss.top());
ss.pop();
}
for (int i = v.size() - 1; i >= 0; i--)
out << v[i] << ' ';
return out;
}