I am making a calculator that accepts multiple operators at once. I have a vector pair that stores the position of operators, and the type of operator. The vector must be updated after each loop as the previous positions are no longer valid due to the result replacing a part of the input string.
I've attempted to use clear() so that it starts from the beginning again, but that results in Expression: vector iterators incompatible. I don't think I can use std::replace since the number of operators in the string changes after each loop. Is there a way to just have it start from the beginning again and overwrite any existing elements?
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
int main()
{
std::cout << "C++ Calculator" << std::endl;
while (true) //runs forever with loop
{
std::string input;
std::getline(std::cin, input);
//erases whitespace
int inp_length = input.length();
for (int i = inp_length - 1; i >= 0; --i)
{
if (input[i] == ' ')
input.erase(i, 1);
}
std::vector<std::pair<int, char>> oper_pvec;
int vec_pos = 0;
//finds the position of operators and type
for (std::string::iterator i = input.begin(); i != input.end(); ++vec_pos, ++i)
{
switch (*i)
{
case 'x':
{
oper_pvec.push_back(std::pair<int, char>(vec_pos, 'x'));
break;
}
case '/':
{
oper_pvec.push_back(std::pair<int, char>(vec_pos, '/'));
break;
}
case '+':
{
oper_pvec.push_back(std::pair<int, char>(vec_pos, '+'));
break;
}
case '-':
{
oper_pvec.push_back(std::pair<int, char>(vec_pos, '-'));
break;
}
}
}
//declarations before loop to make sure they're all able to be accessed, will probably change later
int loper_pos = 0; //must be set 0 since there's no left operator at first
int roper_pos;
double lnum; //left number
double rnum; //right number
char loper; //left operator
char roper; //right operator
int pos = -1; //position of loop, needs to be -1 since it increments it each time
std::string holder = input; //copy of input
auto op = oper_pvec.begin();
while (op != oper_pvec.end())
{
op = oper_pvec.begin();
++pos; //position of loop
int key = std::get<0>(*op); //gets first value from vector pair
char val = std::get<1>(*op); //gets second value from vector pair
//gets previous/next vector pairs
std::vector<std::pair<int, char>>::iterator prev_op = oper_pvec.begin();
std::vector<std::pair<int, char>>::iterator next_op = oper_pvec.end();
if (op != oper_pvec.begin()) prev_op = std::prev(op);
if (op != oper_pvec.end()) next_op = std::next(op);
//extracts the value of pairs
if (pos > 0)
{
loper_pos = std::get<0>(*prev_op);
loper = std::get<1>(*prev_op);
}
if (pos == oper_pvec.size() - 1) roper_pos = oper_pvec.size();
else
{
roper_pos = std::get<0>(*next_op);
roper = std::get<1>(*next_op);
}
//replaces numbers and etc with product, only multiplication for now
switch (val)
{
case 'x':
{
int lnum_start = loper_pos + 1;
if (loper_pos == 0) lnum_start = 0;
int lnum_len = key - (loper_pos + 1);
if (loper_pos == 0) lnum_len = key;
lnum = std::stod(input.substr(lnum_start, lnum_len));
int rnum_start = key + 1;
int rnum_len = (roper_pos - 1) - key;
rnum = std::stod(input.substr(rnum_start, rnum_len));
double prod = lnum * rnum;
std::string to_string = std::to_string(prod);
input.replace(loper_pos, roper_pos, to_string);
break;
}
}
/////////////////////////////////problem area////////////////////////////////////////
//clears the vector and then finds the operators again
oper_pvec.clear();
int vpos = 0;
for (std::string::iterator it = input.begin(); it != input.end(); ++vpos, ++it)
{
if (vpos == input.length())
{
vpos = 0;
break;
}
switch (*it)
{
case 'x':
{
oper_pvec.push_back(std::pair<int, char>(vpos, 'x'));
break;
}
case '/':
{
oper_pvec.push_back(std::pair<int, char>(vpos, '/'));
break;
}
case '+':
{
oper_pvec.push_back(std::pair<int, char>(vpos, '+'));
break;
}
case '-':
{
oper_pvec.push_back(std::pair<int, char>(vpos, '-'));
break;
}
}
}
/////////////////////////////////////////////////////////////////////////////////////
}
//converts to double then prints on screen
double out = std::stod(input);
std::cout << out << std::endl;
}
}
Not really the answer to the title but instead of trying to overwrite the vector I just continued to use clear() and changed the while loop to oper_pvec.size() != 0.
(Full code since I had to change quite a few things to get it to work properly)
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
int main()
{
std::cout << "C++ Calculator" << std::endl;
while (true) //runs forever with loop
{
std::string input;
std::getline(std::cin, input);
//erases whitespace
int inp_length = input.length();
for (int i = inp_length - 1; i >= 0; --i)
{
if (input[i] == ' ')
input.erase(i, 1);
}
std::vector<std::pair<int, char>> oper_pvec;
int vec_pos = 0;
//finds the position of operators and type
for (std::string::iterator i = input.begin(); i != input.end(); ++vec_pos, ++i)
{
switch (*i)
{
case 'x':
{
oper_pvec.push_back(std::pair<int, char>(vec_pos, 'x'));
break;
}
case '/':
{
oper_pvec.push_back(std::pair<int, char>(vec_pos, '/'));
break;
}
case '+':
{
oper_pvec.push_back(std::pair<int, char>(vec_pos, '+'));
break;
}
case '-':
{
oper_pvec.push_back(std::pair<int, char>(vec_pos, '-'));
break;
}
}
}
//declarations before loop to make sure they're all able to be accessed, will probably change later
int loper_pos = 0; //must be set 0 since there's no left operator at first
int roper_pos;
double lnum; //left number
double rnum; //right number
char loper; //left operator
char roper; //right operator
int pos = -1; //position of loop, needs to be -1 since it increments it each time
std::string holder = input; //copy of input
std::vector<std::pair<int, char>> vec_copy = oper_pvec;
auto op = oper_pvec.begin();
while (oper_pvec.size() != 0)
{
op = oper_pvec.begin();
++pos; //position of loop
int key = std::get<0>(*op); //gets first value from vector pair
char val = std::get<1>(*op); //gets second value from vector pair
//gets previous/next vector pairs
std::vector<std::pair<int, char>>::iterator prev_op = oper_pvec.begin();
std::vector<std::pair<int, char>>::iterator next_op = oper_pvec.end();
if (op != oper_pvec.begin()) prev_op = std::prev(op);
if (op != oper_pvec.end()) next_op = std::next(op);
//extracts the value of pairs
if (pos > 0)
{
loper_pos = std::get<0>(*prev_op);
loper = std::get<1>(*prev_op);
}
if (op == oper_pvec.begin()) loper_pos = 0;
if (oper_pvec.size() == 1) roper_pos = input.length();
else
{
roper_pos = std::get<0>(*next_op);
roper = std::get<1>(*next_op);
}
//replaces numbers and etc with product, only multiplication for now
switch (val)
{
case 'x':
{
int lnum_start = loper_pos + 1;
if (lnum_start == 1) lnum_start = 0;
if (loper_pos > 0)
if (isdigit(input[loper_pos - 1])) lnum_start = 0;
int lnum_len = key - (loper_pos + 1);
if (lnum_start == 0)
{
if (key == 1) lnum_len = 1;
else lnum_len = key - 1;
}
lnum = std::stod(input.substr(lnum_start, lnum_len));
int rnum_start = key + 1;
int rnum_len = (roper_pos - 1) - key;
rnum = std::stod(input.substr(rnum_start, rnum_len));
double prod = lnum * rnum;
std::string to_string = std::to_string(prod);
input.replace(loper_pos, roper_pos, to_string);
break;
}
}
//clears the vector and then finds the operators again
oper_pvec.clear();
int vpos = 0;
for (std::string::iterator it = input.begin(); it != input.end(); ++vpos, ++it)
{
if (vpos == input.length())
{
vpos = 0;
break;
}
switch (*it)
{
case 'x':
{
oper_pvec.push_back(std::pair<int, char>(vpos, 'x'));
break;
}
case '/':
{
oper_pvec.push_back(std::pair<int, char>(vpos, '/'));
break;
}
case '+':
{
oper_pvec.push_back(std::pair<int, char>(vpos, '+'));
break;
}
case '-':
{
oper_pvec.push_back(std::pair<int, char>(vpos, '-'));
break;
}
}
}
}
//converts to double then prints on screen
double out = std::stod(input);
std::cout << out << std::endl;
}
}