string Expression::addevaluate(string x){
stringrep = x; //Stringrep is the string that user typed in,
//it might be 5+6+7-8-9*3/(2+5)
int totalnum = stringrep.length();
for(int i=0;i < totalnum;i++){ //This for loop will seperate the
//string by "+" and output a vector
//with seperate string
int addop = stringrep.find("+");
addvector.push_back(stringrep.substr(0,addop));
string a =stringrep.substr(0,addop);
totalnum=totalnum-(a.length());
stringrep = stringrep.substr(addop+1,totalnum);
}
int vectorlength = addvector.size();
for(int i = 0;i<vectorlength;i++){
cout << i+1<<":"<<addvector[i]<<",";
}
subevaluate(addvector);
return stringrep;
}
string Expression::subevaluate(vector<string> &v){
int totalnum = v.size();
//This is the question, I have no idea how can i set the value totalnum
//If it's the size of vector,it's too small. If it's the last totalnum
//from last function. Then there is a error. In addition,I do not know
//what the error is.
for(int i=0;i < totalnum;i++){
int addop = v[i].find("-");
if(addop > 0){
subtvector.push_back(v[i].substr(0,addop));
string a =v[i].substr(0,addop);
totalnum=totalnum-a.length();
v[i] = v[i].substr(addop+1,totalnum);
}
}
int vectorlength = subtvector.size();
for(int i = 0;i<vectorlength;i++){
cout << i+1<<":"<<subtvector[i]<<",";
}
return stringrep;
}
I do not know why I did wrong for the second for loop. Please help me solve the for loop. In addition,how can i seperate all the string by."+","-","*","/". Then calculate the answer like a calculator. Thanks.
This implementation will not work... suppose you have
"1+2*(3+4)"
the first split (even when written correctly) will get
"1"
"2*(3"
"4)"
What are you going to do with "2*(3" ?
At the very minimum to write a calculator with this approach you need:
add "(" front and add ")" at the end (i.e. change to "(1+2*(3+4))"
look for last OPEN parenthesis
move from there to the first CLOSED parenthesis
process what is in-between (i.e. "3+4") and replace the whole parenthesized expression it in the original string with the result (i.e. get from "1+2*(3+4)" to "(1+2*7)")
repeat until there are no more parenthesis
For splitting a string on a given character you should write a dedicated function, for example:
std::vector<std::string> split(const std::string& text, char sep) {
std::vector<std::string> result;
size_t pos = text.find(sep);
while(pos != std::string::npos) {
result.push_back(text.substr(0, pos));
text = text.substr(pos + 1);
}
result.push_back(text);
return result;
}
then you can write
std::vector<std::string> res = split(text, '+');
to get from "1+2*3+4" to a vector containing "1", "2*3", "4".
PS: Note that this way of computing expression is not what normally is done, but it can be made working so you should in my opinion keep working on it until it's done.
I think it will be difficult to make the code work when you split the string into a vector. The operator precedence will be too hard to handle, I think.
How about a recursive process?
In this way you can simplify the original string step by step. You just keep calling the evaluate function with substrings until they are turned into simple expressions.
Example:
exp = 12/(5+1)
call 1: call f("12/(5+1)")
call 1: f identifies the substring "5+1" and call itself (recursive)
call 2: call f("5+1")
call 2: simple expression calculates into "6" which is returned
call 1: The substring "(5+1)" is replaced by the returned "6"
call 1: exp now looks "12/6"
call 1: simple expression calculates into "2" which is returned
More complex expressions like "48/(5 + (2*3/(3-1))) would just result in more calls so that the string is simplified step by step.
The code could look like the code below. Only the structure is include - it is for OP to fill in the actual code.
bool isSimpleExpression(string& s)
{
// Return true if s is simple, i.e. X+Y, X-Y, X*Y, X/Y
// Otherwise false
}
string evaluateString(string& exp)
{
while(!isSimpleExpression(exp))
{
// exp must be broken into smaller part as it isn't simple yet
if (ExpContainsParanthesis() )
{
// Example: exp is "12/(5+1)"
string s1 = FindSubstringInMostInnerMatchingParanthesis(exp);
// Example: s1 is "5+1"
// Example: call evaluateString("5+1")
strint s2 = evaluateString(s1); // Recursive call
// Example: s2 is 6
ReplaceS1WithS2(exp, s1, s2);
// Example: exp is "12/6"
}
else if (ExpContainsMultiplication())
{
// Find the substring with multiplication
// Call this function with the substring
// Replace the substring with the returned result
}
else if ....
{
// division
}
// ... and so on
}
// Calculate the simple expression
string result;
// ..
// ..
return result;
}
Related
My function is a simple palindrome algorithm that is supposed to take an input integer (test), turn it into a string (test_s), then reverse that string in a new variable (test_s_reverse) then return true if test_s is equal to test_s_reverse.
bool is_palindrome(int test){
test_s = to_string(test);
test_length = test_s.length();
for (int i=(test_length + 1); i>=0; i--){
test_s_reverse += test_s[i];
}
if (test_s_reverse == test_s){
return true;
}
else {
return false;
}
}
I created a main function to test results and using input 12321 the program returns false, even though cout for test_s_reverse is = 12321 and test_s = 12321.
What am I doing wrong?
You should use test_length - 1 instead of test_lenght + 1, because in the new reversed string you have some extra characters which you can't see if you print them.
The .length() function returns you exactly the number of characters in the string. So you either go with test_length, but you do i>0, or if you go in the loop with i>=0 you go with test_length - 1, so you will be sure that there are no blank characters at the end of new string.
But, if you start with test_length, you will need to edit, since there is no character at test_length position:
test_s_reverse += test_s[i-1];
If you use pure C++ code, it should look like this:
bool is_palindrome(int test){
string test_s = to_string(test);
int test_length = test_s.length();
string test_s_reverse;
for (int i=(test_length); i>0; i--){
test_s_reverse += test_s[i-1];
}
if (test_s_reverse == test_s){
return true;
}
else {
return false;
}
}
If it were up to me, I'd do the job rather differently.
std::string has a constructor to build a string for a pair of iterators. It also supports reverse iterators. As such, you can construct a reversed string a bit more easily like this:
std::string test_s_reversed(test_s.rbegin(), test_s.rend());
...but given how much that simplifies the task, you can boil most of it down to something like this:
bool is_palindrome(std::string const &s) {
return s == std::string(s.rbegin(), s.rend());
}
Implementation for an integer would then be something like this:
bool is_palindrome(int val) {
return is_palindrome(std::to_string(val));
}
Given an input string A, is there a concise way to generate a string B that is lexicographically larger than A, i.e. A < B == true?
My raw solution would be to say:
B = A;
++B.back();
but in general this won't work because:
A might be empty
The last character of A may be close to wraparound, in which case the resulting character will have a smaller value i.e. B < A.
Adding an extra character every time is wasteful and will quickly in unreasonably large strings.
So I was wondering whether there's a standard library function that can help me here, or if there's a strategy that scales nicely when I want to start from an arbitrary string.
You can duplicate A into B then look at the final character. If the final character isn't the final character in your range, then you can simply increment it by one.
Otherwise you can look at last-1, last-2, last-3. If you get to the front of the list of chars, then append to the length.
Here is my dummy solution:
std::string make_greater_string(std::string const &input)
{
std::string ret{std::numeric_limits<
std::string::value_type>::min()};
if (!input.empty())
{
if (std::numeric_limits<std::string::value_type>::max()
== input.back())
{
ret = input + ret;
}
else
{
ret = input;
++ret.back();
}
}
return ret;
}
Ideally I'd hope to avoid the explicit handling of all special cases, and use some facility that can more naturally handle them. Already looking at the answer by #JosephLarson I see that I could increment more that the last character which would improve the range achievable without adding more characters.
And here's the refinement after the suggestions in this post:
std::string make_greater_string(std::string const &input)
{
constexpr char minC = ' ', maxC = '~';
// Working with limits was a pain,
// using ASCII typical limit values instead.
std::string ret{minC};
auto rit = input.rbegin();
while (rit != input.rend())
{
if (maxC == *rit)
{
++rit;
if (rit == input.rend())
{
ret = input + ret;
break;
}
}
else
{
ret = input;
++(*(ret.rbegin() + std::distance(input.rbegin(), rit)));
break;
}
}
return ret;
}
Demo
You can copy the string and append some letters - this will produce a lexicographically larger result.
B = A + "a"
My program aim is to convert given mathematical expression into polish notation.
Mathematical expression is stored in vectors and recursion is used.
I am getting run time error when compiling it.
Using debugger it shows segmentation fault (SIGSEGV).
It is assumed that input is of the form (a+b) and corresponding output should be ab+. (brackets mandatory)
Actually it is this problem of SPOJ.
My approach is as follows -
Use vectors to store expression
If expression size is greater than 1
then remove brackets which are always present at end of string using pop_back and vector::erase .
divide the vector into three parts - string 1 , operator and string 2 (string 1 and string 2 are vectors )
simplify vectors string 1 and string 2 to their polish notation using recursion
Clear original vector passed to function and add string 1 , operator and string 2 to it.
change() is used to accomplish all these things
Here is my program which converts a single statement to it's polish notation.
#include <iostream>
#include <vector>
void convert(std::vector <char> &vect)
{
if(vect.size()!=1&&vect.size()!=0)
{
vect.pop_back();
vect.erase(vect.begin());
std::vector<char> str1;
std::vector<char> str2;
char op;
int i=0;
int reqcount = 0;
while(1)
{
if(vect[i]=='(')
reqcount++ ;
if(vect[i]==')')
reqcount-- ;
str1.push_back(vect[i]);
if(!reqcount)
break;
i++;
}
op=vect[i];
while((i+1)!=vect.size())
{
str2.push_back(vect[i+1]);
i++;
}
vect.clear();
convert(str1);
convert(str2);
for(int i=0; i<str1.size(); i++ )
{
vect.push_back(str1[i]);
}
vect.push_back(op);
for(int i=0;i<str2.size();i++)
{
vect.push_back(str2[i]);
}
}
}
int main( void )
{
std::vector<char> testinput;;
char c;
std::cin>>std::noskipws;
while(std::cin>>c&&c!='\n')
testinput.push_back(c);
convert(testinput);
for(int i=0;i<testinput.size();i++)
std::cout<<testinput[i];
return 0;
}
You start with int i = 0 which will point to the overall brackets (). It will read the entire string as str1 minus the closing ')'. op will be ')' if I'm not mistaking and str2 will remain empty.
Then you recursively call 'convert(str1)' which doesn't have a closing braket so you'll try to go past the end of the array and get your SIGSEGV.
Make sure your split works correctly and double-check the boundary conditions to prevent it.
P.S. Your solution is O(N^2). You can do O(N).
Given the following function:
If I execute setColor("R:0,G:0,B:255,");
I'm expecting the red, grn, blu values to be:
0 0 255 except I'm getting 0 0 0
It's working fine for R:255,G:0,B:0, or R:0,G:255,B:0, though.
int setColor(String command) {
//Parse the incoming command string
//Example command R:123,G:100,B:50,
//RGB values should be between 0 to 255
int red = getColorValue(command, "R:", "G");
int grn = getColorValue(command, "G:", "B");
int blu = getColorValue(command, "B:", ",");
// Set the color of the entire Neopixel ring.
uint16_t i;
for (i = 0; i < strip.numPixels(); i++) {
strip.setPixelColor(i, strip.Color(red, grn, blu));
}
strip.show();
return 1;
}
int getColorValue(String command, String first, String second) {
int rgbValue;
String val = command.substring(command.indexOf(first)+2, command.indexOf(second));
val.trim();
rgbValue = val.toInt();
return rgbValue;
}
Without knowing your String implementation, I can only make an educated guess.
What happens is that indexOf(second) doesn't give you what you think.
"R:0,G:0,B:255,"
^ ^- indexOf("B:")
|- indexOf(",")
It works for your other cases as none of the things they look for occur more than once in the string.
Looking at the SparkCore Docs we find the documentation for both indexOf and substring.
indexOf()
Locates a character or String within another String. By default, searches from the beginning of the String, but can also start from a given index, allowing for the locating of all instances of the character or String.
string.indexOf(val)
string.indexOf(val, from)
substring()
string.substring(from)
string.substring(from, to)
So now to fix your problem you can use the second variant of indexOf and pass that the index you found from your first search.
int getColorValue(String command, String first, String second) {
int rgbValue;
int beg = command.indexOf(first)+2;
int end = command.indexOf(second, beg);
String val = command.substring(beg, end);
val.trim();
rgbValue = val.toInt();
return rgbValue;
}
In this instance, I would split the string using a comma as the delimiter then parse each substring into a key-value pair. You could use a vector of value for the second part if you always have the sequence "R,G,B" in which case why have the "R:", "G:" or "B:" at all?
I can suppose that command.indexOf(second) will always find you the first comma, therefore for B the val becomes empty string.
Assuming that indexOf is something similar to .Net's, maybe try
int start = command.indexOf(first)+2;
int end = command.indexOf(second, start)
String val = command.substring(start+2, end);
Note the second argument for the second call to indexOf, I think it will make indexOf to look for matches after start. I also think you'd better pass a "," as a second for all calls, and add +1 or -1 to end to compensate for this passing "," instead of "G" and "B".
Or just use another limiter for B part, like R:0,G:0,B:0. (dot instead of comma).
I ended up just modifying my code:
int setColor(String command) {
int commaIndex = command.indexOf(',');
int secondCommaIndex = command.indexOf(',', commaIndex+1);
int lastCommaIndex = command.lastIndexOf(',');
String red = command.substring(0, commaIndex);
String grn = command.substring(commaIndex+1, secondCommaIndex);
String blu = command.substring(lastCommaIndex+1);
// Set the color of the entire Neopixel ring.
uint16_t i;
for (i = 0; i < strip.numPixels(); i++) {
strip.setPixelColor(i, strip.Color(red.toInt(), grn.toInt(), blu.toInt()));
}
strip.show();
return 1;
}
I simply just do: 255,0,0 and it works a treat.
i want to generate block comment using eclipse-Indigo like this. I'm C++ programmer.
/**
*
* #param bar
* #return
*/
int foo(int bar);
how can i do like this.
IF your input is pretty much static, you can write a simplified lexer that will work, requires simple string mungeing. string has lots of nice editing capabilities in it with .substr() and .find() in it. all you have to do is figure out where the perens are. you know you can optionally process this as a stringstream, which makes this FAR easier (don't forget to use std::skipws to skip whitespace.
http://www.cplusplus.com/reference/string/string/substr/
http://www.cplusplus.com/reference/string/string/find/
#include <vector>
#include <string>
typedef STRUCT arg_s {
string sVarArgDataType, sVarArg;
} arg_s ARG;
ARG a;
vector<ARG> va;
char line[65000];
filein.getline(line, 65000);
line[65000-1]='\0'; //force null termination if it hasn't happened
get line and store in string sline0
size_t firstSpacePos=sline.find(' ');
size_t nextSpacePos = sline.find(' ',firstSpacePos+1);
size_t prevCommaPos = string::npos;
size_t nextCommaPos = sline.find(',');
size_t openPerenPos=sline.find('(');
size_t closePerenPos=sline.find(");");
string sReturnDataType, sFuncName;
if (
string::npos==firstSpacePos||
string::npos==semicolonPos||
string::npos==openPerenPos||
string::npos==closePerenPos) {
return false; //failure
}
while (string::npos != nextSpacePos) {
if (string::npos != nextCommaPos) {
//found another comma, a next argument. use next comma as a string terminator and prevCommaPos as an arg beginning.
//assume all keywords are globs of text
a.sVarArgDataType=sline.substr(prevCommaPos+1,nextSpacePos-(prevCommaPos+1));
a.sVarArg=sline.substr(nextSpacePos+1,nextCommaPos-(nextSpacePos+1));
} else {
//didn't find another comma. use ) as a string terminator and prevCommaPos as an arg beginning.
//assume all keywords are globs of text
a.sVarArgDataType=sline.substr(prevCommaPos+1,nextSpacePos-(prevCommaPos+1));
a.sVarArg=sline.substr(nextSpacePos+1,closePerenPos-(nextSpacePos+1));
}
va.push_back(a); //add structure to list
//move indices to next argument
nextCommaPos = sline.find(',', secondSpacePos+1);
nextSpacePos = sline.find(' ', secondSpacePos+1);
}
int i;
fileout<<"/**
*
";
for (i=0; i < va.size(); i++) {
fileout<<" * #param "<<va[i].sVarArg;
}
fileout<<"
* #return
*/
"<<sReturnDataType<<" "<<sFuncName<<'(';
for (i=0; i < va.size(); i++) {
fileout<<va[i].sArgDataType<<" "<<va[i].sVarArg;
if (i != va.size()-1) {
fileout<<", "; //don;t show a comma-space for the last item
}
}
fileout<<");"<<std::endl;
this will handle any number of arguments EXCEPT ... the variable argument type. but you can put in your own detection code for that and the if statement that switches out between ... and the 2-keyword argument types. here I am only supporting 2 keywords in my struct. you can support more by using a while to search for all the spaces before the next , comma or ) right peren in inside the while loop add your variable number of strings to a vector<string> inside the struct you are going to replace - nah, just make a vector<vector<string> >. or, just one vector and do a va.clear() after every function is done.
I just noticed the eclipse tag. I don't know much about eclipse. I can't even get it to work. some program.