Question

I have written this function in C++. A string of integers is passed in as dString and an integer is passed in as count. each dString integer is then indexed and passed into strTolnt integer array. each strTolnt integer is then passed through an equation and stored in temp array. I have printed each index in the array and the correct values are displayed, but when I print the entire array the value is displayed as hex. temp array is then passed back into dString, then dString is returned via stoi), when I print dString boxes are displayed. I want the result, in this case 5555 to be returned via dString as an integer 6 ainclude <string» 10 /Function that iapl ements modulo equation encryption 11 dstring passed in fron deteString and 13 int equatien encrypt istring dString, iet count) / /Declere temp erray of length count int tesp[count J: int str int bi cout << "Fron equation-dstrirg: " くく dstring 《< endl; //For Debugging for. (irt í "@g 1 < count; İ++) /1 Convert each indeed string into a. integer strToInt[1)- stoiCa); for loop cycling froe count to a and1 from count to 1,1 represents posit1on of digit read from right to lef for (int í =e, j = count; ǐくcourt, j 1;i++, j-) cou << "aqisation dStrlng[" <c ΐ << ")" << dString[1] ead1; //For Debuggl ng /Result of equation for each indexed dstring stored in temo array tmp,[1]-(MrTutnt[3].J) % 10İ //et.uetlun rur mtrypt lur cout << "equation i- i endl; //For Debugging e4 endl .temp 11 くく end1; //Far Debugging cout << ation teap array: << tespくく endL; //For Debugg for (int i-e;i < count; i++) "eString is ass1gned the encrspted value stored In tenp array and converted to string"/ b - temp[i dstringri1 b t e "equation encrypted string: "< dString < endl;yFor Debuggin return stol(dstring 11) Enter data for encryptaf Decrypt stored data Please mter data Cother than 0): 1234 Fron nain datastri equation dstring]i equation dstringti] equationi equation dStrinac temp [2] 5 equationj-1 quation temp array: ox61fedc erminete called after throwing an instance ef stdiaimalid ument what O: stoi 0 128

Answer 1

You cannot store directly int number in string[0]. In dString when you are assign to b value that is not storing.So in dString string is empty when you are converting stoi(dSting) thats why that is showing that error. now i changed that.can you check modify your code in 3rd for loop.

code:

#include<iostream>

#include<string>

using namespace std;

int equation_encrypt(string dString,int count)

{

int temp[count];

int strToInt[count];

string a;

int b;

cout <<"from equation - dString" << dString <<endl;

for(int i=0;i<count;i++)

{

a=dString[i];

strToInt[i]=stoi(a);

}

for(int i=0,j=count;i<count,j>=1;i++,j--)

{

cout << "equation dString [" << i << "]" << dString[i] << endl;

temp[i] = (strToInt[i]+j)%10;

cout << "temp[" << i <<"]"<<temp[i] <<endl;

}

cout << "equation temp array " << temp << endl;

for(int i=0;i<count;i++)

{

string c = to_string(temp[i]);

cout << c<<endl;

dString[i] =c[0];

cout << "equation encrypted char :" <<dString[i] << endl;

}

cout << "equation encrypted string :" << dString << endl;

return stoi(dString);

}

int main(void)

{

cout <<"after conversion string is = " <<equation_encrypt("1234",4) <<endl;//testing purpose

}

output: