3

I'm trying to count the cats and dogs in a string. So,

for example:

if the string is "cc+dd-cd",:

I want to count 2 positive cats, 2 positive dogs, 1 negative cat, 1 negative dog (yes I know this is a weird way to count, but it's part of my assignment).

I thought about doing a for loop where I iterated over the string, then nested inside the for loop, I'd have a while loop that would run until a '-', then run until a '+' or the end. I started the code, and without even getting that far into it, it created an infinite loop.

Question: How can I fix it?

Thanks!

string animalparkString = "cc+dd-cd" 
for (int k = 0; k != animalparkString.size(); k++)
   {
       while (k != '-'){
           pdc = pdc + 1; //positive dog count
           pcc = pcc + 1; //positive cat count
       }
       while (k != '+') {
           ndc = ndc + 1;
           ncc = ncc + 1;
       }
       
   }
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3
  • 1
    Those while loops don't do anything with k, so if the condition is true the first time, it will always be true. Besides, the condition doesn't make much sense anyway, instead of checking chars againt k, you should check them against animalparkString[k]. Commented Jan 30, 2020 at 8:39
  • 1
    For starters, k is an index, so to compare with '-' or '+' you should be using animalparkString[k]. Commented Jan 30, 2020 at 8:42
  • don't forget to check K in the will loop as well, the answer by @Blaze is correct and add checks Commented Jan 30, 2020 at 8:47

4 Answers 4

Reset to default
4

The immediate issue is that the while loops check k's value, but don't modify it. So once you entered one of them you'll be stuck there infinitely as k doesn't change inside the loop's body.

I wouldn't bother with nested loops - I'd just go over the entire string in one loop, and evaluate each character in its turn. A neat way to accomplish this is to keep a state of whether you're adding or subtracting (according to whether you last encountered a + or a - sign:

bool positive = true;
string animalparkString = "cc+dd-cd";
for (int k = 0; k < animalparkString.size(); k++) {
     char ch = animalparkString[k];

     switch (ch) {
         case '+':
             positive = true;
             break;
         case '-':
             positive = false;
             break;
         case 'c':
             if (positive) {
                 pcc++;
             } else {
                 ncc++
             }
             break;
         case 'd':
             if (positive) {
                 pdc++;
             } else {
                 ndc++
             }
             break;
     }
}
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Comments

1

This post describes iterating all characters is a string.

This is a simple solution using modern C++: 

int sign = 1; 
int dogs_count = 0, cats_count = 0;  
std::string animalparkString = "-cccccc+dddd-cd"; 
for (const char& ch : animalparkString)
{
     switch (ch) {
         case '+':
             sign = 1;
             continue;
         case '-':
             sign = -1;
             continue;
         case 'c':
             cats_count += sign;
             continue;
         case 'd':
             dogs_count += sign;
             continue;
     } 
}

std::cout <<"dogs: " << dogs_count << " cats: " <<cats_count;

A couple of suggestions to help you get started:
1. Use online c++ compilers to quickly test code
2. If your code doesn't behave as expected, use step-by-step debugging in your IDE or print our variables as you go using std::cout
3. Explicitly stating namespace is considered good practice. i.e: 

// preferable:  
std::string myString; 
// avoid:  
using namespace std;  
string myString
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Comments

0

To make your code work without too many changes , you can simply replace the while() condition with an if(). Moreover, instead of checking the iterator value k, you should compare the kth string element animalparkString[k].

Then you might start wondering if the code you wrote is actually doing what you expect. Possible questions you could try to answer is "how do I distinguish between positive or negative counts" and, then, "how do I distinguish between cats and dogs "? You will probably need to check also for cs and ds, not only for the operation sign! 

string animalparkString = "cc+dd-cd" 
for (int k = 0; k != animalparkString.size(); k++)
   {
       if(animalparkStrink[k] != '-'){
           // Now you know, there will be a pos count. Dog or Cat?
       }
       if(animalparkString[k] != '+') {
          // Now you know, there will be a neg count. Dog or Cat?
       }

   }

Note that if you write while( k != '-'), it will always evaluate true and, therefore, you will be stuck there. If it is the first time working with for-loops, consider printing the iterator value, to understand when and where you are stuck. 

string animalparkString = "cc+dd-cd" 
for (int k = 0; k != animalparkString.size(); k++)
   {

       std::cout << "for-loop iteration number: " << k << std::endl;

       if(animalparkStrink[k] != '-'){
           // Now you know, there will be a pos count. Dog or Cat?
       }
       if(animalparkString[k] != '+') {
          // Now you know, there will be a neg count. Dog or Cat?
       }

   }
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0

for and while together approach is unnecessarily complicated. Here's a simpler solution:

#include <concepts>
#include <iostream>

int main() {
  auto const& str{"cc+dd-cd"};
  std::boolean auto isPositive = 1;
  std::integral auto pdc{0}, pcc{0}, ndc{0}, ncc{0};
  for (char const ch : str) {
    switch (ch) {
      case 'c': {
        pcc += isPositive;
        ncc += !isPositive;
        break;
      }
      case 'd': {
        pdc += isPositive;
        ndc += !isPositive;
        break;
      }
      case '+': {
        isPositive = true;
        break;
      }
      case '-': {
        isPositive = false;
        break;
      }
    }
  }

  std::cout << "pcc: " << pcc << '\n'
            << "ncc: " << ncc << '\n'
            << "pdc: " << pdc << '\n'
            << "ndc: " << ndc << '\n';
}

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