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    <p>So there are a number of problems with your code. I'll post what (should be) a corrected solution, which has copious comments to explain what's happening and where you've made mistakes. A few things up front:</p> <ol> <li><p>I'll use <code>std::string</code> instead of <code>char *</code> because it makes things much cleaner, and honestly, you should be using it in <code>C++</code> unless you have a very good reason not to (such as interoperability with a <code>C</code> library). This version also returns a <code>string</code> instead of taking a <code>char *</code> as a parameter.</p></li> <li><p>I'm using the stack from the standard library, <code>&lt;stack&gt;</code>, which is slightly different to your home-rolled one. <code>top()</code> shows you the next element <em>without</em> removing it from the stack, and <code>pop()</code> returns <code>void</code>, but removes the top element from the stack.</p></li> <li><p>It's a free function, not part of a class, but that should be easy to modify - it's simply easier for me to test this way.</p></li> <li><p>I'm not convinced your operator precedence tables are correct, however, I'll let you double check that.</p></li> </ol> <hr> <pre><code>#include &lt;stack&gt; #include &lt;cctype&gt; #include &lt;iostream&gt; std::string convertToPostfix(std::string&amp; infix) { std::string postfix; //Our return string std::stack&lt;char&gt; stack; stack.push('('); //Push a left parenthesis ‘(‘ onto the stack. infix.push_back(')'); //We know we need to process every element in the string, //so let's do that instead of having to worry about //hardcoded numbers and i, j indecies for(std::size_t i = 0; i &lt; infix.size(); ++i) { //If it's a digit, add it to the output //Also, if it's a space, add it to the output //this makes it look a bit nicer if(isdigit(infix[i]) || isspace(infix[i])) { postfix.push_back(infix[i]); } //Already iterating over i, so //don't need to worry about i++ //Also, these options are all mutually exclusive, //so they should be else if instead of if. //(Mutually exclusive in that if something is a digit, //it can't be a parens or an operator or anything else). else if(infix[i] == '(') { stack.push(infix[i]); } //This is farily similar to your code, but cleaned up. //With strings we can simply push_back instead of having //to worry about incrementing some counter. else if(isOperator(infix[i])) { char operator1 = infix[i]; if(isOperator(stack.top())) { while(!stack.empty() &amp;&amp; precedence(operator1,stack.top())) { postfix.push_back(stack.top()); stack.pop(); } } //This shouldn't be in an else - we always want to push the //operator onto the stack stack.push(operator1); } //We've hit a right parens - Why you had a for loop //here originally I don't know else if(infix[i] == ')') { //While top of stack is not a right parens while(stack.top() != '(') { //Insert into postfix and pop the stack postfix.push_back(stack.top()); stack.pop(); } // Discard the left parens - you'd forgotten to do this stack.pop(); } } //Remove any remaining operators from the stack while(!stack.empty()) { postfix.push_back(stack.top()); stack.pop(); } } </code></pre>
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    1. POConvert from an infix expression to postfix (C++) using Stacks
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    1. USYuushi
    1. USYuushi
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    1. COI added somecomments as answer (needed more space and format stuff) - perhaps you want to improve your answer :)
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      1. USStefan
    2. CO@Stefan Thanks, all good catches. I've updated my answer, hopefully it's error free this time (and this is a lesson in why I should write tests!)
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