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copied!<ol> <li><p>You're thinking of the stack backwards - the top of the stack is in the highest valid index, not in <code>val[0]</code>. This behaviour is evident when you look at the pushes of the operands. Your output:</p> <pre><code>3 4 + pass #0 val[0]: 3.00 pass #1 val[0]: 3.00 val[1]: 4.00 </code></pre> <p>First, the <code>3</code> is pushed - going onto the top of the (previously empty) stack - it's in slot 0. Next <code>4</code> is pushed. As you can see, it goes into <code>val[1]</code>, clearly showing that <code>val[0]</code> is not the top of the stack in this case.</p></li> <li><p>You're printing the stack incorrectly, which is confusing you further. Change your print loop to:</p> <pre><code>while (i < sp) { printf("val[%d]: %.2f\n",i,val[i]); ++i; } </code></pre> <p>That is, print only the valid entries in the stack, and you'll see your error.</p> <p>Your current comparison is looking for a <code>0</code> entry on the stack, which isn't how the program is identifying the free entries. That's what the <code>sp</code> variable is used for. In addition to looking for the wrong thing, it's doing it in a bizarro way - <code>val</code> is a an array of floating-point numbers - why are you comparing to a character literal <code>\0</code>?</p> <p>Here's the complete corrected output:</p> <pre><code>3 4 + pass #0 val[0]: 3.00 pass #1 val[0]: 3.00 val[1]: 4.00 pass #2 val[0]: 7.00 7 </code></pre> <p>Now, you see the correct output - both the <code>3.00</code> and <code>4.00</code> are popped, and <code>7.00</code> is pushed back onto the stack. It's now the only valid entry.</p></li> </ol>

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