无循环无数组从6个整数中找最小奇偶整数的代码问题排查

Hey there! Let's work through fixing your code to correctly find the smallest even and odd numbers from 6 input integers (3 even, 3 odd). I spotted a couple key issues that are throwing off your output—let's break them down and fix them step by step.

Key Issues in Your Current Code #

• Incorrect initialization: Setting smallEven=0 and smallOdd=0 causes problems if your input includes negative numbers (e.g., if the smallest even is -6, your code would keep 0 instead). It also leads to wrong logic when checking if we've set the initial value.
• Assignment instead of comparison: You're using = (assignment) in your if conditions like if (smallEven=0) instead of == (equality check). This overwrites your variable and evaluates to false every time, so your code never updates the minimum values correctly.
• Incomplete input handling: You only process the first two numbers (a and b)—you need to handle all 6 inputs!

Corrected Code #

First, we'll initialize our min values to a very large number (using INT_MAX from the <climits> header) so any valid input number will be smaller than the initial value. Then, we'll process each input with proper equality checks and comparisons.

#include <iostream>
#include <climits> // For INT_MAX

using namespace std;

int main() {
    int a, b, c, d, e, f;
    int smallEven = INT_MAX; // Initialize to largest possible int
    int smallOdd = INT_MAX;

    // Process first number
    cout << "Enter a number:\n";
    cin >> a;
    if (a % 2 == 0) {
        smallEven = a;
    } else {
        smallOdd = a;
    }

    // Process second number
    cout << "Enter a number:\n";
    cin >> b;
    if (b % 2 == 0) {
        if (b < smallEven) {
            smallEven = b;
        }
    } else {
        if (b < smallOdd) {
            smallOdd = b;
        }
    }

    // Process third number
    cout << "Enter a number:\n";
    cin >> c;
    if (c % 2 == 0) {
        if (c < smallEven) {
            smallEven = c;
        }
    } else {
        if (c < smallOdd) {
            smallOdd = c;
        }
    }

    // Process fourth number
    cout << "Enter a number:\n";
    cin >> d;
    if (d % 2 == 0) {
        if (d < smallEven) {
            smallEven = d;
        }
    } else {
        if (d < smallOdd) {
            smallOdd = d;
        }
    }

    // Process fifth number
    cout << "Enter a number:\n";
    cin >> e;
    if (e % 2 == 0) {
        if (e < smallEven) {
            smallEven = e;
        }
    } else {
        if (e < smallOdd) {
            smallOdd = e;
        }
    }

    // Process sixth number
    cout << "Enter a number:\n";
    cin >> f;
    if (f % 2 == 0) {
        if (f < smallEven) {
            smallEven = f;
        }
    } else {
        if (f < smallOdd) {
            smallOdd = f;
        }
    }

    // Output results
    cout << "Smallest even number: " << smallEven << endl;
    cout << "Smallest odd number: " << smallOdd << endl;

    return 0;
}

How This Works #

1. Initialization: Using INT_MAX ensures that the first even/odd number we input will always replace the initial value, regardless of whether it's positive or negative.
2. Proper comparisons: For each input, we check if it's even or odd, then compare it to the current minimum value. If it's smaller, we update the minimum.
3. Full input handling: We process all 6 required inputs, as specified in your assignment.

Test Case Example #

If you input the numbers: -6, 3, -2, -5, 4, 1

• The smallest even number will be -6
• The smallest odd number will be -5

This code handles both positive and negative values correctly, and follows your assignment rules (no loops, no arrays).

内容的提问来源于stack exchange，提问作者edmonda7