陣列中的

Question

考慮查找最大和最小數：

#include <iostream> // Include header file 

using namespace std; 

int main() //start of main function 
{ 

    int values[20]; // Declares array and how many elements 
    int small, big; // Declares integer 
    big = small = values[0]; // Assigns element to be highest or lowest value 

    for (int i = 0; i < 20; i++) // Counts to 20 and prompts the user for a value and stores it 
    { 
     cout << "Enter value " << i << ": "; 
     cin >> values[i]; 
    } 

    for (int i = 0; i < 20; i++) // Works out the biggest number 
    { 
     if(values[i] > big) // Compare biggest value with current element 
     { 
      big = values[i]; 
     } 
    } 

    for (int i = 0; i < 20; i++) // Works out the smallest number 
    { 
     if (values[i] < small) // Compares smallest value with current element 
     { 
      small = values[i]; 
     } 
    } 

    cout << "The biggest number is " << big << endl; // Prints outs the biggest number 
    cout << "The smallest number is " << small << endl; // Prints out the smallest number 
} 

這是到目前爲止我的代碼。我遇到的問題是打印出最大數量的數組。將第一個元素分配給最高值和最低值有關。它是有效的，如果我分開做它們。有什麼建議麼？

Answer 1

big=small=values[0]; //assigns element to be highest or lowest value 

應該AFTER填充循環

//counts to 20 and prompts user for value and stores it 
for (int i = 0; i < 20; i++) 
{ 
    cout << "Enter value " << i << ": "; 
    cin >> values[i]; 
} 
big=small=values[0]; //assigns element to be highest or lowest value 

，因爲當你聲明陣列 - 這是unintialized（存儲一些未定義的值），因此，您big和small分配將存儲undefined值過之後。

當然，您可以使用std::min_element，std::max_element或std::minmax_element從C++11，而不是編寫循環。

Answer 2

除非您確實需要實施您自己的解決方案，否則您可以使用std::minmax_element。這將返回一對迭代器，一個到最小的元素，一個到最大的元素。

#include <algorithm> 

auto minmax = std::minmax_element(std::begin(values), std::end(values)); 

std::cout << "min element " << *(minmax.first) << "\n"; 
std::cout << "max element " << *(minmax.second) << "\n"; 

Answer 3

您在數組初始化之前分配給大和小，即大和小在此處假設堆棧上的任何值。由於它們只是簡單的值類型，並且沒有引用，所以一旦通過cin >>寫入值[0]，它們就不會假設新值。

只需在第一次循環後移動賦值，它應該沒問題。

Answer 4

int main() //start of main fcn 
{ 

    int values[ 20 ]; //delcares array and how many elements 
    int small,big; //declares integer 
    for (int i = 0; i < 20; i++) //counts to 20 and prompts user for value and stores it 
    { 
     cout << "Enter value " << i << ": "; 
     cin >> values[i]; 
    } 
    big=small=values[0]; //assigns element to be highest or lowest value 
    for (int i = 0; i < 20; i++) //works out bigggest number 
    { 
     if(values[i]>big) //compare biggest value with current element 
     { 
      big=values[i]; 
     } 
     if(values[i]<small) //compares smallest value with current element 
     { 
      small=values[i]; 
     } 
    } 
    cout << "The biggest number is " << big << endl; //prints outs biggest no 
    cout << "The smallest number is " << small << endl; //prints out smalles no 
} 

Answer 5

可以填充數組初始化之後，或者你可以寫：

small =~ unsigned(0)/2; // Using the bit-wise complement to flip 0's bits and dividing by 2 because unsigned can hold twice the +ve value an 

整數可以容納。代替

big =- 1*(small) - 1; 

：

big = small = values[0] 

因爲當你填充所述陣列之前寫這條線，大和小的值將等於一個隨機剩餘值（如整數爲POD）從存儲器和如果這些數字大於或小於數組中的任何其他值，則會將它們作爲輸出。