Question

我编写了以下程序，我知道它的作用，需要10个数字才能将它自己增加，然后在最后将它们加在一起。

    #include <iostream>
    using namespace std;

    main()
    {
    int a[10];//1
    int sumOfSquares = 0 ;//2
    int i =0; //3`enter code here`

    cout << "Please enter the ten numbers one by one " << endl;//4

    for (i = 0 ; i < 10 ; i++)//5 dont get what this does,   
                              //  obviously its a for loop, 
                              //  but why does it increment i by one
    {
    cin >> a [i];//6 this means store the 10 numbers
                 //  in array a and refer to it by the variable i
    }
    for (i = 0 ; i < 10 ; i++) //7 again i dont get why this is here
    {
    sumOfSquares = sumOfSquares + a[i]*a[i];//8
    }

   cout << "The sum of squares is "<< sumOfSquares << endl; //9

   }

Answer 1

为什么它会增加一个

数组索引从0运行到N-1，其中N是数组中元素的数量。 i++将i的值增加1（相当于i = i + 1;）。在i循环中递增for是用于访问数组a的每个元素的构造（按顺序）：

for (int i = 0; i < 10; i++)
{
    a[i] = 2; /* just example */
}

相当于：

a[0] = 2;
a[1] = 2;
...
a[9] = 2;

正如其他人所评论的那样，获得一本C ++书籍（see this SO question for a list of C++ books）。

Answer 2

#include <iostream>
using namespace std;

main()
{
    //declare space in memory for 10 numbers to be stored sequentially
    //this is like having ten variables, a1, a2, a3, a4 but rather than having
    //hardcoded names, you can say a[4] or rather i=4, a[i], to get variable a4.
    int a[10];  
    int sumOfSquares = 0 ;  //space for a number to be stored called sumOfSquares
    int i =0;               //space for a number to be stored called i

    cout << "Please enter the ten numbers one by one " << endl; //print msg to screen

    for (i = 0 ; i < 10 ; i++) //make i = 0; while i < 10 run this code! increase i by 1 each time
    {
        //a stores 10 numbers. put the number the user entered in space
        //a[0] the first time, a[1] the second time, a[2] the third time etc etc.
        cin >> a [i];

    }

    //run this code 10 times, with i=0 the first time, i=1 the second time,
    // i=3 the third time etc etc.
    for (i = 0 ; i < 10 ; i++)
    {
        //get the number at a[0] multiple it by the number at a[0]
        //add it to value already in sumOfSquares so this number goes up and up and up.
        //the second time through the loop get a[1] and multiply it by a[1].
        sumOfSquares = sumOfSquares + a[i]*a[i]; 
    }

    //print answer to screen
    cout << "The sum of squares is "<< sumOfSquares << endl; //9

}

与c ++程序混淆？

2 个答案: