Question

我一直试图将为该特定日期计算的总收入计入阵列。因此，最后我可以将数组中的所有值总计为总计。

到目前为止，我有2个阵列需要馅饼和挑选的苹果数量。为了计算那天馅饼，苹果托盘和总收入的收入，我把它放到了for循环中。

到目前为止，我已经得到了这个:(这是为了输入数组中的计算值）

float total[30];

int i, incmPie, numPie, rApples, applesLeft, FTray, incmFTray, PFTray;
float totalincm, incmApples, incmRApples, incmPFTray, totalincome;
**float total[30];**
int pieDemand[30]={4, 4, 2, 7, 1, 6, 7, 8, 9, 12, 2,13,13, 5, 3, 9, 9, 9, 8, 7,
                   12, 1, 3, 3,10,12, 3, 6, 9, 3}; 
int applesPicked[30]={330,123,110,245,321,999,0,100,77,89,100,200,300,390,700,20,701,6,800,90,
                   600,45,690,700,719,790,800,1000,66,666};
int date[30] = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30};
printf("\n==============================================================================");
printf("\n Date         Income from Pie         Income from Apples          Total income");
printf("\n==============================================================================");


for (i = 0 ; i <30; i++)
{
    if (applesPicked[i] == 0)
    {
        incmPie = 0;
        incmApples = 0;
        totalincm = 0;
        **total[i] = totalincm;**
    }

    else if (applesPicked[i] < (pieDemand[i]*8))
    {
        numPie = applesPicked[i]/8;
        incmPie = numPie * 14;
        rApples = applesPicked[i]%8;
        incmRApples = rApples * 0.5;
        incmApples = incmRApples;
        totalincm = incmPie + incmRApples;
        **total[i] = totalincm;**
    }

    else
    {
        incmPie = pieDemand[i] * 14;
        applesLeft = applesPicked[i] - (pieDemand[i]*8);
        FTray = applesLeft/20;
        incmFTray = FTray * 12;
        PFTray = applesLeft%20;
        incmPFTray = PFTray * 0.5;
        incmApples = incmFTray + incmPFTray;
        totalincm = incmApples + incmPie;
        **total[i] = totalincm;**
    }

    **totalincome** = total[1] + total[2] + total[3] + total[4] + total[5] + total[6] + total[7] + total[8] + total[9] + total[10] + total[11] + total[12] + total[13] + total[14] + total[15] + total[16] + total[17] + total[18] + total[19] + total[20] + total[21] + total[22] + total[23] + total[24] + total[25] + total[26] + total[27] + total[28] + total[29] + total[30];
    printf("\n"); //prints onto the next line.

    printf("%d/04/2013",date[i]); // prints the date.
    printf("%15d", incmPie); // prints the income from pies for each value in the arrays.
    printf("%20g", incmApples); // prints the income from apples from both full trays and remaining apples for each value in the arrays.
    printf("%28g", totalincm);
}
printf("\n==============================================================================");
**printf("\n Total income for the entire month: $%g", totalincome);**
printf("\n------------------------------------------------------------------------------");

_getch();
}

并且我使用此代码来汇总数组的总和：

totalincome = total[1] + total[2] + ... + total[30];

任何帮助将不胜感激！ :)

Answer 1

在C ++（几乎所有编程语言）中，数组索引从0开始，而不是1！查看Zero-based numbering以获取更多信息。

将其更改为

totalincome = total[0] + total[1] + ... + total[29];

或者简单地说，为了让您的生活更轻松，请使用循环：

totalincome = 0;
for (int i = 0; i < sizeof(total)/sizeof(total[0]); ++i)
     totalincome += total[i];

Answer 2

totalincome = 0;
for (int i = 0; i < sizeof(total)/sizeof(total[0]); ++i)
     totalincome += total[i];

对于静态arrray，这将起作用。如果数组是动态分配或作为指针传递的，则必须跟踪元素的数量。

totalincome = 0;
for (int i = 0; i < numelements; ++i)
     totalincome += total[i];

您需要将totalincome放在循环之外。

将计算值分配给float数组 - c ++

2 个答案: