Question

我使用Visual Studio 2013在C中编写了一个程序，它根据三个数值计算平均时间和油漆干燥。

程序运行正常没有错误但是在调试时，控制台应用程序只显示光标闪烁。
首先我认为这只是这个程序，但后来我打开了我过去的项目，同样的事情发生时调试。

当我在另一台机器上调试时，控制台应用程序会显示程序应该打印的内容，但我的计算机却没有。这是编译器的问题吗？

#include <stdio.h>
#include <stdlib.h>

/ 代码的这个起始部分是打印出提示的地方使用户输入程序所需的信息打印输出结果基于输入。 /

int main(){
int numofpaints;
char paintname[30][10];
int time[3][10];
int meantime;
int var = 0;
int var2 = 0;
int hours;
int minutes;

printf("Input the number of paints you want:\n\n");
scanf("%d", &numofpaints);

/ 这个for循环是平均值的计算发生的地方计算以确定油漆是否符合快速干燥状态。 /

for (var = 0; var < numofpaints; var++)

{
    printf("please enter the name of the paint:\n\n");
    scanf("%s", paintname[var]);

    printf("How long did %s take to dry first the first time:\n\n", paintname[var]);
    scanf("%d", &time[0][var]);

printf("How long did %s take to dry for the second time:\n\n", paintname[var]);
    scanf("%d", &time[1][var]);

    printf("How did it take for %s to dry for the third time:\n\n", paintname[var]);
    scanf("%d", &time[2][var]);
}



for (var2 = 0; var2 < numofpaints; var2++)
{
    meantime = (time[0][var2] + time[1][var2] + time[2][var2]) / 3;
    hours = meantime / 60;
    minutes = meantime % 60;

/ 这些if语句用于根据所做的计算打印输出结果使用用户通过键盘输入的值。例如，如果值输入的用户输入5然后将使用适当的语句将打印“5小时”而不是“5小时0分钟”。 /

if (hours > 1 && minutes == 0)
    {
        printf("\n%s finished drying in %d hours. ", paintname[var2], hours);
    }
    else if (hours > 1 && minutes > 1)
    {
        printf("\n%s finished drying in %d hours and %d minutes. ", paintname[var2], hours, minutes);
    }
    else if (hours > 1 && minutes == 1)
    {
        printf("\n%s finished drying in %d hours and %d minute. ", paintname[var2], hours, minutes);
    }
    else if (hours == 0 && minutes != 1)
    {
        printf("\n%s finished drying in %d minutes. ", paintname[var2], minutes);
    }
    else if (hours == 1 && minutes == 1)
    {
        printf("\n%s finished drying in 1 hour and 1 minute. ", paintname[var2]);
    }
    else if (hours == 1 && minutes > 1)
    {
        printf("\n%s finished drying in 1 hour and %d minutes. ", paintname[var2], minutes);
    }
    else if (hours == 0 && minutes == 1)
    {
        printf("\n%s finished drying in %d minute. ", paintname[var2], minutes);
    }
    else if (hours == 1 && minutes == 0)
    {
        printf("\n%s finished drying in an hour. ", paintname[var2]);
    }

/ 这些if语句可以确定油漆是否符合即时，快速或快速的条件。第一个如果说明是否计算了3个输入的平均时间值小于或等于1小时/ 60然后它适合即时油漆类别。所以每一个在3个陈述中，将确定油漆是否适合即时，快速或快速。 / 如果（同时<360） printf（“这意味着％s符合条件”，paintname [var2]）; 其他

        printf("This means %s did not qualify.\n\n\n", paintname[var2]);

if (meantime <= 60)
    {
        printf("and\nit'll fit into the category of instant paint.\n\n\n", paintname[var2]);
    }
    else if (meantime <= 119){
        printf("and\nit'll fit into the category of fast paint.\n\n\n", paintname[var2]);
    }
    else if (meantime <= 359){
        printf("and\nit'll fit into the category of quick paint.\n\n\n", paintname[var2]);
    }
        else{

        }


}

}

控制台应用程序不显示任何内容

0 个答案: