Question

请帮助我需要阻止传入的字母和负数像错误处理或其他东西这是代码

{
printf("\nEnter number of Processes:");
scanf("%d", &n);
for(i=0;i<n;i++)


{
    printf("\nEnter Process ID %d :",i+1);
    scanf("%d", &process[i]);
    printf("\nEnter Process Time %d:",i+1);
    scanf("%d",&ptime[i]);
}

for(i=0;i<n-1;i++)
{
    for(j=i+1;j<n;j++)
    {
        if(ptime[i]>ptime[j])
        {
            temp = ptime[i];
            ptime[i] = ptime[j];
            ptime[j] = temp;
            ptemp = process[i];
            process[i] = process[j];
            process[j] = ptemp;
        }
    }
}
    wtime[0]=0;
for(i=1;i<n;i++)
{
    wtime[i]=wtime[i-1]+ptime[i-1];
    total=total+wtime[i];
}
avg=(float)total/n;
printf("\nProcess ID\t Process TIME\t Waiting TIME\n");
for(i=0;i<n;i++)
    printf("    %d\t           %d\t          %d\n",  process[i],  ptime[i],  wtime[i]);
printf("\nTotal Waiting Time: %d \nAverage Waiting Time: %f", total, avg);

先谢谢你需要完成它hehehe

Answer 1

简单方法：检查错误并在出现错误时终止。

listview.setAdapter(mydapter)

<强>更新

要完成所有退出adapter.notifyDataSetChanged()，使用旗帜是一个好主意。通过这种方式，可以更轻松地执行其他操作，例如打印错误消息。

// add #include <cstdlib> at the beginning of your code to use exit()

{
printf("\nEnter number of Processes:");
if(scanf("%d", &n)!=1 || n<0)exit(1);
for(i=0;i<n;i++)


{
    printf("\nEnter Process ID %d :",i+1);
    if(scanf("%d", &process[i])!=1 || process[i]<0)exit(1);
    printf("\nEnter Process Time %d:",i+1);
    if(scanf("%d",&ptime[i])!=1 || ptime[i]<0)exit(1);
}

Answer 2

尝试使用字符串类型获取输入，然后将字符串转换为您想要的结果。例如：

#include <string>
#include <iostream>
#include <sstream>
#include <exception>
using namespace std;

int main() {
    int processed_input;
    string input;
    cin >> input;
    stringstream ss(input);
    try {
        ss >> processed_input;
        if(processed_input < 0) throw "Negative number";
    } catch(exception &e) {
        cout << "Error : non-numeric characters inputted";
    } catch(const char* msg) {
        cout << "Error : " << msg << " inputted";
    }
    return 0;
}

我怎么能错误处理字母和负数？

2 个答案: