C语言 如何从头开始重复 ac 程序并清理屏幕和第一个输入值？

how to repeat a c program from the beginning and clean the screen and 1st input values?

提问by MaxySpark

i m new in programing. i've written a simple program. i want to repeat the program again and again and it can only exit when user wants to exit. here is my program

我是编程新手。我写了一个简单的程序。我想一次又一次地重复这个程序，它只能在用户想要退出时退出。这是我的程序

#include<stdio.h>
#include<conio.h>
main()
{
    char ch;
    int num1, num2, a, m, s, choice;
    float d;
    printf("\nEnter The First Number: ");
    scanf("%d", &num1);
    printf("\nEnter The Second Number: ");
    scanf("%d", &num2);
    a=num1+num2;
    m=num1*num2;
    s=num1-num2;
    d=(float)(num1/num2);
    printf("\nEnter Your Choice \nFor Addition Type A \nFor Multipication Type M \nFor Division Type D \nFor Substraction Type S : ");
    scanf(" %c", &ch);
    switch(ch)
        {
            case 'A': printf("\nThe Addition Of The Number Is= %d", a);
                break;
            case 'M': printf("\nThe Multipication Of The Numbers Is= %d", m);
                break;
            case 'S': printf("\nThe Substraction Of THe Numbers Is= %d", s);
                break;
            case 'D': printf("\nThe Division Of The Two Numbers Is= %f", d);
                break;
            default : printf("\nInvalid Entry");
                break;
        }
    printf("\nPress Any Key To Exit");
    getch();
    return 0;
}

and here is the output

这是输出

"Enter The First Number: 10

"输入第一个数字：10

Enter The Second Number: 10

输入第二个数字：10

Enter Your Choice

输入您的选择

For Addition Type A

对于添加类型 A

For Multipication Type M

对于乘法类型 M

For Division Type D

对于 D 型分区

For Substraction Type S : A

对于减法类型 S : A

The Addition Of The Number Is= 20

数字的加法是= 20

Press Any Key To Exit"

按任意键退出“

I want a line before the line Press Any Key To Exit

我想要一行前一行按任意键退出

"If You Want To Calculate Again Press Y

"如果您想再次计算，请按 Y

or

或者

Press Any Key To Exit"

按任意键退出“

when press Ythen the program should start from the beginning.

当按Y 时，程序应该从头开始。

How can i do this???

我怎样才能做到这一点？？？

回答by Unix von Bash

wrap the code inside a do{} while()?

将代码包裹在do{} while()?

char answer;
do{
printf("\nEnter The First Number: ");
scanf("%d", &num1);
printf("\nEnter The Second Number: ");
scanf("%d", &num2);
a=num1+num2;
m=num1*num2;
s=num1-num2;
d=(float)(num1/num2);
printf("\nEnter Your Choice \nFor Addition Type A \nFor Multipication Type M \nFor Division Type D \nFor Substraction Type S : ");
scanf(" %c", &ch);
switch(ch)
    {
        case 'A': printf("\nThe Addition Of The Number Is= %d", a);
            break;
        case 'M': printf("\nThe Multipication Of The Numbers Is= %d", m);
            break;
        case 'S': printf("\nThe Substraction Of THe Numbers Is= %d", s);
            break;
        case 'D': printf("\nThe Division Of The Two Numbers Is= %f", d);
            break;
        default : printf("\nInvalid Entry");
            break;
    }
printf("\nPress Y to continue. Press any Key To Exit");
scanf(" %c",&answer); // dont forget type &
}
while(answer == 'y' || answer == 'Y');

Declare a variable, let's say answer, which will store the user answer when you ask for "Press Y to continue. Press any Key To Exit". Check to see what value has that variable. If is 'y' or 'Y', the loop will repeat. If the user pressed other key, the loop is over.

声明一个变量，比如 answer，当您要求“按 Y 继续。按任意键退出”时，它将存储用户的回答。检查以查看该变量的值。如果是 'y' 或 'Y'，循环将重复。如果用户按下其他键，则循环结束。

回答by Jens Mühlenhoff

I would move the calculation stuff in it's own function and then use while()in main.

我会在它自己的函数中移动计算内容，然后while()在main.

I have tried to fix other problems as well (this solution only uses standard C functions).

我也尝试解决其他问题（此解决方案仅使用标准 C 函数）。

#include <stdio.h> // puts, printf, fprintf, scanf, getchar, stderr, EOF
#include <stdlib.h> // exit, EXIT_SUCCESS, EXIT_FAILURE
#include <ctype.h> // toupper

char fail_on_eof (int c)
{
    if (c == EOF)
        exit (EXIT_FAILURE);
    // In case of fail_on_eof (scanf (...)) the return value of this this
    // function is not useful
    // scanf () returns the number of chars read or EOF
    // getchar () returns a char or EOF
    return (char) c;
}

void skip_to_next_line (void)
{
    char c;
    do
    {
        c = fail_on_eof (getchar ());
    } while (c != '\n');
}

char read_upcase_char_line (char* prompt)
{
    char c;
    printf ("%s ", prompt);
    c = fail_on_eof (toupper (getchar ()));
    skip_to_next_line ();
    return c;
}

int read_num_line (char* prompt)
{
    int num;
    printf ("%s ", prompt);
    fail_on_eof (scanf ("%d", &num));
    skip_to_next_line ();
    return num;
}

int calculate (void)
{
    char choice;
    int num1, num2, cont;
    num1 = read_num_line ("Enter the first number:");
    num2 = read_num_line ("Enter the second number:");
    puts("A - addition");
    puts("S - subtraction");
    puts("M - multiplication");
    puts("D - division");
    choice = read_upcase_char_line ("[ASMD]?");
    puts("");
    switch(choice)
    {
        case 'A':
            printf("%d + %d = %d", num1, num2, num1 + num2);
            break;
        case 'S':
            printf("%d - %d = %d", num1, num2, num1 - num2);
            break;
        case 'M':
            printf("%d * %d = %d", num1, num2, num1 * num2);
            break;
        case 'D':
            if (num2 == 0)
                // Better use stderr for error messages
                fprintf(stderr, "The divisor can not be zero");
            else
            {
                printf("%d / %d = %f", num1, num2, (double)num1 / num2);
            }
            break;
        default :
            // Better use stderr for error messages
            fprintf(stderr, "Invalid entry");
            break;
    }
    printf("\n");
    for (;;)
    {
      cont = read_upcase_char_line ("Continue [YN]?");
      if (cont == 'Y')
          return -1;
      else if (cont == 'N')
          return 0;
    }
}

int main(void)
{
    while (calculate ());
    return EXIT_SUCCESS; // Use this constant to avoid platform specific issues with the return code
}

回答by Vektorweg

You can also use recursion, which is often used in more functional oriented programming languages.

您还可以使用recursion，它通常用于更多面向函数的编程语言。

Pseudo-code:

伪代码：

myfunction = do 
    ...
    b <- somethingtodo
    ...
    if b
      then myfunction 
      else return ()

Relative to Jens's solution, it would look like:

相对于Jens 的解决方案，它看起来像：

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

int main (void)
{
    char choice;
    int num1, num2, cont;
    printf("Enter the first number: ");
    scanf("%d", &num1);
    getchar ();
    printf("Enter the second number: ");
    scanf("%d", &num2);
    getchar ();
    printf("A - addition\n");
    printf("S - subtraction\n");
    printf("M - multiplication\n");
    printf("D - division\n");
    printf("[ASMD]? ");
    choice = (char)toupper(getchar());
    getchar (); 
    printf("\n");
    switch(choice)
    {
        case 'A':
            printf("%d + %d = %d", num1, num2, num1 + num2);
            break;
        case 'S':
            printf("%d - %d = %d", num1, num2, num1 - num2);
            break;
        case 'M':
            printf("%d * %d = %d", num1, num2, num1 * num2);
            break;
        case 'D':
            if (num2 == 0)
                fprintf(stderr, "The divisor can not be zero");
            else
            {
                printf("%d / %d = %f", num1, num2, (double)num1 / num2);
            }
            break;
        default :
            fprintf(stderr, "Invalid entry");
            break;
    }
    printf("\n");
    for (;;)
    {
      printf("Continue [YN]? ");
      cont = toupper(getchar());
      getchar ();
      if (cont == 'Y')
          return main(); // the difference.
      else if (cont == 'N')
          return EXIT_SUCCESS;
    }
}