c - 如何将int转换为char类型并用它来停止循环？

Question

我是 C 编程的新手。我写了一个带有“for”循环的程序，它将接受两个 int 类型的输入 K 和 M 并打印出计算(除法中的散列函数)。如果我输入 q 或 Q 作为 K 或 M 的输入，那么它将退出。我将如何做？任何人都可以帮助我。

int main(){

int K;
int M;

for(;;)
{
    printf("Enter the value of K: \n");
    scanf("%d",&K);

    printf("Enter the value of M: \n");
    scanf("%d",&M);

    printf("The Hash address of %d and %d is %d",K,M,K%M);
}
return system("pause");}

Answer 1

这会检查 scanf() 的返回值以查看扫描是否成功。如果不是，则清除缓冲区并检查指示程序应该退出的“q”。
getint() 接受提示消息和指向标志的指针。将标志设置为 -1 告诉调用者退出。

#include <stdio.h>
#include <math.h>

int getint ( char *prompt, int *result);

int main ( int argc, char* argv[])
{
    int K, M;
    int ok = 0;

    do {
        K = getint ( "\nEnter the value of K ( or q to quit)\n", &ok);
        if ( ok == -1) {
            break;
        }
        M = getint ( "\nEnter the value of M ( or q to quit)\n", &ok);
        if ( ok == -1) {
            break;
        }
        printf("\nThe Hash address of %d and %d is %d\n", K, M, K % M);
    } while ( ok != -1);

    return 0;
}

//the function can return only one value.
//int *result allows setting a flag:
//  0 the function is looping
//  1 the function is returning a valid int
// -1 the function returns and the program should exit
// the caller can see the flag of 1 or -1
int getint ( char *prompt, int *result)
{
    int i = 0;
    int n = 0;

    *result = 0;

    do {
        printf("%s", prompt);
        if ( scanf("%d",&n) != 1) {// scan one int
            while ( ( i = getchar ( )) != '\n' && i != 'q') {
                //clear buffer on scanf failure
                //stop on newline
                //quit if a q is found
            }
            if ( i != 'q') {
                printf ( "problem with input, try again\n");
            }
            else {//found q. return and exit
                *result = -1;
                n = 0;
            }
        }
        else {//scanf success
            *result = 1;//return a valid int
        }
    } while ( *result == 0);

    return n;
}