下面的代码应该接受初始输入，并且应该继续请求输入，直到两个输入都从 0 到 2000

Question

下面的代码应该接受初始输入，并且应该继续请求输入，直到两个输入都从 0 到 2000，但只询问两者是否小于 0 或大于 2000

int main(){

    float x;
    int y;
    scanf("%f",&x);
    scanf("%d",&y);
while(!(x>0 && x<=2000) && !(y>0 && y<=2000)){
    printf("invalid balance or initial amount\n");
    printf("enter initial balance\n");
    scanf("%f",&x);
    printf("enter withdrawal amount\n");
    scanf("%d",&y);
    continue;
     }
return 0;

}

Answer 1

简单的答案：把中间的&&改成||：

while (!(x>0 && x<=2000) || !(y>0 && y<=2000)) {

或者，您可以使用等价物：

while (! ((x>0 && x<=2000) && (y>0 && y<=2000))) {

但是，我会稍微清理一下，因为 initialscanf 调用 [在循环之前] 不会提示用户。所以，我稍微颠倒一下逻辑：

#include <stdio.h>

int
main(void)
{
    float x;
    int y;

    while (1) {
        printf("enter initial balance\n");
        scanf("%f", &x);

        printf("enter withdrawal amount\n");
        scanf("%d", &y);

        if (((x > 0) && (x <= 2000)) && ((y > 0) && (y <= 2000)))
            break;

        printf("invalid balance or initial amount\n");
    }

    printf("x=%g y=%d\n",x,y);

    return 0;
}

下面的代码应该接受初始输入，并且应该继续请求输入，直到两个输入都从 0 到 2000

the code below is supposed to take initial inputs and is supposed to keep on asking for inputs until the both inputs are from 0 to 2000

c

logic

loops

continue

while-loop