C 编程 - 重复 scanf 除非用户输入“0”

C Programming - Repeat scanf unless user enters "0"

你好,我是 C 的新手,正在尝试制作一个程序,要求用户输入一大堆数字。我想循环 scanf 以便它不断询问,当用户输入“0”时,它会停止,读取输入的偶数和奇数,并分别计算它们。现在,我可以在用户按下 "Enter" 后继续询问新号码,但是当我键入“0”时,它只会不断询问更多号码并且不会停止。我究竟做错了什么?就像我之前说的,我很新,所以婴儿的话是最好的。

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

int main (void) {

    int number_of_integers, sum = 0, i, integer;
    char user_name[128];

    printf("What is your name?\n");
    scanf("%s", user_name);

    printf("\nEnter any real numbers followed by ENTER.\n");
    while (integer != 0) {
        scanf("%s", &integer);
        if (integer == 0)
            break;
    }

    printf("%s, the numbers you entered are broken down as follows:\n", user_name);

    return 0;
}

正如评论者所指出的,我们不是家庭作业或辅导服务,但我通过制作一个我刚刚编造的实际工作示例来帮您一个忙。现在由您来完全根据您的需要进行定制。

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

int main (void) {
    int max=100;
    int integer;
    int even[max];
    int odd[max];
    int index=0;
    int evencount=0;
    int oddcount=0;
    char user_name[128];

    printf("What is your name?\n");
    scanf("%s", user_name);

    printf("\nEnter any real numbers followed by ENTER.\n");
    while (index < max){
        scanf("%d", &integer);
        if (integer == 0)
        break;
        if ((integer % 2) == 0){
            evencount++;
            even[evencount]=integer;
        }else{
            oddcount++;
            odd[oddcount]=integer;
        }
        index++;
    }

    printf("%s, the %d numbers you entered are broken down as follows:\n", user_name,index);
    printf("%d odd integer(s):\n",oddcount);
    while (oddcount > 0){
    printf("%d\n",odd[oddcount]);
    oddcount--;
    }
    printf("%d even integer(s):\n",evencount);
    while (evencount > 0){
    printf("%d\n",even[evencount]);
    evencount--;
    }
    return 0;
}