c - 有没有可能在c语言中循环strcmp函数？

Question

我是一名大学新生，目前正在学习C语言。昨晚我刚刚发现我的代码存在很大的问题，但我以我糟糕的技能无法解决任何问题...你能帮我解决这个问题吗？

(条件)

• 您正在尝试使用 ID/密码登录，但只有 3 次机会。
• 每个都有 5 个 ID 和密码。
• 如果id不存在，则打印id不存在。
• 如果 id 存在但 pw 不匹配，则打印 pw 不匹配。
• 如果 id 和 pw 匹配，则打印登录成功并退出程序。

有没有办法循环标记的strcmp函数？

下面是代码。

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

int main(void) {
    char id0[] = "user1";
    char id1[] = "user2";
    char id2[] = "user3";
    char id3[] = "user4";
    char id4[] = "user5";

    char pw0[] = "pass1";
    char pw1[] = "pass2";
    char pw2[] = "pass3";
    char pw3[] = "pass4";
    char pw4[] = "pass5";

    char idsol[100] = { 0 };
    char pwsol[100] = { 0 };

    int idmat = 1;
    int pwmat = 1;
    int logstack = 0;
    for (int i = 0; i <= 2;) {
        logstack = i;
        rewind(stdin);
        printf("ID: ");
        scanf_s("%s", idsol, sizeof(idsol));
        printf("PW: ");
        scanf_s("%s", pwsol, sizeof(pwsol));

        if (strcmp(idsol, id0) == 0) { //i want to make these 5 blocks in one command.
            idmat = 0;
            if (strcmp(pwsol, pw0) == 0) {
                pwmat = 0;
            }
        }
        if (strcmp(idsol, id1) == 0) {
            idmat = 0;
            if (strcmp(pwsol, pw1) == 0) {
                pwmat = 0;
            }
        }
        if (strcmp(idsol, id2) == 0) {
            idmat = 0;
            if (strcmp(pwsol, pw2) == 0) {
                pwmat = 0;
            }
        }
        if (strcmp(idsol, id3) == 0) {
            idmat = 0;
            if (strcmp(pwsol, pw3) == 0) {
                pwmat = 0;
            }
        }
        if (strcmp(idsol, id4) == 0) { //these 5 blocks!!!
            idmat = 0;
            if (strcmp(pwsol, pw4) == 0) {
                pwmat = 0;
            }
        }
        if (idmat != 0) {
            printf("ID does not exist.\n");
            i++;
            idmat = 1;
        }
        else if (idmat == 0 && pwmat != 0) {
            printf("PW does not match.\n");
            i++;
            idmat = 1;
        } else {
            logstack = 0;
            break;
        }
    }
    if (logstack >= 2) {
        printf("login failed.\n");
    } else {
        printf("login successful!\n");
    }
    return 0;
}

感谢您的任何评论!祝你有美好的一天

Answer 1

您应该使用数组来收集要在循环中使用的值。

#include <stdio.h>
#include <string.h>
int main(void) {
    char id0[] = "user1";
    char id1[] = "user2";
    char id2[] = "user3";
    char id3[] = "user4";
    char id4[] = "user5";

    char pw0[] = "pass1";
    char pw1[] = "pass2";
    char pw2[] = "pass3";
    char pw3[] = "pass4";
    char pw4[] = "pass5";

    /* gather strings in arrays */
    char* ids[] = { id0, id1, id2, id3, id4 };
    char* pws[] = { pw0, pw1, pw2, pw3, pw4 };

    char idsol[100] = { 0 };
    char pwsol[100] = { 0 };

    int idmat = 1;
    int pwmat = 1;
    int logstack = 0;
    for (int i = 0; i <= 2;) {
        logstack = i;
        rewind(stdin);
        printf("ID: ");
        scanf_s("%s", idsol, sizeof(idsol));
        printf("PW: ");
        scanf_s("%s", pwsol, sizeof(pwsol));

        /* use the arrays for looping */
        for (int j = 0; j < 5; j++) {
            if (strcmp(idsol, ids[j]) == 0) {
                idmat = 0;
                if (strcmp(pwsol, pws[j]) == 0) {
                    pwmat = 0;
                }
            }
        }
        if (idmat != 0) {
            printf("ID does not exist.\n");
            i++;
            idmat = 1;
        }
        else if (idmat == 0 && pwmat != 0) {
            printf("PW does not match.\n");
            i++;
            idmat = 1;
        }
        else { logstack = 0; break; }
    }
    if (logstack >= 2) {
        printf("login failed.\n");
    }
    else { printf("login successful!\n"); }

    return 0;
}

另一种方法是将字符串直接放入数组中:

#include <stdio.h>
#include <string.h>
int main(void) {
    char ids[][6] = {
        "user1",
        "user2",
        "user3",
        "user4",
        "user5"
    };

    char pws[][6] = {
        "pass1",
        "pass2",
        "pass3",
        "pass4",
        "pass5"
    };

    char idsol[100] = { 0 };
    char pwsol[100] = { 0 };

    /* omit: same as the first code */
}