我是一名大一新生，目前正在学习C语言。昨晚我刚刚发现我的代码有很大的问题，但我的糟糕技能无法解决。。。你能帮我解决这个问题吗？
(条件(
- 您尝试使用id/pw登录，但只有3次机会
- 每个上都有5个id和pw
- 如果id不存在，请打印
id does not exist - 如果id存在但pw不匹配，则打印
pw does not match - 如果id和pw匹配，则打印
login successful并退出程序
有什么方法可以循环标记的strcmp函数吗？
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;
}
感谢您的评论！今天过得愉快
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：
#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 */
}
我刚刚定义了一个函数来执行循环，并将数组定义更改为字符串列表。代码如下(按照代码中的解释(：
#include <stdio.h>
#include <string.h>
#define NTIMES 3 /* ask three times for a password */
/* this table organization allows you to iterate for each pair of ids[i]
* and pws[i]. */
char *ids[] = {
"user1",
"user2",
"user3",
"user4",
"user5",
NULL, /* to indicate we are finished this controls how the end of
* the list is reached below */
};
char *pws[] = {
"pass1",
"pass2",
"pass3",
"pass4",
"pass5", /* don't need to add a NULL here, but there
* must be as many entries as for used ids */
};
/* this function searches the two tables above for a valid user/pass match */
int check_user(char *user, char *pass)
{
int id;
for (id = 0; ids[id] != NULL; id++) { /* iterate the list */
if (0 == strcmp(ids[id], user)) { /* if user found */
if (0 == strcmp(pws[id], pass)) { /* if pass matches */
printf("Found user %d, id=%sn", id, ids[id]);
return id; /* found user, return value of id to know
* outside which user was selected. */
} else {
printf("PW does not matchn");
/* we continue for the next pair in the loop */
}
}
}
/* we ended the loop without a valid user/pass match */
printf("ID %s does not exist, or password incorrectn", user);
return -1; /* not found, we return a number out of the valid range */
}
int main(void) {
/* initializing these variables is not neccessary as you assign them
* (by reading into them from stdin) before using. But it is a good
* practice, so it is more difficult that you forget initializing
* variables that should have a default value. */
char idsol[100] = { 0 };
char pwsol[100] = { 0 };
// int idmat = 1; // unnecessary
// int pwmat = 1; // unnecessary
// int logstack = 0; // unnecessary
/* we need to declare times outside the for loop, because we are using
* the last value assigned to it in the loop, so in case we get out of
* the loop prematurely, we know we got a valid user. */
int times = -1; /* we initialize, for the same reason explained above */
for (times = 0; times < NTIMES; times++) {
// logstack = i; // unnecessary
// rewind(stdin); // unnecessary
printf("ID: ");
if (!fgets(idsol, sizeof idsol, stdin)) {
break; /* EOF on input, get out of the loop */
}
char *id_pointer = strtok(idsol, "n"); /* take off the last
* newline */
// scanf_s("%s", idsol, sizeof(idsol)); // scanf_s is not standard
printf("PW: ");
if (!fgets(pwsol, sizeof pwsol, stdin)) {
break; /* EOF while reading input, get out of the loop */
}
char *pass_pointer = strtok(pwsol, "n"); /* take off the newline */
// scanf_s("%s", pwsol, sizeof(pwsol)); // scanf_s is not standard
/* this calls the function above that tests a username/password
* and prints the messages in the block below */
if (check_user(id_pointer, pass_pointer) >= 0)
break; /* get out of the loop */
/* this was the code you wanted to eliminate, so here it is, commented */
// 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;
// }
}
/* i have changed slightly this code also to use a single constant
* (by using its name, so i can modify the number of times to ask for
* a password, by editing only in one place (at the top). */
if (times < NTIMES) {
printf("login successful!n");
return 0; /* successful exit code, see exit(3) for details */
} else {
printf("login failed.n");
return 1; /* some error happened */
}
}