Linux登录和strcmp

Question

我正在为Linux编写一个简单的程序.我有一个功能"addonlyonce"的问题.它比较字符串并添加到没有重复的数组(如果字符串不存在于数组中).该函数运行良好,但不适用于utmp结构.当比较utmp结构的每个字符串总是返回1(这意味着字符串已经存在于数组中 - 并且它是错误的:().您可以编译此代码gcc thiscode.c -o test(它仅适用于Linux).

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

int addonlyonce(char **array, char *str) {
    int i=0;
    printf("I got %s\n",str);
    while(*array != '\0') {
        if(strcmp(*array,str)==0) {
            printf("Already exists!\n");
            return 1;
        }
        i++;
        array++;
    }
    *array=str;
    printf("Sit down on the number: %d\n",i);
    return 0;
}

void printarray(char **array) {
    printf("Array looks like: ");
    while(*array != '\0') {
        printf("%s ",*array);
        array++;
    }
    printf("\n");
}

int main(void) {
    char *users[20]={0};
        char exuser2[]="exuser2";
    struct utmp current_record;
    FILE *fp = fopen(UTMP_FILE, "r");

    addonlyonce(users,"exuser1");
    printarray(users);
    addonlyonce(users,exuser2);
    printarray(users);
    addonlyonce(users,"exuser1");
    printarray(users);
    addonlyonce(users,"exuser3");
    printarray(users);
    while (fread(&current_record, sizeof(struct utmp), 1, fp) == 1) {
        addonlyonce(users,current_record.ut_name); //HERE DON'T WORK
        printarray(users);
    }
    addonlyonce(users,"exuser4");
    printarray(users);
    fclose(fp);
    return 0;
}

我想这样做(正确运作):

I got exuser1
Sit down on the number: 0
Array looks like: exuser1 
I got exuser2
Sit down on the number: 1
Array looks like: exuser1 exuser2 
I got exuser1
Already exists!
Array looks like: exuser1 exuser2 
I got exuser3
Sit down on the number: 2
Array looks like: exuser1 exuser2 exuser3 
I got reboot
Sit down on the number: 3
Array looks like: exuser1 exuser2 exuser3 reboot 
I got kelloco2
Sit down on the number: 4
Array looks like: exuser1 exuser2 exuser3 reboot kelloco2 
I got przemek
Sit down on the number: 5
Array looks like: exuser1 exuser2 exuser3 reboot kelloco2 przemek
I got guest
Sit down on the number: 6
Array looks like: exuser1 exuser2 exuser3 reboot kelloco2 przemek guest 
I got exuser4
Sit down on the number: 7
Array looks like: exuser1 exuser2 exuser3 reboot kelloco2 przemek guest exuser4

但是,执行后我得到这样的东西:

I got exuser1
Sit down on the number: 0
Array looks like: exuser1 
I got exuser2
Sit down on the number: 1
Array looks like: exuser1 exuser2 
I got exuser1
Already exists!
Array looks like: exuser1 exuser2 
I got exuser3
Sit down on the number: 2
Array looks like: exuser1 exuser2 exuser3 
I got reboot
Sit down on the number: 3
Array looks like: exuser1 exuser2 exuser3 reboot 
I got kelloco2
Already exists!   //Here a problem arises
Array looks like: exuser1 exuser2 exuser3 kelloco2 
I got przemek
Already exists!
Array looks like: exuser1 exuser2 exuser3 przemek 
I got guest
Already exists!
Array looks like: exuser1 exuser2 exuser3 guest 
I got exuser4
Sit down on the number: 4
Array looks like: exuser1 exuser2 exuser3 guest exuser4 

.

可能有什么不对？问候K.

Answer 1

没有读过你的所有代码,但这让我感到困惑:

while(*array != '\0') {

数组是一个**char.这意味着*数组是一个*char.但是你将它与char('\0')进行比较.你可能意味着NULL？

现在你的错误:

*array=str;

这是坏线.你没有复制字符串!你只是指着内存中的同一个地方!

这意味着如果更改指向str的字符串,指向的字符串*array也会改变!

所以一旦你添加一个字符串ut_name然后再改变ut_name它并再添加它 - 它已经找到了它,因为你拥有相同的内存ut_name和*array.

无论如何 - 使用strcpy(分配内存!!!)而不是那条线*array=str,一切都会正常工作.