标志验证不起作用。 使用数组，scanf函数和strcmp

Question

I'm a beginner. I'm trying to write a program which will validate some words which I will enter with a given set of available words and then compare them. What I'm using in this code is what I've learned so far. Please help me to understand what is wrong with this code.

So, when I enter a word, like "flag", it prints an error

"...line 71: 1872 Sigmentation fault sh"${SHFILE}...

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

int main(void) {
    system("COLOR B0");

    char *enter_flags[3];

    char*available_flags[3]={"print","scan","flag"};

    printf("\r\nEnther your flags here please:  ");

    for(int i=0;i<3;i++){

        scanf("%s",&enter_flags[i]);

        for(int j=0;j<3;j++){
            if(strcmp(enter_flags[i],available_flags[j])==0)
            {
                printf("---%s---|---%s--- MATCH", enter_flags[i], available_flags[j]);
            }

            else printf("---%s---|---%s--- INCORRECT", enter_flags[i], available_flags[j]);
        }


    }
    return 0;
}

I Understood my mistake. Thank you all!

Answer 1

The main problem with your code is that you do not reserve any memory for storing the values (aka strings) entered by the user.

This line

char *enter_flags[3];

reserves memory for 3 char pointers but no memory for the strings (aka char arrays).

Instead you can do something like:

char enter_flags[3][32];

This will allow you to store 3 strings with max length 31 (plus the required zero termination).

To avoid overflow your scanf should then be:

scanf("%31s", enter_flags[i]);

btw....

Now you may wonder why this code char *available_flags[3]= "print","scan","flag"}; is okay when char *enter_flags[3]; is wrong.

The reason is that available_flags is 3 char pointers to (constant) string literals. The compiler will place the 3 strings somewhere in memory and make the pointers in available_flags point to the strings. Your code does not change these strings (and isn't allowed to).

enter_flags is different because you want it to hold 3 strings that your program can change (ie via user input). Therefore enter_flags can't be 3 char pointers. There must be memory for storing the user input. You achieve that by making enter_flags a 2 dimension char-array. In C, a 2 dimensional char array can be used as an array of strings.

Answer 2

you are using the enter_flags array ,this var is array of pointer and it's hold unknown addresses because you just declare it without any value.you have to specify that this this varibles is hold at least 6 chars because the largest keyword or flag is "print" and have 5 chars .
the solution is declaring

char enter_flags[3][6];

or you should use malloc function.