错误：指针与整数之间的比较错误

Question

My code seems fine, but I get this warning (warning: comparison between pointer and integer), what is the best solution to solve this problem? I have already used double notation marks for ( char exit = "E" ), also used the same thing with while but the same problem.

#include <stdio.h>

int main()
{
    char c[1];
    char exit = 'E';

    while (c != exit)
    {
        printf("Enter a character\n\n");
        scanf("%s", c);
        printf("your character is : %s\n-------------------\n", c);
    }
}

Answer 1

 #include <stdio.h> int main() { char c[1]; char exit = 'E'; while (c != exit) // here ... { printf("Enter a character\\n\\n"); scanf("%s", c); printf("your character is : %s\\n-------------------\\n", c); } } 

you are trying to compare a char to the pointer the array c decays to. What you perhaps wanted to do is to compare the first character of the array to the character exit :

while (c[0] != exit)

But that still doesn't make much sense since c is uninitialized and the user not yet had a chance to make any input. Better use a do ... while -loop:

#include <stdio.h>

int main()
{
    char c[1];
    char exit = 'E';

    do {
        printf("Enter a character\n\n");
        scanf("%s", c);
        printf("your character is : %s\n-------------------\n", c);
    } while (c[0] != exit);
}

Next thing is, that scanf("%s", c); could fail (yes unlikely, but possible). And the user could enter more characters than there is room for in the array c . You should never use scanf() whithout checking the return value nor "%s" without specifying a WIDTH for the conversion specifier to limit the characters put into the array.

When reading a string you need memory for WIDTH characters + a terminating '\\0' . So if you want to read a string of one character, the array has to have at least 2 elements:

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

int main()
{
    char c[2];
    char exit = 'E';

    do {
        printf("Enter a character\n\n");
        while (scanf("%1s", c) != 1 ) {
            fputs("Input error!\n");
            return EXIT_FAILURE;
        }
        printf("your character is : %s\n-------------------\n", c);
    } while (c[0] != exit);
}

But if you only want to read one character you are better off with getchar() :

#include <stdio.h>
{
    int ch;
    while (printf("Enter a character\n\n"),
           (ch = getchar()) != EOF && ch != 'E')
    {
        printf("your character is: %c\n-------------------\n", (char) ch);
    }    
}

Answer 2

i believe this is what you are trying to do.

warning is because you have not initialzed your character and also you were comparing address of character to character value.

#include <stdio.h>

int main()
{
    char c[1];
    char exit = 'E';

    while ((c[0]=getchar()) != exit)
    {
        if(c[0]==EOF)break;
        printf("your character is : %c\n",c[0]) ;
    }
    printf("ended");
}