CODEWITHSUNDEEP
c
Anik Shahriar
Anik Shahriar

Reputation: 151

Passing argument 2 of strcmp makes pointer from integer without a cast

void viewonechar(){

char name[25], c[25];
int n;

fp = fopen("Phonebook.txt","r");

printf ("\n\n Enter Character : ");
scanf ("%s",c);

fscanf (fp, "%s %d", name, &n);

while (!feof(fp)){

    if ((strcmp(c, name[0])) == 0){  \\ Warning in here

        printf (" %s +880%d\n",name, n);

    }

    fscanf (fp, "%s %d", name, &n);

}

printf ("\n\n");

fclose(fp);

menu();

}

When i compile the code, on the marked line this warning appears, "Passing argument 2 of strcmp makes pointer from integer without a cast". What exactly am i doing wrong?

Upvotes: 0

Views: 4383

Answers (2)

P.P
P.P

Reputation: 121387

There are a number of problems.

The below code fixes the above issues:

void viewonechar(void) 
{
    char name[25], c[25];
    int n;

    FILE    fp = fopen("Phonebook.txt","r");

    if (!fp) {
        perror("fopen");
        exit(1);
    }

    printf ("\n\n Enter Character : ");
    if (fgets(c, sizeof c, stdin) == NULL) {
        fprintf(stderr, "Input error\n");
        exit(1);
    }

    while (fscanf (fp, "%24s %d", name, &n) == 2) {
        if (c[0] == name[0]) {
            printf (" %s +880%d\n",name, n);
        }  
    }
    printf ("\n\n");
    fclose(fp);
    menu();
}

Upvotes: 1

Ari0nhh
Ari0nhh

Reputation: 5920

int strcmp ( const char * str1, const char * str2 );

Since name is an array of char, name[0] is a char. strcmp takes char pointers as an arguments, so the char you have supplied is implicitly cast to an int-type and then - to the pointer, which produces an undefined behavior and most likely will lead to the segfault.

Upvotes: 1

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