CODEWITHSUNDEEP
c
francis kyle flores
francis kyle flores

Reputation: 71

How do I deal with string input in C?

I'm just practicing C and I would like to make a simple Login program with a username and a password... here is my code:

int main(){
   char uname, passwd;

   printf("Enter your username: ");
   scanf("%c", uname);
   printf("Enter your password: ");
   scanf("%c", passwd);

   printf(" \n");

   if (uname == "waw" && passwd == "wow"){
      printf("You have logged in\n");
   } else {
      printf("Failed, please try again\n");
   }
   return 0;
}

And then I got this error when I try to compile it

log.c: In function ‘main’:
log.c:7:10: warning: format ‘%c’ expects argument of type ‘char *’, but     argument 2 has type ‘int’ [-Wformat=]
    scanf("%c", uname);
          ^
log.c:9:10: warning: format ‘%c’ expects argument of type ‘char *’, but     argument 2 has type ‘int’ [-Wformat=]
    scanf("%c", passwd);
          ^
log.c:13:14: warning: comparison between pointer and integer
    if (uname == "waw" && passwd == "wow"){
              ^
log.c:13:33: warning: comparison between pointer and integer
    if (uname == "waw" && passwd == "wow"){
                                 ^

Upvotes: 2

Views: 4025

Answers (4)

archimedes
archimedes

Reputation: 89

This is how you can do it with milkstrings

tXt ask(tXt prmp) {
  printf("%s ?",prmp) ;
  return(txtFromFile(stdin)) ; }



void main(void) {
  tXt user = ask("user") ;
  if ( strcmp(txtConcat(user,"/",ask("password"),NULL),"waw/wow") == 0)
    printf("You have logged in\n"); 
  else
    printf("Failed, please try again\n");
}

Upvotes: 1

abhiarora
abhiarora

Reputation: 10450

There is no string data type in C programming language. Strings in C are represented as array of characters.

In C, char is a data type to represent a character. So, all you need to do is declare a array of char data type to represent a string in C. Each element in that array will hold a character of your string.

Also, operators in C like ==, !=, +=, + are defined for build-in data types in C and since, there is no operator overloading in C, you can't use these operators with your C-String as C-Strings are not build-in data type in C programming language.

Note: C-Strings are actually Null-terminated array of char data types. That means, last character in any C-String in C will be used to store a Null Character ('\0') which marks the end of the string.

The Header file <string.h> has predefined functions that you can use to operate on C-String(Null-terminated array of char data type). You can read more about it over here.

So, your program should look like:

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

#define CSTR_MAX_LEN 100
int main()
{
    char uname[CSTR_MAX_LEN+1];   //An extra position to store Null Character if 100 characters are to be stored in it.
    char passwd[CSTR_MAX_LEN+1];

    printf("Enter your username: ");
    scanf("%s", uname); 


    printf("Enter your password: ");
    scanf("%s", passwd);

    // Use strcmp to compare values of two strings.
    // If strcmp returns zero, the strings are identical
    if ((strcmp(uname, "waw") == 0) && (strcmp(passwd, "wow") == 0)){
         printf("You have logged in\n");
    } else{
         printf("Failed, please try again\n");
    }
    return 0;
 }

Remember, the correct format specifier to handle C-Strings are %s in C programming language. So, i have to change your scanf() function call. Also, i have used strcmp() function which is declared in header file <string.h> (i have included this header file in the above program). It returns numeric zero if strings match. You can read more about it above here. Also, have a look at strncmp() which is safer version of strcmp(). Also, remember that it can lead to undefined behavior if you try to access any array out of it's bound. So, it would be good idea to include checking in your program. As, someone included in their answer you can change my scanf() function call to:

scanf("%100s", uname);

It avoids to read more than 100 characters in array uname as it is allocated only to store 100 characters(excluding the Null character).

Upvotes: 1

4386427
4386427

Reputation: 44340

1) Don't read (scan) into a char. You need a string. Use format specifier %s instead of %c

2) Use strcmp to compare string values

Try this:

#include <stdio.h>

int main(){
   int res;
   char uname[40];   // char arrays instead of single char
   char passwd[40];

   printf("Enter your username: ");
   res = scanf("%39s", uname);  // Set a maximum number of chars to read to
                                // 39 to avoid overflow
   if (res != 1) exit(1);       // Error on stdin. End program

   printf("Enter your password: ");
   res = scanf("%39s", passwd);
   if (res != 1) exit(1); // Error on stdin

   printf(" \n");

   // Use strcmp to compare values of two strings.
   // If strcmp returns zero, the strings are identical
   if ((strcmp(uname, "waw") == 0) && (strcmp(passwd, "wow") == 0)){
       printf("You have logged in\n");
   } else{
       printf("Failed, please try again\n");
   }
   return 0;
}

Upvotes: 4

Marievi
Marievi

Reputation: 5009

First of all, you have declared uname and passwd as single characters, but you want to read strings. So you need to define the maximum length for each one and declare them as following:

char [X] uname;     //where X is the maximum length for uname, for instance 25
char [Y] passwd;   //where Y is the maximum length for passwd, for instance 20

Accordingly, your scanf() function should then be modified to :

scanf("%s", uname);
scanf("%s", passwd);

Apart from that, you need strcmp() to compare a string with a value. Change your if statement to :

if ( (strcmp(uname, "waw") == 0) && (strcmp(passwd, "wow") == 0) ) {
   printf("You have logged in\n");

Upvotes: 0

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