CODEWITHSUNDEEP
cpointersstrcmp
jotape
jotape

Reputation: 319

strcmp with pointers not working in C

Why this code isn't working. Just trying to check if the user input is the same as a password

char *pass;

printf("Write the password: ");
scanf("%s", pass); // Because is a pointer the & is out ?


if( strcmp( pass , "acopio") == 0)

Upvotes: 8

Views: 23141

Answers (7)

x y
x y

Reputation: 1

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

int main()
{
    
     char str1[] = "Hello" ;
     char str2[] = "HeLLo" ;
     char str3[10];
    
     int *s;  // make "s" an integer pointer to string 
    
     /* if you want to scanf in a string to compare */
     printf("Enter your password:  ");
     scanf("%9s", str3); //don't overrun the string size!
    
     s = strcmp(str3, str1); 
    
     /* anything other than zero returned is not a match */
     if( s == 0 )  {
         printf("   The two strings are the same\n"); }
         else printf("   The two strings are different\n");
     /* print the value of s */
   
     printf("   The value of s is %d", s);
  
     return 0;
}

Upvotes: -2

user5841469
user5841469

Reputation: 

#include<stdio.h>
main()
{
    int mystrcmp(char *,char *);

    char s1[100],s2[100];
    char *p1,*p2;
    p1=s1;
    p2=s2;
    printf("Enter the first string..?\n");
    scanf("%s",p1);
    printf("Enter the second string..?\n");
    scanf("%s",p2);
    int x=mystrcmp(p1,p2);
    if(x==0)
        printf("Strings are same\n");
    else
        printf("Strings are not same..\n");


}
int mystrcmp(char *p1,char *p2)
{
    while(*p1==*p2)
    {
        if(*p1=='\0' || *p2=='\0')
            break;
        p1++;
        p2++;
    }
    if(*p1=='\0' &&as *p2=='\0')
        return(0);
    else
        return(1);
}

simple code for beginners....

Upvotes: -2

Joe
Joe

Reputation: 7798

You've not actually allocated any space to put data. Defining a pointer just defines a variable that can hold the address of a block of data, it doesn't allocate the block.

You have a couple of options, allocate dynamic memory off the heap to write into and make the pointer point to it. Or use statically allocated memory on the stack and pass the address of it to your calls. There's little benefit to dynamic memory in this case (because it's temporary in use and small). You would have more work to do if you used dynamic memory - you have to make sure you got what you asked for when allocating it and make sure you've given it back when you're done AND make sure you don't use it after you've given it back (tricky in a big app, trust me!) It's just more work, and you don't seem to need that extra effort.

The examples below would also need significant error checking, but give you the general idea.

e.g.

char *pass = malloc (SOMESIZE);

printf("Write the password: ");
scanf("%s", pass);


if( strcmp( pass , "acopio") == 0)

or

char pass[SOMESIZE];

printf("Write the password: ");
scanf("%s", pass);


if( strcmp( pass , "acopio") == 0)

Upvotes: 12

John Bode
John Bode

Reputation: 123458

You haven't initialized pass to point to a buffer or other location to store the input.

For something simple like this, you can declare pass as an array of char instead of a pointer:

char pass[N]; // where N is large enough to hold the password plus a 0 terminator

scanf("%s", pass);
if (strcmp(pass, "acopio") == 0)
{
  ...
}

Except when it is the operand of the sizeof, _Alignof, or unary & operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array.

When you pass pass as an argument to scanf and strcmp, the type of the expression pass is converted from "N-element array of char" to "pointer to char", and the value of the expression is the address of the first element of pass, or &pass[0]. That's why you don't need to use the & operator in the scanf call.

Similarly, in the strcmp call, the string literal "acopio" is converted from an expression of type "7-element array of char" (const char in C++) to "pointer to char".

Upvotes: 0

Sravan
Sravan

Reputation: 39

Yes the pointer was not initialized. If you debug it you will get a access violation or segmentation fault. The code can be changed as follows.

 char pass[22];//22 can be replaced with other number

    printf("Write the password: ");
    scanf("%s", pass); 
    if( strcmp( pass , "acopio") == 0)
    printf("fu");//just to check

Upvotes: 0

SomeWittyUsername
SomeWittyUsername

Reputation: 18338

You need to allocate the pass so the scanf will have a place to store the input. Otherwise you have memory corruption.

Upvotes: 1

md5
md5

Reputation: 23699

pass is an unitialized pointer, and you attempt to write into it. You have to allocate enough memory to hold a string. For example, char pass[SIZE] will work better.

Upvotes: 5

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