Thread: Help with strings

Check out the strcpy() function.

Or maybe you could do something with pointers!

Code:

#include <stdio.h>

int main()
{
    char *string;
    
    scanf("%s", &string);
    printf("%s", &string);
    
    return 0;
}

Originally Posted by xxxrugby

Or maybe you could do something with pointers!

Code:

#include <stdio.h>

int main()
{
    char *string;
    
    scanf("%s", &string);
    printf("%s", &string);
    
    return 0;
}

OMFG No! You're joking right?! Please tell me you're joking!

Ups. I think that he wont something else. Sorry. I now take a dictionary and see that my translation was not wery good!
I translate that he want a string of let say five char. And then the array is make by five places, after the string was insert!
My mistake!

xxxrugby, your code gave me hives.

Originally Posted by xxxrugby

Or maybe you could do something with pointers!

Code:

#include <stdio.h>

int main()
{
    char *string;
    
    scanf("%s", &string);
    printf("%s", &string);
    
    return 0;
}

Since everyone likes a helpful reply, here's mine:

Code:

#include <stdio.h>

int main()
{
    char *string;  /* 1 */ 
    
    scanf("%s", &string);  /* 2 */ 
    printf("%s", &string);  /* 3 */ 
    
    return 0;
}

1) Here you declare a pointer. Where is it pointing now? In other words, your pointer has never been set to point at anything, so it's pointing at some random spot in memory. It has no memory allocated for anything other than the pointer itself.

2) Here you scan into where the pointer is pointing. Where is it pointing again? Oh, also, your scanf call is wrong. You don't need to give the address-of the pointer, because scanf expects a pointer, not the address of a pointer. So remove the & when you're using a pointer already.

3) Your printf call is also wrong. Even if this weren't a pointer, you wouldn't use the address-of operator. You never use the address-of operator with printf, unless you are in fact trying to display the address-of a variable.

Quzah.

Yes you need in scanf and printf &
If I remove them then I get

Code:

The instruction at "0x78...." referenced memory at "0x00....". This memory could not be "read"

>>Yes you need in scanf and printf &
>>If I remove them then I get
You missed the point. It's broken either way if the pointer doesn't point to anything except a random address.

Originally Posted by xxxrugby

Yes you need in scanf and printf

Please don't act like you know what you're talking about.

Quzah.

Originally Posted by xxxrugby

Yes you need in scanf and printf &
If I remove them then I get

This is likely because of the first point quzah made, i.e.:

Originally Posted by quzah

1) Here you declare a pointer. Where is it pointing now? In other words, your pointer has never been set to point at anything, so it's pointing at some random spot in memory. It has no memory allocated for anything other than the pointer itself.

You need to malloc some space for the string pointer to point to before you start saving data to that location.

Actually,

Code:

#include <stdio.h>

int main(void){

	char myString[] = "I can do this";

	printf("%s\n", myString);

	return 0;
}

This is what is known as a string literal, and the compiler automatically assigns the terminating character \0 to the array.

Originally Posted by nizbit

Just to clear anything up, I'm not trying to get input from the user. I have the string that should be put in the array. I know you can't put a string into an array with a simple assisgnment statement. My question is then, how do you do it?

strcpy