Thread: Printing a char.

Originally Posted by itsme86

Well, you're lucky it works because str is pointing to a random memory address. You haven't allocated any memory for it. Try declaring str as an array instead of a pointer.

While I was reading this.. I figured because I'm not allocating any memory on it, I don't have to put a pointer on it. So without the pointer it also works .

My code:

Code:

#include <stdio.h>

int main(void)
{
      char str;
      printf("Name: ");
      scanf("%s", &str);
      printf("Hello %s\n", &str); 
      system("pause");
      return 0;
}

Jst.

I've never worked with the maloc function before, but I guess it's time that I should...

Jst.

While I was reading this.. I figured because I'm not allocating any memory on it, I don't have to put a pointer on it. So without the pointer it also works .

There is a fundamental mistake with the code you have in this post. You are expecting user to enter the name as string. However, you are just reading the entire string into a single character. It will just store the first character of your name. Not the entire name.

Read my last post. It tell you how to handle this problem with pointers.

Originally Posted by nkhambal

However, you are just reading the entire string into a single character. It will just store the first character of your name. Not the entire name.

Well, the last piece of code I posted did actually print my whole name or any other text I used for input.


I don't know allot about these pointers yet, but when I declare the "%s" at a printf or scanf line to get a string, I don't have to dereference a variable I just have to retrieve the original value with the & pointer and then the printf function will print the string. I could be totally wrong, it's just a small theory .

Jst.

I guess you were just lucky.

anyway, in C, strings are nothing but array of chars. i.e.

Code:

char name[10];

when passing this array to printf() and scanf() requires you to pass the base address of the array without & operator. i.e.

Code:

scanf ("%s",name);
printf("Name: %s\n", name);

Another important thing to remember when dealing with strings in C is that, it is MUST to terminate all the strings with NULL character or ' \0 '. If this is not done then your prgram may crash at runtime, with string operating functions trying to read or write passed the mamory location allocated to it. When you allocate the memory to store a string, its always adviced to allocate an extra byte than the required size. Some string operting functions such as strcpy(), strdup() and strcat() take care of terminating the new string with null character. Nevertheless you need to take care of allocating an extra byte for it.

Originally Posted by Micko

Well, the arguments you pass to scanf must always be pointers, since str is a pointer there is no need to use &.
Just a little correction

- Micko

But Jst, said that there's an error with the program, before the '&' was used. So, wouldn't it be neccessary to include a '&'? Anyway, I don't have a compiler to verify that now, so if the '&' makes no differences, what could be wrong with the original code(below), then?

Code:

#include <stdio.h>

main()
{
      char *str;
      printf("Name: ");
      scanf("%s", str);
      printf("Hello %s", str); 
      system("pause");
      return 0;
}

How would you create it?

Tnx
Jst.

Originally Posted by Jst

How would you create it?

Tnx
Jst.

well you could do this, "char name[100];" instead of "char *name;" or if you want you can leave it as a pointer and allocate memory to it like this "name = malloc(100);". If you choose to allocate the memory, remember to free it again like this "free(name)". Of course these example are if you want it to be 100 chars long.

K, tnx.. Well I sure have allot to learn about pointers, although I do know allot more about it before I started this topic.

Jst.

Originally Posted by nkhambal

That is because, array and pointers are treated differently. When arrays are declared with their size, a memory of that size if pre allocated at run time. Pointers on the other hand do not have any memory allocated at run time. They point to no where unless they are told to point at some variables address or some memory is allocated using malloc() for them to point at.

The reason, jst you got the winxp error is because of core dump. When you read the name into str variable , it attempted writing into some other programs memory space, a space not owned by you. This is not allowed. Hence your program crashed. This is a programmer mistake. Compiler will not give any warning as there is no problem with the syntax anywhere. Its programmer's responsibilty to malloc the memory to pointer before storing something in it. Following should be the modified version of your program.

Code:

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

int main(void)
{
      char *str;
      printf("Name: ");
       str=(char *) malloc (10 * sizeof(char)); /*allocated 10 bytes of memory to store the name . This step is a MUST.*/
      scanf("%s", &str);
      printf("Hello %s\n", &str); 
      system("pause");
      return 0;
}

Why are you using the 'address of operator' on str in the scanf function?. It should be scanf("%s", str); ...in your example, you're giving the scanf call a pointer to a pointer (scanf("%s", &str)) . That's not right. str already exposes the address it points to so there's no need for &str.

xeddiex.