Possible Duplicate:
Segmentation Fault - C
#include<stdio.h>
#include<string.h>
int main()
{
char *p;
printf("enter some thing:");
gets(p);
printf("you have typed:%s\n",p);
}
Why doesn't this program work? i can't use pointer as a string.
Output is:
enter some thing:raihan
Segmentation fault (core dumped)
I get this error every time when I use a char pointer. How can I solve this problem? I am using code-blocks on Linux mint13 KDE.
You have not allocated memory. You just declared a pointer, p, but didn't make it point at anything. That explains the segmentation fault. You will need to allocate memory for your buffer.
What's more, gets does not allow you to specify how big the buffer is. So you are at risk of running over the end of the buffer. So use fgets instead.
int main(void)
{
char buffer[1024];//allocates a buffer to receive the input
printf("enter some thing: ");
fgets(buffer, sizeof(buffer), stdin);
printf("you have typed: %s\n", buffer);
return 0;
}
I also corrected your declaration of main and made sure that it returns a value.
You haven't allocated any memory for p. Also, use fgets instead of gets which may overflow the input buffer.
char *p;
printf("enter some thing:");
gets(p);
Wrong. Gets() tries to fill in the array pointed to by the supplied pointer - and it segfaults, because that pointer hasn't been initialized, so it might (and does) point to some garbage/invalid memory location. Use
char p[256];
or something like this instead - you still have to worry about a buffer overflow in if the user enters a string longer than 255 characters. You can solve that one using
fgets(p, sizeof(p), stdin);
Your pointer is declared but you have not initialised it and so its value will be some arbitrary memory location that you may not have access to write to. Thus anytime you read or write to this you run the risk of segfault. Allocate some heap memory for the pointer using a call to malloc then you wont get segfaults when writing to it.
You have just defined a pointer - no memory for the characters have been allocated!
Use either an array or malloc.
A pointer is just a memory address. It says "you have some data here". But it doesn't actually reserve that data.
In your case the problem was two-fold. The pointer didn't point to valid memory and you never even set it to anything (so it pointed to somewhere random).
You can fix this in different ways. The easiest is to just use an array (it's sort of implicitly a pointer):
char something[100];
printf("enter some thing:");
gets(something);
That gives you 100 chars on the stack. You can also point to it if you want, but in this case it's a bit redundant:
char *p = something;
The other way is dynamic allocation, where you ask the operating system at runtime to give you some number of bytes. This way you have to give it back when you're finished using it.
char *something = (char*)malloc( 100 * sizeof(char) ); // Ask for 100 chars
printf("enter some thing:");
gets(something);
free(something); // Do this when you don't need that memory anymore.
PS: Remember when you have strings, you always need one extra byte than the number of characters you intend to store. That byte is for the string terminator, and the value of it is 0.
