simple pointer

[cheeta]

#include<stdio.h>
#include<stdlib.h>
int main(void)
{

char sentence[256];
printf("\nEnter the sentence : ");
fgets(sentence,255,stdin);
fputs(sentence,stdout);
}

this code works fine...

but when i tried with pointers,say

 
char *sentence;
printf("\nEnter the sentence : ");
fgets(sentence,255,stdin);
fputs(sentence,stdout);

it isnt working...don think my question as silly...i am new to c and pointers...help me plz..

[claudiu]

#include<stdio.h>
#include<stdlib.h>
int main(void)
{

char sentence[256];
printf("\nEnter the sentence : ");
fgets(sentence,255,stdin);
fputs(sentence,stdout);
}

this code works fine...

but when i tried with pointers,say

 
char *sentence;
printf("\nEnter the sentence : ");
fgets(sentence,255,stdin);
fputs(sentence,stdout);

it isnt working...don think my question as silly...i am new to c and pointers...help me plz..

A pointer is an address. If you don't have the pointer pointing to a valid address you are going to get a segfault. You either make the pointer point to a statically allocated space or to a dynamically allocated space.

Like so:

/*find a chunk of free memory on the heap and make this pointer point to it*/
char *ptr = malloc(sizeof(char) * 101); 

fgets(ptr,100,stdin);

[cas]

Just for reference, you do not need to subtract one from your array size when you pass it to fgets(). That function understands the need for null termination, so you can just tell it the whole size of your array.

[claudiu]

Cool, thanks for clarifying that cas, I didn't remember if that was how fgets worked or not. Hopefully, that will clear the problem for the OP.