I am new to pointers, please let me know how can i print the entered character.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char *ptr;
ptr = malloc(32 * sizeof(char));
*ptr = 'h';
ptr++;
*ptr = 'e';
ptr++;
*ptr = 'l';
ptr++;
*ptr = 'l';
ptr++;
*ptr = 'o';
ptr++;
*ptr = '\n';
printf("value entered is %s\n", ptr);
return 0;
}
I want to print hello
You forgot the null-terminator. Add this:
ptr++;
*ptr = '\0';
Also, the pointer is now pointing to the null-terminator (or previously the newline character). You have to set it back to point to the 'h' again:
ptr -= 6;
And when you're done, you should free the memory:
free(ptr);
You should fix your code like this, with a temporary pointer:
#include <stdio.h>
#include <stdlib.h>
int main (void)
{
char* ptr;
ptr = malloc(32 * sizeof(char));
if(ptr == NULL)
{
puts("Allocation failed");
return EXIT_FAILURE;
}
char* tmp = ptr;
*tmp = 'h';
tmp++;
*tmp = 'e';
tmp++;
*tmp = 'l';
tmp++;
*tmp = 'l';
tmp++;
*tmp = 'o';
tmp++;
*tmp = '\0'; // NOTE: null termination not \n
printf("value entered is %s\n", ptr);
free(ptr);
return 0;
}
A proper version without messy pointer arithmetic looks like this:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main (void)
{
char* ptr;
ptr = malloc(32 * sizeof(char));
if(ptr == NULL)
{
puts("Allocation failed");
return EXIT_FAILURE;
}
strcpy(ptr, "hello");
printf("value entered is %s\n", ptr);
free(ptr);
return 0;
}
free(ptr); ptr = malloc(...) that free() up old data and can be called at any point, regardless of whether the pointer is pointing to allocated memory or not. This is taking advantage of free(NULL) being a no-op. Anyway, this isn't the place for this discussion.You can, rather using the malloc() function, use the calloc() function, which achieves the same goal of malloc(), but fills the memory with '\0'. This makes it easier to play with non-fixed length strings.
You can find the documentation of this function here.
Here is the code I've made :
#include <stdio.h>
#include <stdlib.h>
int main()
{
char *ptr;
ptr = calloc(32,sizeof(char));
*ptr = 'h';
ptr++;
*ptr = 'e';
ptr++;
*ptr = 'l';
ptr++;
*ptr = 'l';
ptr++;
*ptr = 'o';
ptr++;
*ptr = '\0'; //It should be null terminated
ptr -= 5;
printf("value entered is %s\n", ptr);
free(ptr);
return 0;
}
ptris no longer pointing to the original location; Andcharstrings in C are really called null-terminated byte strings. This null-terminator is incredibly important for all string handling (including usingprintfwith the%sformat) to know where the string ends.ptrright after malloc, then do all arithmetic on the temporary pointer.