Storing user inputs without declaring an arbitrarily large array. - c

Question

I am taking character input and storing it without declaring an arbitrarily large array. Problem is that the code does not print the values stored (although it perfectly prints the number of elements that I enter). The working principle is: in the first for loop execution "b" is created and "c" is copied to it(c right now contains something arbitrary),then user overwrites whatever is there in "b" then the updated "b" is copied to "c". In the second and following loop executions "c" is basically the old "b" and "b" is constantly updated by copying "c" to it and entering new element at the end.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
char e = 'a';
char *b,*c = &e;
printf("start entering the characters and enter Z to terminate:\n");
char d;
int i,m;
for(i=0;(d=getchar()) != 'Z';i++)
{
b=malloc(sizeof(char)*(i+1));
strcpy(b,c);
scanf("%c",b+i);
c=malloc(sizeof(char)*(i+1));
strcpy(c,b);
}
printf("-----------------------------------------------------------------\n");
int q=strlen(b);
printf("%d\n",q);
//printf("%s\n",b);
for(m=0;m<q;m++)
printf("%c",b[m]);
return 0;
}

Answer 1

I'm not sure why the question code uses both 'getchar()' as well as 'scanf()'; perhaps I am missing something?
And, as mentioned by 'BLUEPIXY', realloc() is better.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
char *b = NULL; /* This will be the reference to the self-growing array. */
size_t bLen = 0; /* This is the length of the string in 'b' (not counting the string termination character.) */
for(;;)
{
char *x; /* Used to safely grow the array 'b' larger. */
int d; /* Use an 'int'. 'getchar()' returns an 'int', (not a 'char'). */
/* Get a character from stdin. */
d=getchar();
if('Z' == d)
break;
/* Safely grow the array 'b' large enough to hold the one more character. */
x=realloc(b, (bLen+1) * sizeof(*b));
if(NULL == x)
{
fprintf(stderr, "realloc() failed.\n");
exit(1);
}
b=x;
/* Store the character in the array and re-terminate the string. */
b[bLen++] = d;
b[bLen] = '\0';
}
printf("-----------------------------------------------------------------\n");
printf("%s\n",b);
if(b)
free(b);
return 0;
}

Answer 2

Some impressions:
Giving all of your variables single-character names makes it a lot harder than it should be for someone reading this code to figure out what you're trying to do.
When you're done with a block of memory that you have allocated using malloc(), you need to release it using free().
strcpy() is for use with null-terminated strings, so I don't think it's going to do what you expect here. Consider memcpy() instead. Your use of strlen() has the same problem. You shouldn't need to replace that with anything, since i should already give you the character count.
Is your scanf() statement trying to copy a character into your buffer? Just use a simple assignment, like b[i] = d.
If Z is the user's first keypress, b will never be initialized, and Bad Things will happen when you try to access it in the code following the loop.
Reallocating memory during every iteration of the loop is very inefficient. Instead, consider allocating a small amount of space at the outset—say, 20 characters' worth. Then in the body of your loop, if your buffer has enough space to accommodate a new character, all you need to do is copy that one character. If your buffer lacks space, then reallocate a larger block of memory, but not just for 1 extra character; reserve space for another 20 (or whatever).
Each reallocation should only require one call to malloc() (or, even better, realloc() as suggested by BLUEPIXY). I'm not sure why you're using two.
Null-terminate the input at the end so you can treat it like a string when you want to display it.
I'm happy to help if you have specific questions about any of this.