Reading a string without wasting memory - c

Question

I have to read a string from stdin, allocating memory dinamically without wasting it.
I've done this, but i'm not convinced about it,because in this way i think i waste memory!
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
char *alloc_memory(int n)
{
char *p;
p=malloc(n*sizeof(char));
if(p==NULL)
{
fprintf(stderr,"Error in malloc\n");
exit(EXIT_FAILURE);
}
return p;
}
int main(int argc, char *argv[])
{
if(argc != 1)
{
fprintf(stderr,"Usage: %s \n",argv[0]);
return EXIT_FAILURE;
}
char string[64];
int lung;
char *p,*s,*w;
printf("Insert string: \n");
p=fgets(string,63,stdin);
if(p==NULL)
{
fprintf(stderr,"Error in fgets\n");
exit(EXIT_FAILURE);
}
printf("You've inserted: %s", string);
lung=strlen(p);
s = alloc_memory(lung+1);
w=strncpy(s,p,lung);
printf("Final string:%s", w);
return EXIT_SUCCESS;
}
any idea? Should i read one character at a time?

Answer 1

To have char str[64] (string is not a good name for a variable, it might lead to ambiguities) declared only temporarily just put it in a local context:
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
char * alloc_memory(size_t n)
{
char * p = malloc(n); /* * sizeof(char) is always 1 */
if (p == NULL)
{
fprintf(stderr, "Error in malloc() when trying to allocate %zu bytes.\n", n);
exit(EXIT_FAILURE);
}
memset(p, 0, n); /* Avoid having strncpy() choke .. later down in this example. */
return p;
}
int main(int argc, char * argv[])
{
if (argc != 1)
{
fprintf(stderr, "Usage: %s \n", argv[0]);
return EXIT_FAILURE;
}
{
char * w = NULL;
printf("Insert string: ");
{
char str[64]; /* here str is allocated */
char * p = fgets(str, 63, stdin);
if (p == NULL)
{
fprintf(stderr, "Error in fgets().\n");
exit(EXIT_FAILURE);
}
printf("You've inserted: '%s'\n", str);
{
size_t lung = strlen(p);
char * s = alloc_memory(lung + 1);
w = strncpy(s, p, lung);
}
} /* here "str" is deallocated */
printf("Final string: '%s'\n", w);
}
return EXIT_SUCCESS;
}