How to dynamically accept a string with spaces? I had tried gets(), but it is not working, and I don't want to take input from file. I want to accept a string without specifying string size.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
char *accept()
{
char *s;
s = (char *)malloc(sizeof(char));
gets(s);
return s;
}
int main()
{
char *str;
printf("\nEnter`enter code here` one string");
str = accept();
}
First of all, never use gets(). Use fgets() instead. gets() can't prevent if you overflow the buffer. It has been removed from the C standard C11.
Here,
s = (char *)malloc(sizeof(char));
you are allocating memory for just a single char which is enough only for the null character to terminate the string. You need to allocate more space to read a line.
s = malloc(256); //sizeof(char) is always 1. So you don't have to multiply it.
if( fgets(s, 256, stdin) == NULL ) { /* handle failure */
fgets() also reads the newline if the buffer has sufficient space. So you may want to remove if it's not desired.
You can remove it using strcspn():
s[strcspn(s,"\n")] = 0;
I want to accept string without specifying sting size.
Here's such a function:
char *accept()
{
// Initial size of the string.
int size = 256;
int c;
int index = 0;
char* s = malloc(size);
if ( s == NULL )
{
return NULL;
}
while ( (c = getchar()) != EOF && c != '\n')
{
// We need one character for the terminating null character
// Hence, if index is equal to (size-1), we need to resize the
// string.
if ( index == size - 1)
{
// Resize the string.
size = size * 2;
s = realloc(s, size);
if ( s == NULL )
{
return NULL;
}
}
s[index] = c;
++index;
}
// NUll terminate the string before returning it.
s[index] = '\0';
return s;
}
PS Don't ever use gets. See Why is the gets function so dangerous that it should not be used?. Use fgets if you need to read a line of text.
s[256] = '\0'; --> UB
realloc() to shrink the string to size. However, that's a veryminor space optimization, and definitely only relevant if the string is big (and I'm not wholly convinced the numbers given are big enough).Define the approximate size of user input by
#define LENGTH 30
char *accept(){
char *s = (char*)malloc(sizeof(char)*LENGTH);
scanf("%29[0-9a-zA-Z ]", s);
//scanf("%[^\n]%*s", s); //or this
return s;
}
Note that the 29 is the maximum number of characters which will be read, so s has to point to a buffer of size LENGTH-1 at least.
"%99[0-9a-zA-Z]" does not "accept a string with spaces". 2) (char*) not needed. 3) What does 99 do with LENGTH? 4) scanf() return value not checked. 5) sizeof(char) is always 1, so why is it there? Recommend fgets()
fgets() better but I just give an alternative.
callocif you want to read strings without specifying size. Do not usegets, instead usefgets.gets()is too dangerous to be used. Seriously look at POSIX functiongetline().accept()is not a good name to use — because there is a system functionaccept()used by networking code (servers, primarily).