The condition is:
I want to input a line from standard input, and I don't know the size of it, maybe very long.
method like scanf, getsneed to know the max length you may input, so that your input size is less than your buffer size.
So Is there any good ways to handle it?
Answer must be only in C, not C++, so c++ string is not what I want. I want is C standard string, something like char* and end with '\0'.
The C standard doesn't define such a function, but POSIX does.
The getline function, documented here (or by typing man getline if you're on a UNIX-like system) does what you're asking for.
It may not be available on non-POSIX systems (such as MS Windows).
A small program that demonstrates its usage:
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char *line = NULL;
size_t n = 0;
ssize_t result = getline(&line, &n, stdin);
printf("result = %zd, n = %zu, line = \"%s\"\n", result, n, line);
free(line);
}
As with fgets, the '\n' newline character is left in the array.
One way is to run a loop with getchar and keep placing the characters into an array. Once the array is full, reallocate it to a larger size.
One of the method is using getchar() function we can get input in a character and transfer it to dynamicall created array. You can see that when it exceeds the default length set by us, we reallocated space for storing character
#include<stdio.h>
#include<stdlib.h>
void main(){
int size = 10;
char* str;
str = (char*) calloc(size,sizeof(char));
char c;
c = getchar();
int t = 0;
int cnt = 0;
int len;
while(c!='\n') {
if(cnt > size) {
str = (char*) realloc(str,2*cnt);
}
str[t] = c;
c = getchar();
t++;
cnt++;
}
str[t]='\0';
printf("The string is %s\n",str);
len = strlen(str);
printf("The size is %d",len);
}
There is an often overlooked conversion specification within scanf that will allocate memory sufficient to hold the string input regardless of length. Newer versions of scanf use m for this purpose. Older versions used a. For example:
#include <stdio.h>
#include <stdlib.h>
int main (void) {
char *str = NULL;
printf (" enter a string of any length, whitespace is OK: ");
scanf ("%m[^\n]%*c", &str);
printf ("\n str: %s\n\n", str);
if (str) free (str);
return 0;
}
Note: scanf requires a char ** pointer argument to receive the allocated string. Also note scanf does not include the '\n' in the stored string. Further note the %*c which receives and discards the '\n' character to prevent the newline from remaining in the input buffer. You may also precede the conversion specifier with whitespace to skip any/all whitepace that may exist in the input buffer.
Lastly Note: there are reports that not all implementations of scanf offer this feature. (which may also be confusion of the m/a change) Check your implementation.
use getchar, malloc and realloc for reading the unlimited input string
Declare String type, you can also use char *
// String type
typedef char *String;
I write this function for joining the char in the end of string
/**
* Join the Char into end of String
*
* @param string - String
* @param c - joined char
*/
void String_joinChar(String *string, const char c)
{
const size_t length = strlen(*string);
(*string) = (String)realloc((*string), sizeof(char) * (length + 2));
(*string)[length] = c;
(*string)[length + 1] = '\0';
}
This function for inputting string, which read the char from keyboard by using getchar and join it in the end of current string.
/**
* Input String
*
* @return Inputed String
*/
String String_input()
{
String string = (String)malloc(sizeof(char));
strcpy(string, "");
char cursor;
fflush(stdin);
while ((cursor = getchar()) != '\n' && cursor != EOF)
{
String_joinChar(&string, cursor);
}
return string;
}
Cause of using char *, malloc and realloc, we must free it
/**
* Destroy String
*
* @param string - Destroyed String
*/
void String_destroy(String string)
{
free(string);
}
And now we just use it !!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
String string = String_input();
printf("\n%s\n", string);
String_destroy(string);
return 0;
}
Hope useful to you!
