Reading string from input with space character? [d

Question

This question already has an answer here:

• How do you allow spaces to be entered using scanf? 12 answers

I'm using Ubuntu and I'm also using Geany and CodeBlock as my IDE. What I'm trying to do is reading a string (like "Barack Obama") and put it in a variable:

#include <stdio.h>

int main(void)
{
    char name[100];

    printf("Enter your name: ");
    scanf("%s", name);
    printf("Your Name is: %s", name);

    return 0;
}

Output:

Enter your name: Barack Obama
Your Name is: Barack

How can I make the program read the whole name?

Answer 1

#include <stdio.h>
// read a line into str, return length
int read_line(char str[]) {
int c, i=0;
c = getchar();
while (c != '\n' && c != EOF) { 
   str[i] = c;
   c = getchar();
   i++;
}
str[i] = '\0';
return i;
}

Answer 2

Here is an example of how you can get input containing spaces by using the fgets function.

#include <stdio.h>

int main()
{
    char name[100];
    printf("Enter your name: ");
    fgets(name, 100, stdin); 
    printf("Your Name is: %s", name);
    return 0;
}

Answer 3

"%s" will read the input until whitespace is reached.

gets might be a good place to start if you want to read a line (i.e. all characters including whitespace until a newline character is reached).

Answer 4

While the above mentioned methods do work, but each one has it's own kind of problems.

You can use getline() or getdelim(), if you are using posix supported platform. If you are using windows and minigw as your compiler, then it should be available.

getline() is defined as :

ssize_t getline(char **lineptr, size_t *n, FILE *stream);

In order to take input, first you need to create a pointer to char type.

#include <stdio.h>
#include<stdlib.h>

// s is a pointer to char type.
char *s;
// size is of size_t type, this number varies based on your guess of 
// how long the input is, even if the number is small, it isn't going 
// to be a problem
size_t size = 10;

int main(){
// allocate s with the necessary memory needed, +1 is added 
// as its input also contains, /n character at the end.
    s = (char *)malloc(size+1);
    getline(&s,&size,stdin);
    printf("%s",s);
    return 0;
}

Sample Input:Hello world to the world!

Output:Hello world to the world!\n

One thing to notice here is, even though allocated memory for s is 11 bytes, where as input size is 26 bytes, getline reallocates s using realloc().

So it doesn't matter how long your input is.

size is updated with no.of bytes read, as per above sample input size will be 27.

getline() also considers \n as input.So your 's' will hold '\n' at the end.

There is also more generic version of getline(), which is getdelim(), which takes one more extra argument, that is delimiter.

getdelim() is defined as:

ssize_t getdelim(char **lineptr, size_t *n, int delim, FILE *stream);

Linux man page

Answer 5

NOTE: When using fgets(), the last character in the array will be '\n' at times when you use fgets() for small inputs in CLI (command line interpreter) , as you end the string with 'Enter'. So when you print the string the compiler will always go to the next line when printing the string. If you want the input string to have null terminated string like behavior, use this simple hack.

#include<stdio.h>
int main()
{
 int i,size;
 char a[100];
 fgets(a,100,stdin);;
 size = strlen(a);
 a[size-1]='\0';

return 0;
}

Update: Updated with help from other users.

Answer 6

"Barack Obama" has a space between 'Barack' and 'Obama'. To accommodate that, use this code;

#include <stdio.h>
int main()
{
    printf("Enter your name\n");
   char a[80];
   gets(a);
   printf("Your name is %s\n", a);
   return 0;
}