I have taken a string from the keyboard using the fgets() function. However, when I print the string using printf(), the cursor goes to a new line.
Below is the code.
#include<stdio.h>
int main()
{
char name[25];
printf("Enter your name: ");
fgets(name, 24, stdin);
printf("%s",name);
return 0;
}
And below is the output.
-bash-4.1$ ./a.out
Enter your name: NJACK1 HERO
NJACK1 HERO
-bash-4.1$
Why is the cursor going to the next line even though I have not added a \n in the printf()?
However, I have noticed that if I read a string using scanf(), and then print it using printf() (without using \n), the cursor does not go to next line.
Does fgets() append a \n in the string ? If it does, will it append \0 first then \n, or \n first and then \0?
The reason printf is outputting a newline is that you have one in your string.
fgets is not "adding" a newline --- it is simply reading it from the input as well. Reading for fgets stops just after the newline (if any).
Excerpt from the manpage, emphasis mine:
The fgets() function reads at most one less than the number of characters specified by size from the given stream and stores them in the string str. Reading stops when a newline character is found, at end-of-file or error. The newline, if any, is retained. If any characters are read and there is no error, a `\0' character is appended to end the string.
An easy way to check if there's a newline is to use the help of one of my favorite little-known functions --- strcspn():
size_t newline_pos = strcspn(name, "\r\n");
if(name[newline_pos])
{
/* we had a newline, so name is complete; do whatever you want here */
//...
/* if this is the only thing you do
you do *not* need the `if` statement above (just this line) */
name[newline_pos] = 0;
}
else
{
/* `name` was truncated (the line was longer than 24 characters) */
}
Or, as an one-liner:
// WARNING: This means you have no way of knowing if the name was truncated!
name[strcspn(name, "\r\n")] = 0;