This question already has answers here:
Removing trailing newline character from fgets() input
(14 answers)
Closed 3 years ago.
I have created a little script in C which displays text in a Linux console, but I found one problem - the script adds a line break at the end of text. I have no idea why, normally I should get line break after /n.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char buf[1024];
char txt[100];
printf("Insert a text: ");
fgets(txt, 100, stdin);
snprintf(buf, sizeof(buf), "echo '%s'", txt);
system(buf);
}
The structure of the code has to stay the same. I need to use system function and snprintf().
I want to make a few things clear. Right when I'm running this script the output looks like this:
root#test:/home# ./app
Insert a text: Hello
Hello
root#test:/home#
How can I remove this newline after Hello?
fgets consumes the \n character in the buffer which you press after entering data. Just add
txt[strcspn(txt,"\n")] = 0;
just after the fgets to replace the \n character at the end of txt with a NUL-terminator. The strcspn function, in your case, counts the number of characters in txt until a \n character. If no newline character is found, then it returns the length of txt(strlen(txt)).
BTW, you need to include string.h if you want to use this function:
#include <string.h>
The problem you're facing is related with how fgets() behave. As per the man page
char *fgets(char *s, int size, FILE *stream);
fgets() reads in at most one less than size characters from stream and stores them into the buffer pointed to by s. Reading stops after an EOF or a newline. If a newline is read, it is stored into the buffer.
So, it reads and stores the trailing \n (newline) generated by pressing ENTER after the input into txt. If you don't want that \n to be present in txt, you need to remove that from the input buffer manually before printing the contents of txt.
You can use strchr() or strcspn() to find out the possible \n and replace that with \0 (null) to solve your issue.
Related
This question already has answers here:
Why does scanf ask twice for input when there's a newline at the end of the format string?
(7 answers)
Closed 5 years ago.
I want to read a single character from the console, but when I do, the program reads characters yet and I must write another character to save the first and finish its execution.
Code:
#include <windows.h>
#include <stdlib.h>
#include <stdio.h>
char peps;
int main(int argc, char const *argv[]) {
printf("write a character:\n");
scanf(" %c\n", &peps);
printf("%c\n", peps);
return 0;
}
Can anyone explain why it does that and how to correct this error?
If you remove the \n from the scanf(), it should work as you want.
ie, do
scanf(" %c", &peps);
instead of
scanf(" %c\n", &peps);
This is because the \n in the scanf() format string is telling the computer to read and ignore all white spaces (including \n) after reading a character.
So all white spaces including the newlines given by typing the enter key, would be ignored. This will stop only when a non-white space character is encountered which won't be read and would thus remain in the input buffer.
So, in your case, a character would first be read and it would wait for a non-white space character before executing the printf() following the scanf(). The non-white space character would remain in the input buffer and was not read and is hence not printed at once. It would be read only upon reading from the stdin again.
If you want to explore this further, consider placing that scanf() and printf() in a loop and examine the output.
Note that replacing that \n with a space would have the same effect.
ie,
scanf(" %c\n", &peps);
and
scanf(" %c ", &peps);
would have the same behavior.
What's the behavior of scanf when the format string ends with a newline?
Behaviour of scanf when newline is in the format string
This question already has answers here:
How do you allow spaces to be entered using scanf?
(11 answers)
Closed 5 years ago.
For Example if I
Input: 'stephen 8108' it outputs
'stephen'
Instead of outputing 'stephen 8108'.
Can someone help me out!
I want the full string to appear in the output.
It reads the string only till the first white space.
Even if i remove the for loop condition it doesn't seem to work it still reads only till the first white space.
#include<fcntl.h>
#include<stdio.h>
#include <unistd.h>
void main()
{
char a[100];
int i,f2,f3,f4;
f2 = creat("a.txt",0666);
f3 = open("a.txt",O_RDWR);
printf("Enter your name & Roll-no\n");
scanf("%s",a);
for(i=0;a[i] != '\0';i++);
write(f3,a,i);
close(f3);
}
This is intended sprintf functionality.
Cite: http://www.cplusplus.com/reference/cstdio/scanf/
Any number of non-whitespace characters, stopping at the first whitespace character found.
One option is to use the negated character matching (quoted from link above):
[^characters] Negated scanset
Any number of characters none of them specified as characters between the brackets.
For example, to match everything excluding a newline:
scanf("%[^\n]", a);
(Full working example below - although please don't take this as necessarily a full and perfect example of reading user input in C++...)
#include<fcntl.h>
#include <stdio.h>
#include <unistd.h>
#include <string.h>
int main()
{
char a[100];
int fp;
fp = open("a.txt", O_CREAT|O_WRONLY|O_TRUNC);
printf("Enter your name & Roll-no\n");
scanf("%[^\n]", a);
write(fp, a, strlen(a));
close(fp);
}
However: I would really encourage you to read the extensive warnings about buffer overflows: https://stackoverflow.com/a/1248017/817132
In short, make sure you don't allow the user to write beyond your (currently 100 character long) memory allocation.
At the current state of your code, the for loop doens't have a body { ... } , so the write and the close operations would be executed only one time.
Also if you want scanf to read a string with spaces you can use %[0-9a-zA-Z ] instead of %s
Regarding the input, there are at least two problems:
The s conversion of scanf() parses until it finds whitespace, it's documented that way.
Without a field width, scanf() will continue parsing when it doesn't find whitespace, overflowing your buffer -> undefined behavior.
The quick fix is to replace scanf("%s",a); with scanf("%99[^\n]",a);. But scanf() is definitely not the best tool to read input, it is for parsing. You seem to just want to read a whole line of input and there is already a function for that: fgets(). Use it in your example like this (include string.h if you want to use this method of stripping the newline character):
fgets(a, 100, stdin);
a[strcspn(a, "\n")] = 0; // remove the newline character if it was read by fgets
This question already has answers here:
Removing trailing newline character from fgets() input
(14 answers)
Closed 6 years ago.
I have a problem. I've tried to see the length of some string after using fgets function. If I enter string under the number of letter which can be in the string (like: the maximum letters in string is 9 and I enter 4 letters), I get length of the string+1. why?
Here's my code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char name[10]={0};
printf("enter your name\n");
fgets(name, 10, stdin);
printf("your name is %s and it is %d letters\n", name, strlen(name)); // length problem
return 0;
}
From fgets manual page (https://linux.die.net/man/3/fgets):
fgets() reads in at most one less than size characters from stream and
stores them into the buffer pointed to by s. Reading stops after an
EOF or a newline. If a newline is read, it is stored into the buffer.
A terminating null byte (aq\0aq) is stored after the last character in
the buffer.
So it adds '\n' after your 4 letters, returning string_length+1.
From Removing trailing newline character from fgets() input you can add #Tim Čas solution to your code.
The line is still read with the fgets() function and after we remove the newline character.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char name[10] = { 0 };
printf("enter your name\n");
fgets(name, 10, stdin);
printf("your name is %s and it is %d letters\n", name, strlen(name)); // length problem
name[strcspn(name, "\n")] = 0;
printf("NEW - your name is %s and it is %d letters\n", name, strlen(name));
return 0;
}
That outputs:
enter your name
Andy
your name is Andy
and it is 5 letters
NEW - your name is Andy and it is 4 letters
Press any key to continue . . .
Because the end of line character '\n' is included in the string copied into name by fgets() .
If character array has enough space then the standard function fgets also includes the new line character in the array that usually corresponds to the entered key Enter.
You can remove this redundant new line character the following way
name[strcspn( name, "\n" )] = '\0';
after that you will get the expected result of applying function strlen.
As is written in man fgets,
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.
Since you are reading from stdin, fgets(name, 10, stdin) reads at most 9 characters from stdin buffer and appends \0 to the end. It just happens that the new line character \n produced when user hit enter is in the buffer too.
As a sidenote, it is customary (and a good practice) to use sizeof() when specifying size of the array passed to fgets.
fgets(name, (int) sizeof(name), stdin);
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;
Because if there is a '\n' in the read text it will be taken by fgets(), the following was extracted from the 1570 draft §7.21.7.2 ¶ 2
The fgets function reads at most one less than the number of characters specified by n
from the stream pointed to by stream into the array pointed to by s. No additional
characters are read after a new-line character (which is retained) or after end-of-file. A
null character is written immediately after the last character read into the array.
I highlighted by making bold the part which says that the '\n' is kept by fgets().
I'm having a weird problem
i'm trying to read a string from a console with scanf()
like this
scanf("%[^\n]",string1);
but it doesnt read anything. it just skips the entire scanf.
I'm trying it in gcc compiler
Trying to use scanf to read strings with spaces can bring unnecessary problems of buffer overflow and stray newlines staying in the input buffer to be read later. gets() is often suggested as a solution to this, however,
From the manpage:
Never use gets(). Because it is
impossible to tell without knowing the
data in advance how many characters
gets() will read, and because gets()
will continue to store characters past
the end of the buffer, it is extremely
dangerous to use. It has been used to
break computer security. Use fgets()
instead.
So instead of using gets, use fgets with the STDIN stream to read strings from the keyboard
That should work fine, so something else is going wrong. As hobbs suggests, you might have a newline on the input, in which case this won't match anything. It also won't consume a newline, so if you do this in a loop, the first call will get up to the newline and then the next call will get nothing. If you want to read the newline, you need another call, or use a space in the format string to skip whitespace. Its also a good idea to check the return value of scanf to see if it actually matched any format specifiers.
Also, you probably want to specify a maximum length in order to avoid overflowing the buffer. So you want something like:
char buffer[100];
if (scanf(" %99[^\n]", buffer) == 1) {
/* read something into buffer */
This will skip (ignore) any blank lines and whitespace on the beginning of a line and read up to 99 characters of input up to and not including a newline. Trailing or embedded whitespace will not be skipped, only leading whitespace.
I'll bet your scanf call is inside a loop. I'll bet it works the first time you call it. I'll bet it only fails on the second and later times.
The first time, it will read until it reaches a newline character. The newline character will remain unread. (Odds are that the library internally does read it and calls ungetc to unread it, but that doesn't matter, because from your program's point of view the newline is unread.)
The second time, it will read until it reaches a newline character. That newline character is still waiting at the front of the line and scanf will read all 0 of the characters that are waiting ahead of it.
The third time ... the same.
You probably want this:
if (scanf("%99[^\n]%*c", buffer) == 1) {
Edit: I accidentally copied and pasted from another answer instead of from the question, before inserting the %*c as intended. This resulting line of code will behave strangely if you have a line of input longer than 100 bytes, because the %*c will eat an ordinary byte instead of the newline.
However, notice how dangerous it would be to do this:
scanf("%[^n]%*c", string1);
because there, if you have a line of input longer than your buffer, the input will walk all over your other variables and stack and everything. This is called buffer overflow (even if the overflow goes onto the stack).
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char *text(int n);
int main()
{
char str[10];
printf("enter username : ");
scanf(text(9),str);
printf("username = %s",str);
return 0;
}
char *text(int n)
{
fflush(stdin);fflush(stdout);
char str[50]="%",buf[50],st2[10]="[^\n]s";
char *s;itoa(n,buf,10);
// n == -1 no buffer protection
if(n != -1) strcat(str,buf);
strcat(str,st2);s=strdup(str);
fflush(stdin);fflush(stdout);
return s;
}