What is the most accurate way to read strings from the keyboard in C, when the string contains spaces in between words? When I use scanf for that purpose then it doesn't read a string with spaces.The second option is to use gets but it is supposed to be harmful(I also want to know why?).Another thing is that I don't want to use any file handling concept like fgets.
These are 2 ways to read strings containing spaces that don't use gets or fgets
You can use getline (POSIX 2008 may not exist in your system) that conveniently manages allocation of the buffer with adequate size to capture the whole line.
char *line = NULL;
size_t bufsize = 0;
size_t n_read; // number of characters read including delimiter
while ((n_read = getline(&line, &bufsize, stdin)) > 1 && line != NULL) {
// do something with line
}
If you absolutely want scanf, in this example it reads to the end of line unless the line has more than the specified number of chars minus 1 for the delimiter. In the later case the line is truncated and you'll get the remaining chars in the next scanf invocation.
char line[1024];
while (scanf("%1023[^\n]\n", line) == 1) {
// do something with line
}
I should also point out that when you read strings from the keyboard with scanf for example, you are actually reading from a file with file pointer stdin. So you can't really avoid "any file handling concept"
#user3623265,
Please find a sample program which Uses fgets to read string from standard input.
Please refer some sample C documents as to how fgets can be used to get strings from a keyboard and what is the purpose of stdin.
#include <stdio.h>
#include <string.h>
int main(void)
{
char str[80];
int i;
printf("Enter a string: ");
fgets(str, sizeof(str), stdin);
i = strlen(str) - 1;
if (str[i] == '\n')
str[i] = '\0';
printf("This is your string: %s", str);
return 0;
}
There is a third option, you can read the raw data from stdin with the read() call:
#include <unistd.h>
int main(void) {
char buf[1024];
ssize_t n_bytes_read;
n_bytes_read = read(STDIN_FILENO, buf, sizeof(buf) - 1);
if (n_bytes_read < 0) {
// error occured
}
buf[n_bytes_read] = '\0'; // terminte string
printf("\'%s\'", buf);
return 0;
}
Please not that every input is copied raw to buf including the trailing return. That is, if you enter Hello World you will get
'Hello World
'
as output. Try online.
If you insist on not having a FILE * in scope, use getchar().
char buff[1024];
int ch;
int i = 0;
while( (ch = getchar()) != '\n' )
if(i < 1023)
buff[i++] = ch;
buff[i] = 0;
/* now move string into a smaller buffer */
Generally however it's accepted that stdout and stdin and FILE * are available. Your requirement is a bit odd and, since you are obviously not an advanced C programmer who has an unusual need to suppress the FILE * symbol, I suspect your understanding of C IO is shaky.
Related
I am a novice in C programming. Suppose I want to read multiple lines of string from stdin. How can I keep reading until a line only containing EOL?
example of input
1+2\n
1+2+3\n
1+2+3+4\n
\n (stop at this line)
It seems that when I hit enter(EOL) directly, scanf won't execute until something other than just EOL has been entered. How can I solve that problem?
I'll be really grateful if someone can help me with this. Thank you.
If you want to learn C, you should avoid scanf. The only use cases where scanf actually makes sense are in problems for which C is the wrong language. Time spent learning the foibles of scanf is not well spent, and it doesn't really teach you much about C. For something like this, just read one character at a time and stop when you see two consecutive newlines. Something like:
#include <stdio.h>
int
main(void)
{
char buf[1024];
int c;
char *s = buf;
while( (c = fgetc(stdin)) != EOF && s < buf + sizeof buf - 1 ){
if( c == '\n' && s > buf && s[-1] == '\n' ){
ungetc(c, stdin);
break;
}
*s++ = c;
}
*s = '\0';
printf("string entered: %s", buf);
return 0;
}
to read multiple lines of string from stdin. How can I keep reading until a line only containing EOL?
Keep track of when reading the beginning of the line. If a '\n' is read at the beginning, stop
getchar() approach:
bool beginning = true;
int ch;
while ((ch = getchar()) != EOF) {
if (beginning) {
if (ch == '\n') break;
}
// Do what ever you want with `ch`
beginning = ch == '\n';
}
fgets() approach - needs more code to handle lines longer than N
#define N 1024
char buf[N+1];
while (fgets(buf, sizeof buf, stdin) && buf[0] != '\n') {
; // Do something with buf
}
If you need to read one character at a time then you can with either getchar or fgetc depending upon whether or not you're reading from stdin or some other stream.
But you said you were reading strings, so I'm assuming fgets is more appropriate.
There are primarily two considerations:
maximum line length
whether or not to handle Windows versus non-Windows line endings
Even if you are a beginner--and I won't go into #2 here--you should know you can defend against it. I will at least say that if you compile on one platform and read from stdin from a redirected file from another platform, then you might have to write a defense.
#include <stdio.h>
#include <string.h>
#include <errno.h>
int main (int argc, char *argv[]) {
char buf[32]; // relatively small buf makes testing easier
int lineContinuation = 0;
// If no characters are read, then fgets returns NULL.
while (fgets(buf, sizeof(buf), stdin) != NULL) {
int l = strlen(buf); // No newline in buf if line len + newline exceeds sizeof(buf)
if (buf[l-1] == '\n') {
if (l == 1 && !lineContinuation) {
break; // errno should indicate no error.
}
printf("send line ending (len=%d) to the parser\n", l);
lineContinuation = 0;
} else {
lineContinuation = 1;
printf("send line part (len=%d) to the parser\n", l);
}
}
printf("check errno (%d) if you must handle unexpected end of input use cases\n", errno);
}
I need to read many strings with a variable length from stdin up to EOF.
The strings are made of consecutive chars, not separated by blanks and there is not a maximum length for the strings.
Using:
char st[101];
while(scanf("%100s",st) != EOF){ //divide the input into parts of 100 chars
int i;
for(i=0; i<strlen(st);i++){
printf("%c",st[i]);
}
}
I can not tell if the current string is part of a previous one or a new one because it is printed all in a single line (and I've no idea where to print "\n").
Sample input:
aaaaaaaaa[...]aaaaa
bbbbbbbbb[...]bbbbb
output:
aaaa[...]aabbbbbb[...]bbbb
Any solution?
P.S. Using getchar() I'm not able to handle the EOF.
Here's how you handle EOF with getchar:
int c;
while ((c = getchar()) != EOF) {
putchar(c);
}
This code will replicate its input on stdout.
If you want to deal with whole lines at a time, the easiest way is to use getline if you have it (it's POSIX, not standard C):
char *line = NULL;
size_t size = 0;
ssize_t nread;
while ((nread = getline(&line, &size, stdin)) != -1) {
// nread characters were read
do_stuff_with(line);
}
free(line);
If you don't have getline, you get to write your own (using getchar and realloc).
I am having the absolute craziest time getting full line input to work. I will explain my problem. I need to get a full line of input, including a space, from the user entered at the keyboard. Simple right? Wrong!
MY GOAL
Store multiple strings, with spaces, into variables. If it makes a difference, I want to make the variables equal to a char pointer. So once I get the input from tempString, I want to set it to a char pointer. Like so:
char *variable1, *variable2;
//get user input
variable1 = tempString;
//get more user input
variable 2 = tempString;
//etc etc etc
Here's what I've tried.
First try
char tempString[100];
scanf("%s", &tempString);
printf("%s", tempString);
Invalid: scanf will stop reading at a white space, so "Example String" would just end up being "Example".
Second try
So I do more research. I thought I found the magic fix.
char tempSTring[100];
fgets(tempString, 100, stdin);
printf("%s", tempString);
Originally this works. However there is a massive problem. I need to get the user to enter about 8 inputs. Meaning I have to use a command like this 8 times. The problem is the program often skips over the fgets command. If I use a scanf previously, somehow the \n character is stuck in the input stream, and automatically feeds into fgets, satisfying its stdin input, and then does not prompt the user for input.
Third try
After thinking fgets was maybe my solution with a work around, I tried some tricks.
char tempSTring[100];
getc(stdin);
fgets(tempString, 100, stdin);
printf("%s", tempString);
I tried adding this getc(stdin) line. It worked for much of my program. It absorbs the \n character left behind in the stream. When it does so, great, it works. But sometimes, for some reason, the \n is NOT left in the stream, and when debugging, it looks like getc(stdin) is requesting input from the user, so it pauses my program to ask for input.
Question
These don't work for me.
How should I be doing this easy task?
To read (up to) 8 lines from a file, you can use either of these solutions. I decline to use variables char *variable1, *variable2, …; — that is an array seeking to escape.
POSIX getline()
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
enum { MAX_LINES = 8 };
char *lines[MAX_LINES];
int index = 0;
char *buffer = 0;
size_t buflen = 0;
while (index < MAX_LINES && getline(&buffer, &buflen, stdin) != -1)
{
lines[index++] = buffer;
buffer = 0;
buflen = 0;
}
free(buffer); // Space may be allocated before EOF is detected
for (int i = 0; i < index; i++)
printf("%d: %s", i, lines[i]);
return 0;
}
If getline() fails to allocate memory, it will report an error, so there is no need to do an explicit error check.
Standard C fgets()
Code using strdup(), another POSIX function. It isn't a part of standard C (though it is widely available). It is trivial to implement.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
enum { MAX_LINES = 8 };
char *lines[MAX_LINES];
int index = 0;
char buffer[4096];
while (index < MAX_LINES && fgets(buffer, sizeof(buffer), stdin) != 0)
{
if ((lines[index] = strdup(buffer)) == 0)
break;
index++;
}
for (int i = 0; i < index; i++)
printf("%d: %s", i, lines[i]);
return 0;
}
The test in the loop allows for the possibility of strdup() failing to allocate memory.
Notes
Both the solutions above keep the newline at the end of the input string. If you don't want that, you can zap it with:
lines[i][strcspn(lines[i], "\r\n")] = '\0';
This overwrites a carriage return or newline with a null byte, transforming DOS or Unix line endings. You then need to adjust the printing which assumes the string includes a newline. Note that the expression shown works correctly even if there is no carriage return or newline in the string.
The fgets() solution will break lines at 4095 characters, leaving the rest to be read as 'the next line'. If that's not acceptable, you have a variety of strategies open to you.
You can detect whether there is a newline and arrange to allocate more memory and read the next section of the line into the extra memory, repeating until you come across a newline or EOF.
You can read the remaining characters up to the newline or EOF:
int c;
while ((c = getchar()) != EOF && c != '\n')
;
Implementing strdup()
If for some reason your system doesn't have an implementation of strdup(), you can create a surrogate with:
#include <assert.h>
#include <stdlib.h>
#include <string.h>
char *strdup(const char *old_str)
{
assert(old_str != 0);
size_t old_len = strlen(old_str) + 1;
char *new_str = malloc(old_len);
if (new_str != 0)
memmove(new_str, old_str, old_len);
return new_str;
}
Here's how we old fart C programmers would do it:
#include <stdio.h>
#define MAX_LEN 100
int main( )
{
int c;
char input[MAX_LEN+1];
int i = 0;
while ( (c=getchar()) != '\n' && c != EOF && i < MAX_LEN)
input[i++] = c;
if (c == EOF || c =='\n') {
/* received input that terminated within buffer limit */
input[i] = '\0';
printf("read in your input string of: %s\n", input);
}
else {
printf("don't buffer overflow me dude!\n");
return -1;
}
return 0;
}
But nowadays people will tell you to use one of the library functions. I'm still an old fart though.
EDIT: Fixed my embarrassing mistakes pointed out by the helpful comments below.
You can take care of '\n' left by previous scanf by writing it like this -
scanf("%d%*c", &x); //<-- example to take int input
%*c will read from stdin and then discard it, thus '\n' would be removed from stdin.
You can achieve with scanf like this (a way for your previous attempt)-
char tempString[100];
/* As suggested by chqrile it is essential to check return of scanf */
if(scanf("%99[^\n]", tempString)!=1){
// ^^ & not required
tempString[0]='\0';
}
%99[^\n] this will read 99 characters and will stop only after encountering '\n' , thus would read input with spaces.
I was trying to take a full line input in C. Initially I did,
char line[100] // assume no line is longer than 100 letters.
scanf("%s", line);
Ignoring security flaws and buffer overflows, I knew this could never take more than a word input. I modified it again,
scanf("[^\n]", line);
This, of course, couldn't take more than a line of input. The following code, however was running into infinite loop,
while(fscanf(stdin, "%[^\n]", line) != EOF)
{
printf("%s\n", line);
}
This was because, the \n was never consumed, and would repeatedly stop at the same point and had the same value in line. So I rewrote the code as,
while(fscanf(stdin, "%[^\n]\n", line) != EOF)
{
printf("%s\n", line);
}
This code worked impeccably(or so I thought), for input from a file. But for input from stdin, this produced cryptic, weird, inarticulate behavior. Only after second line was input, the first line would print. I'm unable to understand what is really happening.
All I am doing is this. Note down the string until you encounter a \n, store it in line and then consume the \n from the input buffer. Now print this line and get ready for next line from the input. Or am I being misled?
At the time of posting this question however, I found a better alternative,
while(fscanf(stdin, "%[^\n]%*c", line) != EOF)
{
printf("%s\n", line);
}
This works flawlessly for all cases. But my question still remains. How come this code,
while(fscanf(stdin, "%[^\n]\n", line) != EOF)
{
printf("%s\n", line);
}
worked for inputs from file, but is causing issues for input from standard input?
Use fgets(). #FredK
char buf[N];
while (fgets(buf, sizeof buf, stdin)) {
// crop potential \n if desired.
buf[strcspn(buf, "\n")] = '\0';
...
}
There are to many issues trying to use scanf() for user input that render it prone to mis-use or code attacks.
// Leaves trailing \n in stdin
scanf("%[^\n]", line)
// Does nothing if line begins with \n. \n remains in stdin
// As return value not checked, use of line may be UB.
// If some text read, consumes \n and then all following whitespace: ' ' \n \t etc.
// Then does not return until a non-white-space is entered.
// As stdin is usually buffered, this implies 2 lines of user input.
// Fails to limit input.
scanf("%[^\n]\n", line)
// Does nothing if line begins with \n. \n remains in stdin
// Consumes 1 char after `line`, even if next character is not a \n
scanf("%99[^\n]%*c", line)
Check against EOF is usual the wrong check. #Weather Vane The following, when \n is first entered, returns 0 as line is not populated. As 0 != EOF, code goes on to use an uninitialized line leading to UB.
while(fscanf(stdin, "%[^\n]%*c", line) != EOF)
Consider entering "1234\n" to the following. Likely infinite loop as first fscanf() read "123", tosses the "4" and the next fscanf() call gets stuck on \n.
while(fscanf(stdin, "%3[^\n]%*c", line) != EOF)
When checking the results of *scanf(), check against what you want, not against one of the values you do not want. (But even the following has other troubles)
while(fscanf(stdin, "%[^\n]%*c", line) == 1)
About the closest scanf() to read a line:
char buf[100];
buf[0] = 0;
int cnt = scanf("%99[^\n]", buf);
if (cnt == EOF) Handle_EndOfFile();
// Consume \n if next stdin char is a \n
scanf("%*1[\n]");
// Use buf;
while(fscanf(stdin, "%[^\n]%*c", line) != EOF)
worked for inputs from file, but is causing issues for input from standard input?
Posting sample code and input/data file would be useful. With modest amount of code posted, some potential reasons.
line overrun is UB
Input begins with \n leading to UB
File or stdin not both opened in same mode. \r not translated in one.
Note: The following fails when a line is 100 characters. So meeting the assumption cal still lead to UB.
char line[100] // assume no line is longer than 100 letters.
scanf("%s", line);
Personally, I think fgets() is badly designed. When I read a line, I want to read it in whole regardless of its length (except filling up all RAM). fgets() can't do that in one go. If there is a long line, you have to manually run it multiple times until it reaches the newline. The glibc-specific getline() is more convenient in this regard. Here is a function that mimics GNU's getline():
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
long my_getline(char **buf, long *m_buf, FILE *fp)
{
long tot = 0, max = 0;
char *p;
if (*m_buf == 0) { // empty buffer; allocate
*m_buf = 16; // initial size; could be larger
*buf = (char*)malloc(*m_buf); // FIXME: check NULL
}
for (p = *buf, max = *m_buf;;) {
long l, old_m;
if (fgets(p, max, fp) == NULL)
return tot? tot : EOF; // reach end-of-file
for (l = 0; l < max; ++l)
if (p[l] == '\n') break;
if (l < max) { // a complete line
tot += l, p[l] = 0;
break;
}
old_m = *m_buf;
*m_buf <<= 1; // incomplete line; double the buffer
*buf = (char*)realloc(*buf, *m_buf); // check NULL
max = (*m_buf) - old_m;
p = (*buf) + old_m - 1; // point to the end of partial line
}
return tot;
}
int main(int argc, char *argv[])
{
long l, m_buf = 0;
char *buf = 0;
while ((l = my_getline(&buf, &m_buf, stdin)) != EOF)
puts(buf);
free(buf);
return 0;
}
I usually use my own readline() function. I wrote this my_getline() a moment ago. It has not been thoroughly tested. Please use with caution.
here is my current code:
int num = 0;
char c = '#';
scanf("%d",&num);
do{
for (int i=0;i<num;i++){
printf("%c",c);
}
printf("\n");
}
while (scanf("%d", &num) == 1);
How would I have it so that if the user doesn't enter anything, that the program won't spit out a newline?
Any help is appreciated, thank you!
This code should work for what you want to do :
#include <stdio.h>
int main()
{
int num = 0;
char c = '#';
char readLine[50];
while ((fgets(readLine, sizeof readLine, stdin) != NULL) && sscanf(readLine, "%d", &num) == 1)
{
for (int i=0;i<num;i++){
printf("%c",c);
}
printf("\n");
fflush(stdout);
}
return 0;
}
The behaviour of this code is the following : fgets will read anything you enter in the standard stream (stdin), and put it in the readLine array. The program will then try to read the number which is in your readLine variable and put it in your num variable with the sscanf function. If a number is read, the program will execute the behaviour you did present in your question (writing a # character "num" times), and go back to the beginning of the loop. If anything else than a number has been read, the loop is stopped.
In general, avoid scanf. It's very easy to leave yourself with unprocessed cruft on the input stream. Instead, read the whole line and then use sscanf (or something else) to process it. This guarantees that you won't get stuck with a partially read line, those are hard to debug.
I prefer getline to fgets to read lines. fgets requires you to guess how long the input might be, and input might get truncated. getline will allocate the memory to read the line for you avoiding buffer overflow or truncation problems.
NOTE: getline is it's not a C standard function, but a POSIX one and fairly recent (2008), though it was a GNU extension well before that. Some older compilers may not have it.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char c = '#';
char *line = NULL;
size_t linelen = 0;
/* First read the whole line */
while( getline(&line, &linelen, stdin) > 0 ) {
/* Then figure out what's in it */
long num = 0;
if( sscanf(line, "%ld", &num) > 0 ) {
for( int i = 0; i < num; i++ ) {
printf("%c", c);
}
printf("\n");
}
}
free(line);
return 0;
}
if( sscanf(line, "%ld", &num) > 0 ) { will ignore any line that does not match any part of the pattern, such as a blank line or a line full of words, by checking how many things matched. Yet it will still handle 0 as a valid input.
$ ./test
foo
bar
foo123
12
############
1
#
0
2
##
I also moved num inside the loop to guarantee it's reinitialized each iteration, and on the general principle of putting your variables in minimum scopes to avoid interference. And I upgraded it to a long int better able to handle the unpredictably large numbers users might type in.
Here is how I have done input parsing over the years using the fgets() and sscanf() functions. I don't write c++ much, and if I can I keep code within old style ansi C then I do.
The fgets and sscanf functions from the stdio.h library are universal and are always available on any platform.
For a character array used to read in anything, I generally set LINE_SIZE to 256 or 512 even if I know typically the line to be read is 80 characters or less. With any computer today having over 1GB of RAM, not worth worrying about allocating an extra 500 or so bytes. Obviously, if you have no idea how long the input line is then you either have to:
guess at what LINE_SIZE should be set to and not worry about it
or verify a newline character is present in line[] prior to a null character after calling fgets().
# include <stdio.h>
# define LINE_SIZE 256
int main ( int argc, char *argv[] )
{
FILE *fp;
char line[LINE_SIZE];
int nn;
int value;
fp = fopen( "somefile", "r" );
fgets( line, LINE_SIZE, fp );
/*
this way to read from standard input (i.e. the keyboard)
using fgets with stdin prevents compiler warning when using
deprecated gets function
fgets( line, LINE_SIZE, stdin );
*/
if ( line[0] != '\n' )
{
/* definitely not a blank line */
nn = sscanf( line, "%d", &num );
if ( nn == 1 )
{
/* some number placed into num variable that met the
%d conversion for the sscanf function
*/
}
}
return 0;