I have the following source-code:
#include <stdio.h>
#include <time.h>
#include <string.h>
int main(int argc, char *argv[])
{
char string[100];
printf("Give me some text. \n");
fgets(string, 100, stdin);
char delimiter[]=" ";
char *erg;
erg=strtok(string, delimiter);
while(erg != NULL){
printf("Wort: %s \n", erg);
erg=strtok(NULL, delimiter);
}
return 0;
}
When I for example put in the text "abc def", the program is working like I want it to work. It print out the words "abc" and "def".
But when I put in the text "abc def ", it prints out "abc", "def" and "".
I don't want to print out the last empty "".
Can someone please tell me how to filter that ?
Gruß, Andre
IMO, you need to get rid of the last \n read by fgets() in fgets(string, 100, stdin);
You can do this in many ways, like
include the \n in your delimiter list. " \n", for example.
manually check the length of string, and at the last position, replace the \n with a \0.
fgets gets a whole string along with the \n character which you press after entering the string.So use
string[strlen(string)-1]='\0';
to replace the \n in string with a \0.
Related
#include <stdio.h>
#include <string.h>
int main() {
char a;
char s[100];
char sen[100];
scanf("%c",&a); // take character
printf("%c",a); //print character
scanf("%s",s); //take input as a word
printf("\n%s",s); //print the word
if((gets(sen))=='\n')
gets(sen);// take input as a string
puts(sen); //print that string
}
As gets() takes input from buffer so it will take '\n' as input after that another gets() command should work but that is not working. It doesn't take any input. Why?
gets(sen) returns sen, which is the address of the char array.
Therefore, what you are trying to do should be
if(strcmp(gets(sen), "\n") == 0)
gets(sen);// take input as a string
However, this is error-prone because sen can be a space with a newline, which is not "\n" or so.
#include <stdio.h>
int main(){
char phrase [100]={0};
printf("Write a sentence: \n");
scanf ("%s", phrase);
printf("%s\n", phrase);
}
The purpose of this program is to print a sentence from the user. The problem is that there is something wrong with the ' ' character and i dont know how to solve it.
During the execution, the sentence is only printed until the first empty space, for some reason. but why?
As pointed out in comments, the %s conversion specifier in scanf can't read strings with white spaces; it uses white spaces as delimiters. You can use fgets :
#include <stdio.h>
#include <string.h>
int main() {
char aLine[100];
fgets(aLine, 100, stdin);
aLine[strlen(aLine)-1] = '\0'; // trimming the last \n
printf("%s\n", aLine);
return 0;
}
I am using strtok for converting string into individual words. I have done the following:
int main() {
char target[100];
char *t;
scanf("%s",target);
t = strtok(target," ");
while (t!= NULL)
{
printf("<<%s>>\n", t);
t = strtok (NULL, " ");
}
return 0;
}
The input is a string such as 'this is a string', the output I am getting is<<this>>.
The way you have written scanf it will accept string till white space only
scanf("%s",target);
SO You need to change the way you take input from console
scanf("%99[^\n]",target);
Change:
scanf("%s",target);
to:
fgets(target, 100, stdin);
since the first won't stop when encounters the whitespace in your input.
Output:
this is a string
<<this>>
<<is>>
<<a>>
<<string
>>
Notice how the newline fgets() stores affects the output. You can simply discard it if you want, like this:
fgets(target, 100, stdin);
target[strlen(target) - 1] = '\0';
and now the output is:
this is a string
<<this>>
<<is>>
<<a>>
<<string>>
If you want to continue using scanf(), then you can use the below code snippet:
#include<stdio.h>
#include <string.h>
int main() {
char target[100];
char *t;
//scanf("%s",target);
scanf("%99[0-9a-zA-Z ]", target);
printf("%s\n",target);
t = strtok(target," ");
while (t!= NULL)
{
printf("<<%s>>\n", t);
t = strtok (NULL, " ");
}
return 0;
}
Working code here.
Just writing scanf("%s",target); will read the input only till the first white space; which is why you get only the first word as the output. By writing scanf("%99[0-9a-zA-Z ]", target);, you are reading 99 characters (including numbers 0-9, a-z or A-Z and white space) from the input stream.
Hope this is helpful.
I am trying to copy a sentence into a char array. I have tried using scanf("%[^\n]) and scanf("%[^\n]\n within an if statement but it doesn't work. Can someone please help me figure it out? I am using C language. It works with the first code but not the second.
File #1
#include <stdio.h>
int main ()
{
char c[10];
printf ("Enter text.\n");
scanf("%[^\n]", c);
printf ("text:%s", c);
return 0;
}
File #2
#include <stdio.h>
#include <string.h>
int main(void)
{
char command[10];
char c[10];
printf("cmd> ");
scanf( "%s", command);
if (strcmp(command, "new")==0)
{
printf ("Enter text:\n");
scanf("%[^\n]", c);
printf ("text:%s\n", c);
}
return 0;
}
Put a space before %[^\n] like so:
#include <stdio.h>
#include <string.h>
int main(void)
{
char command[10];
char c[10];
printf("cmd> ");
scanf( "%s", command);
if (strcmp(command, "new")==0)
{
printf ("Enter text:");
scanf(" %[^\n]", c); // note the space
printf ("text:%s", c);
}
return 0;
}
It should work now. The space makes it consume any whitespace of the previous inputs.
Here's my output when I tested it without the space:
cmd> new
Enter text:text:#
------------------
(program exited with code: 0)
And with the space:
cmd> new
Enter text:test
text:test
------------------
(program exited with code: 0)
According to the man page of scanf function, the usual skip of leading white space is suppressed if you use the [ character in the format string. So in the first case, it accepts all the characters until it meets the \n character; In the second case, after the first call of scanf function, the \n character (you press the Enter key in the first call) is still in the input buffer, so if you uses the format string "%[^\n]" in the scanf function, it reads an empty string into the buffer (As already mentioned, it skips the white space in this format case). So you can use the format string " %[^\n]" to force the scanf function to skip the white space.
I need to write a function which takes 2 words and count its length. I wrote below one but this code only woks for 1st word. How can I improve it to count whole sentence?
#include <stdio.h>
int findlen(int *s);
int main(void)
{
char string1[80];
printf("Enter a string: ");
scanf("%s", string1);
printf("Lenght of %s is %d\n", string1, findlen(string1));
}
//find the length of the inputted string
int findlen(char *s)
{
int count = 0;
while (*s != '\0')
{
s++;
count++;
}
return count;
}
scanf will take the one word input only.. (i.e) it breaks when space appears..
Try fgets to read the complete string till \n
You can use fgets to safely get the line from your file:
From here:
char *fgets(char *s, int size, FILE *stream);
so replace your scanf line with:
fgets(string1, sizeof(string1), stdin);
If you used gets instead, it doesn't know how large your buffer is and it would crash when reading too large line.
Next, if you want to know a length of your string you could use strlen function from string.h:
#include<string.h>
...
printf("Lenght of %s is %d\n", string1, strlen(string1));
use [%^\n] format specifier so that it will scan the string till '\n' encounter so the problem in scanning string having space(eg. "hello word") will be solved.
scanf("[%^\n]",straddr);