I'm running a while loop so the user can constantly enter expressions, until they indicate they want to quit the program. I'm using strcmp() to compare two strings so as soon as they enter quit the program will stop. But the program keeps going, any Ideas?
#include <stdio.h>
#include <string.h>
int main()
{
int min12=0;
char opper;
int x=0;
int min13;
char *Repeatprog="cont";
char *Repeatprog1="quit";
while (strcmp(Repeatprog,Repeatprog1))
{
printf("enter the integer number \n");
scanf( "%d %c %d", &min12, &opper, &min13);
printf("%d %c %d\n", min12, opper, min13);
printf("Type the word quit to end program\n");
scanf("%s", Repeatprog);
}
printf("Good Bye");
return 0;
}
Remember always that an Array is a Pointer to the first object of the array.
And secondly, in your call to scanf() you only read a character. Not a whole string (represented by %s in C)
So in conclusion, your call to scanf() shouldn't have a pointer and should have a string instead of a character.
scanf("%s", Repeatprog);
or simply
gets (Repeatprog);
EDIT :
As the commenter #EOF said, gets() is not a good idea since it can lead to Undefined Behaviour. That's because the program can read more characters than it should have and lead to overflow, thus it isn't secure.
So I recommend using char *fgets(char *str, int n, FILE *stream)
Note:
Also, your code is using string literals. So if you make any attempt to change the content of the char pointer then it will lead to Undefined Behaviour.
For this note, please thank the guys below me [comments]. I made a huge mistake and I'm sorry.
This question already has answers here:
fgets instructions gets skipped.Why?
(3 answers)
Closed 6 years ago.
I have written a simple program in C which is as follows:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int length;
printf("Enter the length of the string:\t");
scanf("%d",&length);
char str1[10];
printf("Enter string:\t");
gets(str1);
printf("%s",str1);
return 0;
}
When I execute it - I get an output as:
Enter the length of the string: 5
Enter string:
Process returned 0 (0x0) execution time : 1.740 s
Press any key to continue.
I don't know why it doesn't ask for the string input and simply quits the program.
When you type '5’ followed by the enter key, you are sending two chars to the program - '5' and newline. So your first scanf gets the '5' and the second gets the newline, which it converts to the number zero.
See How to read a line from the console in C?
When you enter 5 and press enter which is "\n" then "\n" remains in stream and gets assigned to str1. You need to take that "\n" out of the input stream for which many choices are there. You can figure that out. :) Perhaps later I will edit this answer to let you know.
Edit 1:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
int length;
char c;
printf("Enter the length of the string:\t");
scanf("%d%c",&length, &c);
char str1[10];
printf("Enter string:\t");
gets(str1);
printf("%s",str1);
return 0;
}
This is incorrect way of doing it but your code will at least start working. You can also simply call getc(stdin) which is slightly better. The scanf regex specified in the other answers where it has been marked as duplicate will also work but is ugly and unnecessarily complicated.
I have not tested this and it may not work.
I was shown this program to read an entire string:
#include <stdio.h>
int main(int argc, char *argv[]){
int something;
printf("Enter something \n");
while (scanf("%c", &something)==1) {
printf("%c", something);
}
return 0;
}
When I enter hello world it outputs hello world.
Can someone please explain why it doesn't output:
h
e
l
l
...
because I thought loops go over one letter at a time I'm confused as to why it didn't output like that. Now, I tried to write a program that does the same thing using getchar instead:
#include <stdio.h>
int main()
{
char c;
printf("Enter character: ");
c = getchar();
printf("Character entered: ");
putchar(c);
return(0);
}
When I enter hello world this program just outputs h. How can I use getchar to do the same thing as the first program? Also, what are the differences between getchar and scanf?
getchar() takes one character input at a time. If you want it to read a whole string or a character array, you may use a while loop or a for loop to store each character one by one. Eg : c[i]=getchar() for i = 0 to string length, or while c[i]!='\n'
In the first program, the problem is the line
printf("%c", something);
It's missing a newline character, it should be
printf("%c\n", something);
if you want to print one character per line.
The problem with the second one is that you are reading only one character with getchar() function. You can't read in the whole string on the same line with getchar(), you can try gets(char* s) instead. And you need to loop over putc() to dump all the characters.
Writing a program for class, restricted to only scanf method. Program receives can receive any number of lines as input. Trouble with receiving input of multiple lines with scanf.
#include <stdio.h>
int main(){
char s[100];
while(scanf("%[^\n]",s)==1){
printf("%s",s);
}
return 0;
}
Example input:
Here is a line.
Here is another line.
This is the current output:
Here is a line.
I want my output to be identical to my input. Using scanf.
I think what you want is something like this (if you're really limited only to scanf):
#include <stdio.h>
int main(){
char s[100];
while(scanf("%[^\n]%*c",s)==1){
printf("%s\n",s);
}
return 0;
}
The %*c is basically going to suppress the last character of input.
From man scanf
An optional '*' assignment-suppression character:
scanf() reads input as directed by the conversion specification,
but discards the input. No corresponding pointer argument is
required, and this specification is not included in the count of
successful assignments returned by scanf().
[ Edit: removed misleading answer as per Chris Dodd's bashing :) ]
try this code and use tab key as delimeter
#include <stdio.h>
int main(){
char s[100];
scanf("%[^\t]",s);
printf("%s",s);
return 0;
}
I'll give you a hint.
You need to repeat the scanf operation until an "EOF" condition is reached.
The way that's usually done is with the
while (!feof(stdin)) {
}
construct.
Try this piece of code..
It works as desired on a GCC compiler with C99 standards..
#include<stdio.h>
int main()
{
int s[100];
printf("Enter multiple line strings\n");
scanf("%[^\r]s",s);
printf("Enterd String is\n");
printf("%s\n",s);
return 0;
}
This question already has answers here:
How do you allow spaces to be entered using scanf?
(11 answers)
Closed 4 years ago.
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?
Use:
fgets (name, 100, stdin);
100 is the max length of the buffer. You should adjust it as per your need.
Use:
scanf ("%[^\n]%*c", name);
The [] is the scanset character. [^\n] tells that while the input is not a newline ('\n') take input. Then with the %*c it reads the newline character from the input buffer (which is not read), and the * indicates that this read in input is discarded (assignment suppression), as you do not need it, and this newline in the buffer does not create any problem for next inputs that you might take.
Read here about the scanset and the assignment suppression operators.
Note you can also use gets but ....
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.
Try this:
scanf("%[^\n]s",name);
\n just sets the delimiter for the scanned string.
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;
}
scanf(" %[^\t\n]s",&str);
str is the variable in which you are getting the string from.
The correct answer is this:
#include <stdio.h>
int main(void)
{
char name[100];
printf("Enter your name: ");
// pay attention to the space in front of the %
//that do all the trick
scanf(" %[^\n]s", name);
printf("Your Name is: %s", name);
return 0;
}
That space in front of % is very important, because if you have in your program another few scanf let's say you have 1 scanf of an integer value and another scanf with a double value... when you reach the scanf for your char (string name) that command will be skipped and you can't enter value for it... but if you put that space in front of % will be ok everything and not skip nothing.
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.
#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;
}
Using this code you can take input till pressing enter of your keyboard.
char ch[100];
int i;
for (i = 0; ch[i] != '\n'; i++)
{
scanf("%c ", &ch[i]);
}
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
If you need to read more than one line, need to clear buffer. Example:
int n;
scanf("%d", &n);
char str[1001];
char temp;
scanf("%c",&temp); // temp statement to clear buffer
scanf("%[^\n]",str);
"%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).
"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;
}
scanf("%s",name);
use & with scanf input