I must write a program in C that can print the middle letter of the string you entered. Spaces () are also calculated, and the number of characters must be odd.
Ex. Input
Hi sussie
--> 9 characters, including space
The output should be s.
I have tried this:
#include <stdio.h>
#include<string.h>
char x[100];
int main(void)
{
printf("Hello World\n");
scanf("%c\n",&x);
long int i = (strlen(x)-1)/2;
printf("the middle letter of the word is %c\n",x[i]);
return 0;
}
and the output always shows the first letter of the word I have entered.
You're only reading the first character from stdin (and incorrectly; you shouldn't be using &).
If you must use scanf, you should use this format:
scanf("%99[^\n]", x);
This is safe and doesn't read past the buffer.
Note that %s wouldn't work here. %s causes scanf to interpret whitespace as the end of the string.
A much better, safer, and easier solution would be to use fgets instead of scanf; fgets is safer and it doesn't require you to change a format string when you change the size of your array:
fgets(x, sizeof(x)-1, stdin);
This eliminates any possible issues with whitespace or buffer overflow.
int main()
{
char arr[1024];
char a;
int i,counter=0;
printf("enter string :: ");
fgets(arr,sizeof(arr),stdin);
for(i=0;i<strlen(arr);i++)
counter++;
for(i=0;i<strlen(arr);i++)
{
if(i==(counter/2))
printf("%c\n",arr[i]);
}
return 0;
}
Related
I was trying to input a string of characters and only output the last and the first character respectively. Below is the code I'm using.
#include<stdio.h>
int main(){
for(int i=0;i<3;i++){
int n; // length of the string
char string[101];
scanf("%d %s", &n, &string);
fflush(stdin); // sometimes I also use getchar();
printf("%c%c", string[n+1], string[0]);
}
printf("\n");
return 0;
}
I'm using for loop because i wanted to input the string 3 times, but when I ran the code the input isn't what I expected. If I input e.g.
5 abcde
output
a //there's space before the a
can you help me tell where I've gone wrong?
input:
5 abcde
6 qwerty
3 ijk
excpeted output:
ea
yq
ki
Few problems in your code:
In this statement
scanf("%d %s", &n, &string);
you don't need to give & operator with string. An array name, when used in an expression, converts to pointer to first element (there are few exceptions to this rule). Also, the size of string array is 101 characters but if you provide input more than 101 characters, the scanf() end up accessing string array beyond its size. You should restrict the scanf() to not to read more than 100 characters in string array when input size is more than that. (keep the remain one character space is for null terminating character that scanf() adds). For this, you can provide width modifier in the format specifier - %100s.
You are not validating the string length input against the input string from user. What happen, if the input string length is greater than or less than the actual length of input string!
fflush(stdin) is undefined behaviour because, as per standard, fflush can only be used with output streams.
I was trying to input a string of characters and only output the last and the first character respectively.
For this, you don't need to take the length of the string as input from user. Use standard library function - strlen(). This will also prevent your program from the problems that can occur due to erroneous length input from user, if that is not validated properly.
Putting these altogether, you can do:
#include <stdio.h>
#include <string.h>
int main (void) {
for (int i = 0; i < 3 ; i++) {
char string[101];
printf ("Enter string:\n");
scanf("%100s", string);
printf("Last character: %c, First character: %c\n", string[strlen(string) - 1], string[0]);
int c;
/*discard the extra characters, if any*/
/*For e.g. if user input is very long this will discard the input beyond 100 characters */
while((c = getchar()) != '\n' && c != EOF)
/* discard the character */;
}
return 0;
}
Note that, scanf(%<width>s, ......) reads up to width or until the first whitespace character, whichever appears first. If you want to include the spaces in input, you can use the appropriate conversion specifier in scanf() or a better alternative is to use fgets() for input from user.
Line 11: string[n+1] -> string[n-1]
I can't take inputs except string.
If I give inputs in code itself, program is working.
But when I try to take inputs, its not working.
How do I get it to accept given inputs?
For example, if s were "codebook", and from == 'o' and to == 'e', s would become "cedebeek".
#include<stdio.h>
#define MAX 50
void replace(char *s, char from, char to)
{
int i=0;
while(s[i]!='\0')
{
if(s[i]==from)
{
s[i]=to;
}
i++;
}
}
int main()
{
char str[MAX];
char from;
char to;
printf("Enter the string");
scanf("%[^\n]s",&str[0]);
printf("\nEnter the character to be replaced");
scanf("%c",&from);
printf("\nEnter the character to be replaced with");
scanf("%c",&to);
replace(str, from, to);
printf("\nThe modified string is %s",str);
return(0);
}
There are a couple of problems in the posted code. Since arrays decay to pointers to their first elements in most expressions, there is no need for the address operator & in the call to scanf(); also, the trailing s is not part of the scanset conversion specifier:
scanf("%[^\n]", str);
As pointed out by #M.M in the comments, it is not incorrect to use &str[0] here instead of str, but it is more idiomatic, and I personally find it more clear, to use the less cluttered str.
When scanf() returns, a newline character will be left in the input stream, so you should add a leading space in the next call to scanf() to skip over this \n character before reading the user input:
scanf(" %c",&from);
And this call to scanf() will also leave a \n character in the input stream, so again:
scanf(" %c",&to);
Note that you should really specify a maximum width to avoid buffer overflow when reading user input into a string; there is no easy way to do this with MAX, but you can do:
scanf("%49[^\n]", str);
You could further improve code and ensure that input is as expected by checking the values returned by the calls to scanf().
using the cprogrammingsimplified tutorial for writing my own stringcompare.
Finished reformatting it and ran it.
works fine for single words,
But
typing space bar skips the second scan and immediately outputs
'words aren't the same'
anyone any idea how to allow the use of even a single space bar?
Thanks in advance.
#include <stdio.h>
int mystrcmp(char s1[], char s2[]);
int main(){
char s1[10], s2[10];
int flag;
printf("Type a string of 10\n\n");
scanf("%s",&s1);
printf("type another string of 10 to compare\n\n");
scanf("%s",&s2);
flag = mystrcmp(s1,s2);
if(flag==0)
printf("the words are the same\n\n");
else
printf("the words are not the same\n\n");
return 0;
}
int mystrcmp(char s1[], char s2[]){
int l=0;
while (s1[l] == s2[l]) {
if (s1[l] == '\0' || s2[l] == '\0')
break;
l++;
}
if (s1[l] == '\0' && s2[l] == '\0')
return 0;
else
return -1;
}
Use fgets() to read full lines, rather than scanf() to read space-separated words.
Remember that fgets() will include the linefeed in the string, though.
It is not strcmp that wouldn't allow space bar, it's scanf with %s format specifier. The input is truncated at the space, so the second string that you read is actually the continuation of the first string.
You can fix this by using %9[^\n] instead of %s in your format specifier:
printf("Type a string of 10\n\n");
scanf("%9[^\n]",s1); //s1 is char [10]
printf("type another string of 10 to compare\n\n");
scanf("%9[^\n]",s2); //s2 is char [10]
9 limits input to nine characters, because you are using a ten-character buffer.
Many answers have told you that scanf("%s",s1) only reads word by word. This is because by default scanf("%s",s1) is delimited by all white spaces, this includes \t, \n, <space>, or any other you can think of.
What scanf("%[^\n]s",s1) does is set the delimiter to \n. So in effect reads all other spaces.
#dasablinklight has also specified a 9 before the '[^\n]' this denotes that scanf() takes 9 values from input buffer.
IMO scanf() is a really nice function due to it's hidden features. I suggest you read more about it in it's documentation.
The problem is that if you type abc def on the first line, the first scanf("%s", s1) (no ampersand required — it should be absent) reads abc and the second reads def. And those are not equal. Type very very and you'd find the words are equal. %s stops reading at a space.
Your buffers of size 10 are too small for comfort.
Fix: read lines (e.g. char s1[1024], s2[1024];) with fgets() or POSIX's getline(), remove trailing newlines (probably: s1[strcspn(s1, "\n")] = '\0'; is a reliable way to do it) and then go ahead compare the lines.
i am a learner of 'C' and written a code, but after i compile it, shows a Debug Error message, here is the code:
#include<stdio.h>
void main()
{
int n,i=1;
char c;
printf("Enter Charecter:\t");
scanf("%s",&c);
printf("Repeat Time\t");
scanf("%d",&n);
n=n;
while (i <= n)
{
printf("%c",c);
i++;
}
}
Pls tell me why this happens and how to solve it
The scanf("%s", &c) is writing to memory it should not as c is a single char but "%s" expects its argument to be an array. As scanf() appends a null character it will at the very least write two char to c (the char read from stdin plus the null terminator), which is one too many.
Use a char[] and restrict the number of char written by scanf():
char data[10];
scanf("%9s", data);
and use printf("%s", data); instead of %c or use "%c" as the format specifier in scanf().
Always check the return value of scanf(), which is the number of successful assignments, to ensure subsequent code is not processing stale or uninitialized variables:
if (1 == scanf("%d", &n))
{
/* 'n' assigned. 'n = n;' is unrequired. */
}
scanf("%s",&c); should be scanf("%c",&c);
The %s format specifier tells scanf you're passing a char array. You're passing a single char so need to use %c instead.
Your current code will behave unpredictably because scanf will try to write an arbitrarily long word followed by a nul terminator to the address you provided. This address has memory allocated (on the stack) for a single char so you end up over-writing memory that may be used by other parts of your program (say for other local variables).
I'm not sure you understood the answer to your other question: Odd loop does not work using %c
These format specifiers are each used for a specific job.
If you want to get a:
character from stdin use %c.
string (a bunch of characters) use %s.
integer use %d.
This code:
char c;
printf("Enter Character:\t");
scanf("%c",&c);
Will read 1 character from stdin and will leave a newline ('\n') character there. So let's say the user entered the letter A in the stdin buffer you have:
A\n
The scanf() will pull 'A' and store it in your char c and will leave the newline character. Next it will ask for your int and the user might input 5. stdin now has:
\n5
The scanf() will take 5 and place it in int n. If you want to consume that '\n' there are a number of options, one would be:
char c;
printf("Enter Character:\t");
scanf("%c",&c); // This gets the 'A' and stores it in c
getchar(); // This gets the \n and trashes it
Here is a working version of your code. Please see inline comments in code for fixes:
#include<stdio.h>
void main()
{
int n,i=1;
char c;
printf("Enter Character:\t");
scanf("%c",&c);//Use %c instead of %s
printf("Repeat Time\t");
scanf("%d",&n);
n=n;//SUGGESTION:This line is not necessary. When you do scanf on 'n' you store the value in 'n'
while (i <= n)//COMMENT:Appears you want to print the same character n times?
{
printf("%c",c);
i++;
}
return;//Just a good practice
}
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