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().
Related
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;
}
why doesn't this code work?
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<string.h>
int main(void)
{
// local declarations
int len;
char* pStr;
// statements
printf(" how many characters you want to enter?\n");
scanf("%d", &len);
pStr=(char*)calloc(len+1,sizeof(char));
printf("\n enter your string: ");
gets(pStr);
*(pStr+len)='\0';
printf("\n your string: ");
puts(pStr);
printf(" oops! last character deleted.");
getch();
return 0;
}
although it runs correct, when i use scanf function to read the string, but
why it does not with gets?
scanf("%s", pStr) skips to the first non-whitespace character while gets doesn't.
After the first scanf the trailing newline is still in the input buffer so that when you call gets the result is an empty line unless you entered extra characters after the number.
Note that gets is marked as obsolete due to serious security flaws.
It is recommended that any use of gets(var) is replaced with fgets(var, length, stdin).
Because arrays are zero based, and (assuming the input is valid and the correct length, assumption which your code ought not to make) *(ptr + len) already contains \0 and you are just overwriting it. You meant to overwrite ptr[len-1]
I am trying to write a function that gets a string of letters, either capital letters or small letters, and prints 2 other strings, one with only the capitals, and one only with the small letters. for example:
input: AaBbCcDD
Output: Capital string is ABCDD, non capital is abc
My code is not working correctly, it seems to skip over the last letter. To test it, I wrote the following code:
int length;
printf("Please enter length of string\n");
scanf("%d",&length);
string=create_string(length);
scan_string(string,length);
printf("The string entered is: \n");
print_string(string,length);
Where create_string is:
char* create_string(int size)
{
char* string;
string=(char*)malloc(size*sizeof(char));
return string;
}
Scan string is:
void scan_string(char* string, int size)
{
int i;
printf("Please enter %d characters\n",size);
for(i=0;i<size;i++)
scanf("%c",string+i);
}
And print string is
void print_string(char* string,int size)
{
int i;
for(i=0;i<size;i++)
printf("%c ",*(string+i));
}
When I try even just to print the string I entered, this is what I get, after I input aaAAB
The output is a a A A.
it skipped over the B.
The problem is with the scanf that reads characters using %c: it follows the scanf that reads the length using %d, which leaves an extra '\n' character in the buffer before the first character that you get.
If you modify the output to put quotes around your characters, you would actually see the \n:
void print_string(char* string,int size)
{
int i;
for(i=0;i<size;i++)
printf("'%c' ",*(string+i));
}
This prints
'
' 'a' 'a' 'A' 'A'
(demo on ideone)
You can change your first scanf to read '\n' as below. This will read the extra '\n'
scanf("%d\n", &length);
I think your code is unnecessarily elaborated. To read a string the function fget() with parameter stdin is a simpler choice.
For example, I wuold not ask to the user for the length of the string.
Perhaps it is better to use a buffer with fixed length, and to restrit the user to enter a string with the length less than which you have been previously stipulated.
#define MAXLEN 1000
char buffer[MAXLEN] = "";
fgets(buffer, MAXLEN, stdin);
If the user attempts to enter a string with more than MAXLEN characters, it would be necessary to handle the end-of-line in some way, but I think this is out of topic.
So, in general, let us suppose that MAXLEN is large enough such that buffer contains the \n mark.
Now, a call to your function print_string() can be done.
However, it would be better to do this:
printf("%s", buffer);
I think that you probably need to take in account the C convention for strings: a string is a char array whose last element is marked with the character '\0' (null character, having always code 0).
Even if you want to insist in your approach, I think that scanf() is a bad choice to read individual characters. it is more easy to use getchar(), instead.
By using scanf() you have to broke your brain figurating out all the stuff around the behaviour of scanf(), or how to handle the read of characters, and so on.
However, getchar() reads one char at a time, and that's (almost) all. (Actually, the console commonly not returns the control to the user until an end-of-line \n has been read).
string[i] = getchar();
The problem is because the scanf does not eat the "\n". Hence there is still one '\n' remaining at your first input. This will be counted at the next scanf.
Try to put an additional getchar() right after your first scanf.
printf("Please enter length of string\n");
scanf("%d",&length);
getchar(); // remove '\n'
string=create_string(length);
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