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
}
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;
}
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().
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]
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
I am trying to take five character and 5 float input.
main()
{
char c[5];
float q[5];
int i;
for(i=0;i<5;i++)
{
printf("\n%d ",i);
scanf("%c",c+i);
scanf("%f",q+i);
}
}
But the output is absurd. After two sequential scans, it skips third scan and then again skips fifth scan.
I am not able to understand why is it showing such a behaviour.
I am working on gcc compiler.
Use of scanf is not recommended because of problems like this.
Instead use fgets to read the entire line and then use sscanf to extract what you want(a char or a float) from the line just read:
char line[MAX];
for(i=0;i<5;i++)
{
if( fgets(line,MAX,stdin) && sscanf(line,"%c", c+i)!=1 )
*(c+i) = 0;
if( fgets(line,MAX,stdin) && sscanf(line,"%f", q+i)!=1 )
*(q+i) = 0.0;
printf("%c %f\n",*(c+i),*(q+i));
}
To directly answer why the 3rd and every other scan "skips", it is the way scanf() and the %c format works. When there is a call to scanf(), you typically have to press enter to "submit" the input. Consequently that inserts a newline character into the stdin stream.
When the previous float scan got inputted, the newline is still left in the stream. When the character scan gets reached, that remaining newline character is read in since it fits effectively "skipping" the call.
You should use fgets() with sscanf() as codaddict suggests.
But as a quick fix, you could try adding a call to getchar() after the float scan to consume that newline character from the stream.
edit:
You're saying this doesn't work? (assuming you input the correct kinds of values, one per scanf call)
main()
{
char c[5];
float q[5];
int i;
for(i=0;i<5;i++)
{
printf("\n%d ",i);
scanf("%c",c+i);
scanf("%f",q+i);
getchar();
}
}
You should try this:
int main(){
char c[6];//char array size must be 6 to store five charecter
//as null-terminator('\0')will use the last one
float q[5];
int i;
for(i=0;i<5;i++){
printf("\n%d\n",i);fflush(stdout);
scanf("%c",&c[i]);fflush(stdin);//fflush(stdin) to clear input stream that
//next scanf() won't skip
scanf("%f",&q[i]);fflush(stdin);//fflush(stdin) to clear input stream that
//scanf() won't skip at new loop
}
return 0;
}
fflush() is not defined on an input stream, like stdin. Don't do it.
If you want to "read and discard until newline", then do:
int ch;
do {
ch = getchar();
} while (ch != EOF && ch != '\n');
Note that %c means "read the next character in the input stream, even if it's whitespace, then stop". %f means "read and discard whitespace, then try to read a float from the input stream, then stop."
Your code should be like this :
main()
{
char c[5];
float q[5];
int i;
for(i=0;i<5;i++)
{
printf("\n%d ",i);
scanf("%c",c+i);
while (getchar()!='\n');
scanf("%f",q+i);
while (getchar()!='\n');
}
}
the sentence while (getchar()!='\n'); search till the end of input, so it would not take '\n' as an input value for q+i.Also another while (getchar()!='\n'); after scanf q+i,since you use loop.
Problem in scanning a char value after a float value.
Solution is very simple!!!
instead of writting your code like this
scanf("%c",&...)
try this,
scanf(" %c",&...)
A Space before the"%c" will resolve the problem by neglecting the value of the Return(Enter) key when pressed after typing the value of the float as an input.
When a float value is scanned before a character value.
The value obtained by pressing the Return(Enter) key is collected in the following char variable. Using a space before(" %c",&...) discards the value collected of the Return(Enter) Key, Causing the Char value to be scanned in the next line. Thus solving The Scanning Float-Char problem.