I write console application which performs several scanf for int
And after it ,I performs getchar :
int x,y;
char c;
printf("x:\n");
scanf("%d",&x);
printf("y:\n");
scanf("%d",&y);
c = getchar();
as a result of this I get c = '\n',despite the input is:
1
2
a
How this problem can be solved?
This is because scanf leaves the newline you type in the input stream. Try
do
c = getchar();
while (isspace(c));
instead of
c = getchar();
Call fflush(stdin); after scanf to discard any unnecessary chars (like \r \n) from input buffer that were left by scanf.
Edit: As guys in comments mentioned fflush solution could have portability issue, so here is my second proposal. Do not use scanf at all and do this work using combination of fgets and sscanf. This is much safer and simpler approach, because allow handling wrong input situations.
int x,y;
char c;
char buffer[80];
printf("x:\n");
if (NULL == fgets(buffer, 80, stdin) || 1 != sscanf(buffer, "%d", &x))
{
printf("wrong input");
}
printf("y:\n");
if (NULL == fgets(buffer, 80, stdin) || 1 != sscanf(buffer, "%d", &y))
{
printf("wrong input");
}
c = getchar();
You can use the fflush function to clear anything left in buffer as a consquence of previous comand line inputs:
fflush(stdin);
A way to clean up anyspace before your desired char and just ignore the remaining chars is
do {
c = getchar();
} while (isspace(c));
while (getchar() != '\n');
For a start the scanf should read scanf("%d\n", &x); or y. That should do the trick.
man scanf
Related
I am trying a simple exercise from K&R to append string2 at the end of string1 using pointers. In case of overflow i.e. buffer of string1 can't contain all of string2 I want to prompt the user to re-enter string2 or exit.
I have written the following code:
#include<stdio.h>
#include<string.h>
#define MAXLINE 1000
int get_input(char *s);
int str_cat(char *s, char *t);
void main()
{
char input1[MAXLINE], input2[MAXLINE], c;
get_input(input1);
check:
get_input(input2);
if((strlen(input1) + strlen(input2) + 2) <= MAXLINE)
{
str_cat(input1, input2);
printf("%s\n", input1);
}
else
{
input2[0] = '\0';
printf("String overflow\n Press: \n 1: Re-enter string. \n 2: Exit.\n");
scanf(" %d", &c);
if(c == 1){
input2[0] = '\0';
get_input(input2);
goto check;
}
}
}
int get_input(char *arr)
{
int c;
printf("Enter the string: \n");
while(fgets(arr, MAXLINE, stdin))
{
break;
}
}
int str_cat(char *s, char *t)
{
while(*s != '\0')
{
s++;
}
while((*s++ = *t++) != '\0')
{
;
}
*s = '\0';
}
Initially, I was using the standard getchar() function mentioned in the book to read the input in get_input() which looked like this:
int get_input(char *arr)
{
int c;
printf("Enter the string: \n");
while((c = getchar()) != '\n' && c != EOF)
{
*arr++ = c;
}
*arr = '\0';
}
I am new and I read this and understood my mistake. I understand that one isn't supposed to use different input functions to read stdin and the '\n' is left in the input stream which is picked by the getchar() causing my condition to fail.
So, I decided to use fgets() to read the input and modified the scanf("%d", &c) as mentioned in the thread with scanf(" %d", c). This does work (kinda) but gives rise to behaviors that I do not want.
So, I have a few questions:
What's a better way to fgets() from reading the input on encountering '\n' than the one I have used?
while(fgets(arr, MAXLINE, stdin))
{
break;
}
fgets() stops reading the line and stores it as an input once it either encounters a '\n' or EOF. But, it ends up storing the '\n' at the end of the string. Is there a way to prevent this or do I have to over-write the '\n' manually?
Even though I used the modified version of scanf(" %d", &c), my output looks like
this: (https://imgur.com/a/RaC2Kc6). Despite that I get Enter the string: twice when prompted to re-enter the second string in case of an overflow situation. Is the modified scanf() messing with my input? And how do I correct it?
In general, do not mix fgets with scanf. Although it may be a bit bloaty, you will avoid many problems by being consistent with reading input with fgets and then parse it with sscanf. (Note the extra s)
A good way to remove the newline is buffer[strcspn(buffer, "\n")] = 0
Example:
// Read line and handle error if it occurs
if(!fgets(buffer, buffersize, stdin)) {
// Handle error
}
// Remove newline (if you want, not necessarily something you need)
buffer[strcspn(buffer, "\n")] = 0;
// Parse and handle error
int val;
if(sscanf(buffer, "%d", &val) != 1) {
// Handle error
}
// Now you can use the variable val
There is one thing here that might be dangerous in certain situations, and that is if buffer is not big enough to hold a complete line. fgets will not read more than buffersize characters. If the line is longer, the remaining part will be left in stdin.
This question already has answers here:
scanf() leaves the newline character in the buffer
(7 answers)
Closed 3 years ago.
In C:
I'm trying to get char from the user with scanf and when I run it the program don't wait for the user to type anything...
This is the code:
char ch;
printf("Enter one char");
scanf("%c", &ch);
printf("%c\n",ch);
Why is not working?
The %c conversion specifier won't automatically skip any leading whitespace, so if there's a stray newline in the input stream (from a previous entry, for example) the scanf call will consume it immediately.
One way around the problem is to put a blank space before the conversion specifier in the format string:
scanf(" %c", &c);
The blank in the format string tells scanf to skip leading whitespace, and the first non-whitespace character will be read with the %c conversion specifier.
First of all, avoid scanf(). Using it is not worth the pain.
See: Why does everyone say not to use scanf? What should I use instead?
Using a whitespace character in scanf() would ignore any number of whitespace characters left in the input stream, what if you need to read more inputs? Consider:
#include <stdio.h>
int main(void)
{
char ch1, ch2;
scanf("%c", &ch1); /* Leaves the newline in the input */
scanf(" %c", &ch2); /* The leading whitespace ensures it's the
previous newline is ignored */
printf("ch1: %c, ch2: %c\n", ch1, ch2);
/* All good so far */
char ch3;
scanf("%c", &ch3); /* Doesn't read input due to the same problem */
printf("ch3: %c\n", ch3);
return 0;
}
While the 3rd scanf() can be fixed in the same way using a leading whitespace, it's not always going to that simple as above.
Another major problem is, scanf() will not discard any input in the input stream if it doesn't match the format. For example, if you input abc for an int such as: scanf("%d", &int_var); then abc will have to read and discarded. Consider:
#include <stdio.h>
int main(void)
{
int i;
while(1) {
if (scanf("%d", &i) != 1) { /* Input "abc" */
printf("Invalid input. Try again\n");
} else {
break;
}
}
printf("Int read: %d\n", i);
return 0;
}
Another common problem is mixing scanf() and fgets(). Consider:
#include <stdio.h>
int main(void)
{
int age;
char name[256];
printf("Input your age:");
scanf("%d", &age); /* Input 10 */
printf("Input your full name [firstname lastname]");
fgets(name, sizeof name, stdin); /* Doesn't read! */
return 0;
}
The call to fgets() doesn't wait for input because the newline left by the previous scanf() call is read and fgets() terminates input reading when it encounters a newline.
There are many other similar problems associated with scanf(). That's why it's generally recommended to avoid it.
So, what's the alternative? Use fgets() function instead in the following fashion to read a single character:
#include <stdio.h>
int main(void)
{
char line[256];
char ch;
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
ch = line[0];
printf("Character read: %c\n", ch);
return 0;
}
One detail to be aware of when using fgets() will read in the newline character if there's enough room in the inut buffer. If it's not desirable then you can remove it:
char line[256];
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
line[strcpsn(line, "\n")] = 0; /* removes the trailing newline, if present */
This works for me try it out
int main(){
char c;
scanf(" %c",&c);
printf("%c",c);
return 0;
}
Here is a similiar thing that I would like to share,
while you're working on Visual Studio you could get an error like:
'scanf': function or variable may be unsafe. Consider using scanf_s instead. To disable deprecation, use _CRT_SECURE_NO_WARNINGS
To prevent this, you should write it in the following format
A single character may be read as follows:
char c;
scanf_s("%c", &c, 1);
When multiple characters for non-null terminated strings are read, integers are used as the width specification and the buffer size.
char c[4];
scanf_s("%4c", &c, _countof(c));
neither fgets nor getchar works to solve the problem.
the only workaround is keeping a space before %c while using scanf
scanf(" %c",ch); // will only work
In the follwing fgets also not work..
char line[256];
char ch;
int i;
printf("Enter a num : ");
scanf("%d",&i);
printf("Enter a char : ");
if (fgets(line, sizeof line, stdin) == NULL) {
printf("Input error.\n");
exit(1);
}
ch = line[0];
printf("Character read: %c\n", ch);
try using getchar(); instead
syntax:
void main() {
char ch;
ch = getchar();
}
Before the scanf put fflush(stdin); to clear buffer.
The only code that worked for me is:
scanf(" %c",&c);
I was having the same problem, and only with single characters. After an hour of random testing I can not report an issue yet. One would think that C would have by now a bullet-proof function to retrieve single characters from the keyboard, and not an array of possible hackarounds... Just saying...
Use string instead of char
like
char c[10];
scanf ("%s", c);
I belive it works nice.
Provides a space before %c conversion specifier so that compiler will ignore white spaces. The program may be written as below:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char ch;
printf("Enter one char");
scanf(" %c", &ch); /*Space is given before %c*/
printf("%c\n",ch);
return 0;
}
You have to use a valid variable. ch is not a valid variable for this program. Use char Aaa;
char aaa;
scanf("%c",&Aaa);
Tested and it works.
Given the following code:
#include <stdio.h>
int main()
{
int testcase;
char arr[30];
int f,F,m;
scanf("%d",&testcase);
while(testcase--)
{
printf("Enter the string\n");
fgets(arr,20,stdin);
printf("Enter a character\n");
F=getchar();
while((f=getchar())!=EOF && f!='\n')
;
putchar(F);
printf("\n");
printf("Enter a number\n");
scanf("%d",&m);
}
return 0;
}
I want a user to enter a string, a character and a number until the testcase becomes zero.
My doubts / questions:
1.User is unable to enter a string. It seems fgets is not working. Why?
2.If i use scanf instead of fgets,then getchar is not working properly, i.e whatever character I input in it just putchar as a new line. Why?
Thanks for the help.
Mixing functions like fgets(), scanf(), and getchar() is error-prone. The scanf() function usually leaves a \n character behind in the input stream, while fgets() usually does not, meaning that the next call to an I/O function may or may not need to cope with what the previous call has left in the input stream.
A better solution is to use one style of I/O function for all user input. fgets() used in conjunction with sscanf() works well for this. Return values from functions should be checked, and fgets() returns a null pointer in the event of an error; sscanf() returns the number of successful assignments made, which can be used to validate that input is as expected.
Here is a modified version of the posted code. fgets() stores input in a generously allocated buffer; note that this function stores input up to and including the \n character if there is enough room. If the input string is not expected to contain spaces, sscanf() can be used to extract the string, leaving no need to worry about the newline character; similarly, using sscanf() to extract character or numeric input relieves code of the burden of further handling of the \n.
#include <stdio.h>
int main(void)
{
int testcase;
char arr[30];
char F;
int m;
char buffer[1000];
do {
puts("Enter number of test cases:");
if (fgets(buffer, sizeof buffer, stdin) == NULL) {
/* handle error */
}
} while (sscanf(buffer, "%d", &testcase) != 1 || testcase < 0);
while(testcase--)
{
puts("Enter the string");
/* if string should not contain spaces... */
if (fgets(buffer, sizeof buffer, stdin) == NULL) {
/* handle error */
}
sscanf(buffer, "%29s", arr);
printf("You entered: %s\n", arr);
putchar('\n');
puts("Enter a character");
if (fgets(buffer, sizeof buffer, stdin) == NULL) {
/* handle error */
}
sscanf(buffer, "%c", &F);
printf("You entered: %c\n", F);
putchar('\n');
do {
puts("Enter a number");
if (fgets(buffer, sizeof buffer, stdin) == NULL) {
/* handle error */
}
} while (sscanf(buffer, "%d", &m) != 1);
printf("You entered: %d\n", m);
putchar('\n');
}
return 0;
}
On the other hand, if the input string may contain spaces, fgets() can read input directly into arr, but then the stored string will contain a \n character, which should probably be removed. One way of doing this is to use the strcspn() function to find the index of the \n:
#include <string.h> // for strcspn()
/* ... */
puts("Enter the string");
/* or, if string may contain spaces */
if (fgets(arr, sizeof arr, stdin) == NULL) {
/* handle error */
}
/* replace newline */
arr[strcspn(arr, "\r\n")] = '\0';
printf("You entered: %s\n", arr);
putchar('\n');
/* ... */
Note that a maximum width should always be specified when using %s with the scanf() functions to avoid buffer overflow. Here, it is %29s when reading into arr, since arr can hold 30 chars, and space must be reserved for the null terminator (\0). Return values from sscanf() are checked to see if user input is invalid, in which case the input is asked for again. If the number of test cases is less than 0, input must be entered again.
Finally got the solution how can we use scanf and fgets together safely.
#include <stdio.h>
int main()
{
int testcase,f,F,m;
char arr[30];
scanf("%d",&testcase);
while((f=getchar())!=EOF && f!='\n')
;
while(testcase--)
{
printf("Enter the string\n");
fgets(arr,30,stdin);
printf("Enter a character\n");
F=getchar();
while((f=getchar())!=EOF && f!='\n')
;
putchar(F);
printf("\n");
printf("Enter a number\n");
scanf("%d",&m);
while((f=getchar())!=EOF && f!='\n')
;
}
}
We need to make sure that before fgets read anything,flushout the buffer with simple while loop.
Thanks to all for the help.
A simple hack is to write a function to interpret the newline character. Call clear() after each scanf's
void clear (void){
int c = 0;
while ((c = getchar()) != '\n' && c != EOF);
}
Refer to this question for further explaination: C: Multiple scanf's, when I enter in a value for one scanf it skips the second scanf
Here first gets() is not working. if I add one more gets() function then from the two last one goes to work. how can I fix it?
CODE
#include<stdio.h>
#include<string.h>
int main(void)
{
short int choice;
char number[15];
do{
printf("\n\nAnswer: ");
scanf("%hd",&choice);
printf("\n");
if(choice==1)
{
printf("Enter the decimal number: ");
gets(number);
}
else
{
printf("Wrong input!.");
system("pause");
system("cls");
}
}while(choice!=1);
return 0;
}
Because the when the user pressed the enter key to give you the input for the scanf call, the enter key added a newline in the input buffer. And the gets call read that newline as an empty line.
One way to solve it is to use fgets to read the first input too, and use sscanf to parse it to a number:
...
printf("\n\nAnswer: ");
char input[64];
fgets(input, sizeof(input), stdin);
sscanf(input, "%hd", &choice);
printf("\n");
...
This make sure that the newline after the input is read and skipped.
Another way is to read one character at a time in a loop after the scanf call, until you have read the newline:
scanf("%hd", &choice);
int ch;
while ((ch = fgetc(stdin)) != EOF && ch != '\n')
{
// Empty
}
And a third way is to simply ask the scanf call to read and ignore white-space after the input:
scanf("%hd ", &choice);
// ^
// |
// Note space here
All of these methods have both pros and cons. You can try them all and use the one that works for you.
You need to skip the whitespace (i.e. the newline) following the number in the input buffer. This can be done by modifying the scanf to:
scanf("%hd ",&choice);
And use fgets(), since gets() is prone to buffer overflows.
basically in codeblocks for windows before each printf I have "fflush(stdin);" which works. When I copied my code to Linux, it doesn't work, nor does any of the alternatives for "fflush(stdin);" that I've found. No matter which way I seem to do it, the input doesn't seem to be clearing in the buffer or something in my code is incorrect.
#include <stdio.h>
#include <math.h>
#include <limits.h>
#include <ctype.h>
int main()
{
char pbuffer[10], qbuffer[10], kbuffer[10];
int p=0, q=0, k=0;
int r, i, Q, count, sum;
char a[3];
a[0]='y';
while(a[0]=='y' || a[0]=='Y')
{
printf("Enter a p value: \n");
fgets(pbuffer, sizeof(pbuffer), stdin);
p = strtol(pbuffer, (char **)NULL, 10);
printf("Enter a q value: \n");
fgets(qbuffer, sizeof(qbuffer), stdin);
q = strtol(qbuffer, (char **)NULL, 10);
printf("Enter a k value: \n");
fgets(kbuffer, sizeof(kbuffer), stdin);
k = strtol(kbuffer, (char **)NULL, 10);
while(p<q+1)
{
Q=p;
sum=0;
count=0;
while(Q>0)
{
count++;
r = Q%10;
sum = sum + pow(r,k);
Q = Q/10;
}
if ( p == sum && i>1 && count==k )
{
printf("%d\n",p);
}
p++;
a[0]='z';
}
while((a[0]!='y') && (a[0]='Y') && (a[0]!='n') && (a[0]!='N'))
{
printf("Would you like to run again? (y/n) ");
fgets(a, sizeof(a), stdin);
}
}
return 0;
}
Calling fflush(stdin) is not standard, so the behavior is undefined (see this answer for more information).
Rather than calling fflush on stdin, you could call scanf, passing a format string instructing the function to read everything up to and including the newline '\n' character, like this:
scanf("%*[^\n]%1*[\n]");
The asterisk tells scanf to ignore the result.
Another problem is calling scanf to read a character into variable a with the format specifier of " %s": when the user enters a non-empty string, null terminator creates buffer overrun, causing undefined behavior (char a is a buffer of one character; string "y" has two characters - {'y', '\0'}, with the second character written past the end of the buffer). You should change a to a buffer that has several characters, and pass that limit to scanf:
char a[2];
do {
printf("Would you like to run again? (y/n) \n")
scanf("%1s", a);
} while(a[0] !='y' && a[0] !='Y' && a[0]!='n' && a[0]!='N' );
}
I think what you are trying to do is more difficult than it seems.
My interpretation of what you are trying to do is disable type ahead so that if the user types some characters while your program is processing other stuff, they don't appear at the prompt. This is actually quite difficult to do because it is an OS level function.
You could do a non blocking read on the device before printing the prompt until you get EWOULDBLOCK in errno. Or the tcsetattr function family might help. It looks like there is a way to drain input for a file descriptor in there, but it might interact badly with fgets/fscanf
A better idea is not to worry about it at all. Unix users are used to having type ahead and what you want would be unexpected behaviour for them.
Drop the need for flushing the input buffer.
OP is on the right track using fgets() rather than scanf() for input, OP should continue that approach with:
char a;
while(a !='y' && a !='Y' && a!='n' && a!='N' ) {
printf("Would you like to run again? (y/n) \n");
if (fgets(kbuffer, sizeof(kbuffer), stdin) == NULL)
Handle_EOForIOerror();
int cnt = sscanf(kbuffer, " %c", &a); // Use %c, not %s
if (cnt == 0)
continue; // Only white-space entered
}
Best to not use scanf() as it tries to handle user IO and parsing in one shot and does neither that well.
Certain present OP's woes stem from fgets() after scanf(" %s", &a); (which is UB as it should be scanf(" %c", &a);. Mixing scanf() with fgets() typically has the problem that the scanf(" %c", &a); leaves the Enter or '\n' in the input buffer obliging the code to want to flsuh the input buffer before the next fgets(). Else that fgets() gets the stale '\n' and not a new line of info.
By only using fgets() for user IO, there need for flushing is negated.
Sample fgets() wrapper
char *prompt_fgets(const char *prompt, char dest, long size) {
fputs(prompt, stdout);
char *retval = fgets(dest, size, stdin);
if (retval != NULL) {
size_t len = strlen(dest);
if (len > 1 && dest[len-1] == '\n') { // Consume trailing \n
dest[--len] = '\0';
}
else if (len + 1 == dest) { // Consume extra char
int ch;
do {
ch == fgetc(stdin);
} while (ch != '\n' && ch != EOF);
}
return retval;
}