fgets() doesn't work as expected in C - c

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

Related

Is there an elegant way to handle the '\n' that gets read by input functions (getchar(), fgets(), scanf()) in C?

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.

2D character array in C [duplicate]

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.

How can I use the "gets" function many times in my C program?

My code:
#include <stdio.h>
#include <math.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
char a[10],b[10];
puts("enter");
gets(a);
puts("enter");
gets(b);
puts("enter");
puts(a);
puts(b);
}
return 0;
}
Output:
1
enter
enter
surya (string entered by user)
enter
surya (last puts function worked)
How can I use “gets” function many times in C program?
You should never ever use gets() in your program. It is deprecated because it is dangerous for causing buffer overflow as it has no possibility to stop consuming at a specific amount of characters - f.e. and mainly important - the amount of characters the buffer, a or b with each 10 characters, is capable to hold.
Also explained here:
Why is the gets function so dangerous that it should not be used?
Specially, in this answer from Jonathan Leffler.
Use fgets() instead.
Also the defintion of a and b inside of the while loop doesn´t make any sense, even tough this is just a toy program and for learning purposes.
Furthermore note, that scanf() leaves the newline character, made by the press to return from the scanf() call in stdin. You have to catch this one, else the first fgets() thereafter will consume this character.
Here is the corrected program:
#include <stdio.h>
int main()
{
int t;
char a[10],b[10];
if(scanf("%d",&t) != 1)
{
printf("Error at scanning!");
return 1;
}
getchar(); // For catching the left newline from scanf().
while(t--)
{
puts("Enter string A: ");
fgets(a,sizeof a, stdin);
puts("Enter string B: ");
fgets(b,sizeof b, stdin);
printf("\n");
puts(a);
puts(b);
printf("\n\n");
}
return 0;
}
Execution:
$PATH/a.out
2
Enter string A:
hello
Enter string B:
world
hello
world
Enter string A:
apple
Enter string B:
banana
apple
banana
The most important message for you is:
Never use gets - it can't protect against buffer overflow. Your buffer can hold 9 characters and the termination character but gets will allow the user to typing in more characters and thereby overwrite other parts of the programs memory. Attackers can utilize that. So no gets in any program.
Use fgets instead!
That said - what goes wrong for you?
The scanf leaves a newline (aka a '\n') in the input stream. So the first gets simply reads an empty string. And the second gets then reads "surya".
Test it like this:
#include <stdio.h>
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
char a[10],b[10];
puts("enter");
gets(a); // !!! Use fgets instead
puts("enter");
gets(b); // !!! Use fgets instead
puts("enter");
printf("|%s| %zu", a, strlen(a));
printf("|%s| %zu", b, strlen(b));
}
return 0;
}
Input:
1
surya
whatever
Output:
enter
enter
enter
|| 0|surya| 5
So here you see that a is just an empty string (length zero) and that b contains the word "surya" (length 5).
If you use fgets you can protect yourself against user-initiated buffer overflow - and that is important.
But fgets will not remove the '\n' left over from the scanf. You'll still have to get rid of that your self.
For that I recommend dropping scanf as well. Use fgets followed by sscanf. Like:
if (fgets(a,sizeof a, stdin) == NULL)
{
// Error
exit(1);
}
if (sscanf(a, "%d", &t) != 1)
{
// Error
exit(1);
}
So the above code will automatically remove '\n' from the input stream when inputtin t and the subsequent fgets will start with the next word.
Putting it all together:
#include <stdio.h>
int main()
{
int t;
char a[10],b[10];
if (fgets(a,sizeof a, stdin) == NULL)
{
// Error
exit(1);
}
if (sscanf(a, "%d", &t) != 1)
{
// Error
exit(1);
}
while(t--)
{
puts("enter");
if (fgets(a,sizeof a, stdin) == NULL)
{
// Error
exit(1);
}
puts("enter");
if (fgets(b,sizeof b, stdin) == NULL)
{
// Error
exit(1);
}
puts("enter");
printf("%s", a);
printf("%s", b);
}
return 0;
}
Input:
1
surya
whatever
Output:
enter
enter
enter
surya
whatever
Final note:
fgets will - unlike gets - also save the '\n' into the destination buffer. Depending on what you want to do, you may have to remove that '\n' from the buffer.

Function fgets skips user input?

When I use the function fgets, the program skips the user input, effecting the rest of the program. An example program with this effect is:
#include <stdio.h>
int main() {
char firstDigit[2];
char secondDigit[2];
printf("Enter your first digit: ");
fgets(firstDigit, 1, stdin);
printf("\nEnter your second digit: ");
fgets(secondDigit, 1, stdin);
printf("\n\nYour first digit is %s and your second digit is %s.\n", firstDigit, secondDigit);
}
I then thought that maybe the problem was that fgets might be writing the newline, so I changed the code to account for that:
#include <stdio.h>
int main() {
char firstDigit[3];
char secondDigit[3];
printf("Enter your first digit: ");
fgets(firstDigit, 2, stdin);
printf("\nEnter your second digit: ");
fgets(secondDigit, 2, stdin);
printf("\n\nYour first digit is %c and your second digit is %c.\n", firstDigit[0], secondDigit[0]);
}
This time, the first input works properly, but the second input is skipped.
What am I doing incorrectly?
char firstDigit[2] and char secondDigit[2] are not large enough to hold a digit, a newline character, and a null-terminator:
char firstDigit[3];
char secondDigit[3];
Then, the calls to fgets() need to specify the size of the buffer arrays:
fgets(firstDigit, sizeof firstDigit, stdin);
/* ... */
fgets(secondDigit, sizeof secondDigit, stdin);
When instead fgets(firstDigit, 2, stdin); is used, fgets() stores at most two characters, including the \0 character, in firstDigit[]. This means that the \n character is still in the input stream, and this interferes with the second call to fgets().
In answer to OP's comment, How would you remove the unread characters from the input stream?, a good start would be to use more generous allocations for firstDigit[] and secondDigit[]. For example, char firstDigit[100], or even char firstDigit[1000] will be large enough that any expected input will be taken in by fgets(), leaving no characters behind in the input stream. To be more certain that the input stream is empty, a portable solution is to use the idiomatic loop:
int c;
while ((c = getchar()) != '\n' && c != EOF) {
continue;
}
Note here that it is necessary to check for EOF, since getchar() may return this value if the user signals end-of-file from the keyboard, or if stdin has been redirected, or in the unlikely event of an input error. But also note that this loop should only be used if there is at least a \n character still in the input stream. Before attempting to clear the input stream with this method, the input buffer should be checked for a newline; if it is present in the buffer, the input stream is empty and the loop should not be executed. In the code below, strchr() is used to check for the newline character. This function returns a null pointer if the sought-for character is not found in the input string.
#include <stdio.h>
#include <string.h> // for strchr()
int main(void)
{
char firstDigit[3]; // more generous allocations would also be good
char secondDigit[3]; // e.g., char firstDigit[1000];
printf("Enter your first digit: ");
fgets(firstDigit, sizeof firstDigit, stdin);
/* Clear input stream if not empty */
if (strchr(firstDigit, '\n') == NULL) {
int c;
while ((c = getchar()) != '\n' && c != EOF) {
continue;
}
}
putchar('\n');
printf("Enter your second digit: ");
fgets(secondDigit, sizeof secondDigit, stdin);
/* Clear input stream if not empty */
if (strchr(secondDigit, '\n') == NULL) {
int c;
while ((c = getchar()) != '\n' && c != EOF) {
continue;
}
}
puts("\n");
printf("Your first digit is %c and your second digit is %c.\n",
firstDigit[0], secondDigit[0]);
return 0;
}
It may be even better to use a single buffer[] to store lines of input, and then to store individual characters in chars. You could also write a function to clear the input stream, instead of rewriting the same loop each time it is needed:
#include <stdio.h>
#include <string.h> // for strchr()
void clear_stdin(void);
int main(void)
{
char buffer[1000];
char firstDigit;
char secondDigit;
printf("Enter your first digit: ");
fgets(buffer, sizeof buffer, stdin);
firstDigit = buffer[0];
/* Clear input stream if not empty */
if (strchr(buffer, '\n') == NULL) {
clear_stdin();
}
putchar('\n');
printf("Enter your second digit: ");
fgets(buffer, sizeof buffer, stdin);
secondDigit = buffer[0];
/* Clear input stream if not empty */
if (strchr(buffer, '\n') == NULL) {
clear_stdin();
}
puts("\n");
printf("Your first digit is %c and your second digit is %c.\n",
firstDigit, secondDigit);
return 0;
}
void clear_stdin(void)
{
int c;
while ((c = getchar()) != '\n' && c != EOF) {
continue;
}
}
For the first case, fgets(firstDigit, 1, stdin); cannot read anything from the input because the buffer has a size of only 1 byte, and fgets() must store a null terminator into the destination.
For the second case: fgets(firstDigit, 2, stdin); reads 1 byte from stdin, the digit that you typed, and cannot read the newline because the destination array is already full, allowing for the null terminator. The second fgets() reads the pending newline from the first entry and returns immediately for the same reason, not letting you type the second input.
You must allow fgets() to read at least 2 bytes by providing a buffer size of at least 3:
#include <stdio.h>
int main(void) {
char firstDigit[3];
char secondDigit[3];
printf("Enter your first digit: ");
if (!fgets(firstDigit, sizeof firstDigit, stdin))
return 1;
printf("\nEnter your second digit: ");
if (!fgets(secondDigit, sizeof secondDigit, stdin))
return 1;
printf("\n\nYour first digit is %s and your second digit is %s.\n",
firstDigit, secondDigit);
return 0;
}
Note that if you type more than a single character before the enter key, the program will still behave in an unexpected way.
This is a buffer problem. When you press enter, don't know why it is saved in the stdin buffer.
After you perform an fgets(...) you must type fflush(stdin); on all circumstances.
Something like this:
printf("Enter your first digit: ");
fgets(firstDigit, 1, stdin);
fflush(stdin);

How to read a integer followed by a string in C? [duplicate]

This question already has an answer here:
How to read / parse input in C? The FAQ
(1 answer)
Closed 6 years ago.
I am trying to write a simple program which will read two input lines, an integer followed by a string. However, it doesn't seem to work for me.
int main()
{
int i;
char str[1024];
scanf("%d", &i);
scanf("%[^\n]", str);
printf("%d\n", i);
printf("%s\n", str);
return 0;
}
Immediately after entering the integer and pressing "Enter", the program prints the integer. It doesn't wait for me to enter the string. Whats wrong? Whats the correct way to program this?
What you need to know
The problem with %[^\n] is that it fails when the first character to be read is the newline character, and pushes it back into the stdin.
The Problem
After you enter a number for the first scanf, you press Enter. %d in the first scanf consumes the number, leaving the newline character ('\n'), generated by the Enter keypress, in the standard input stream (stdin). %[^\n] in the next scanf sees this \n and fails for the reason given in the first paragraph of this answer.
Fixes
Solutions include:
Changing scanf("%d", &i); to scanf("%d%*c", &i);. What %*c does is, it scans and discards a character.
I wouldn't recommend this way because an evil user could trick the scanf by inputting something like <number><a character>\n, ex: 2j\n and you'll face the same problem again.
Adding a space (any whitespace character will do) before %[^\n], i.e, changing scanf("%[^\n]", str); to scanf(" %[^\n]", str); as #Bathsheba mentioned in a comment.
What the whitespace character does is, it scans and discards any number of whitespace characters, including none, until the first non-whitespace character.
This means that any leading whitespace characters will be skipped when inputting for the second scanf.
This is my recommendation: Clear the stdin after every scanf. Create a function:
void flushstdin(void)
{
int c;
while((c = getchar()) != '\n' && c != EOF);
}
and call it after every scanf using flushstdin();.
Other issues:
Issues unrelated to your problem include:
You don't deal with the case if scanf fails. This can be due to a variety of reasons, say, malformed input, like inputting an alphabet for %d.
To do this, check the return value of scanf. It returns the number of items successfully scanned and assigned or -1 if EOF was encountered.
You don't check for buffer overflows. You need to prevent scanning in more than 1023 characters (+1 for the NUL-terminator) into str.
This can be acheived by using a length specifier in scanf.
The standards require main to be declared using either int main(void) or int main(int argc, char* argv[]), not int main().
You forgot to include stdio.h (for printf and scanf)
Fixed, Complete Program
#include <stdio.h>
void flushstdin(void)
{
int c;
while((c = getchar()) != '\n' && c != EOF);
}
int main(void)
{
int i;
char str[1024];
int retVal;
while((retVal = scanf("%d", &i)) != 1)
{
if(retVal == 0)
{
fputs("Invalid input; Try again", stderr);
flushstdin();
}
else
{
fputs("EOF detected; Bailing out!", stderr);
return -1;
}
}
flushstdin();
while((retVal = scanf("%1023[^\n]", str)) != 1)
{
if(retVal == 0)
{
fputs("Empty input; Try again", stderr);
flushstdin();
}
else
{
fputs("EOF detected; Bailing out!", stderr);
return -1;
}
}
flushstdin();
printf("%d\n", i);
printf("%s\n", str);
return 0;
}
This simply, will work:
scanf("%d %[^\n]s", &i, str);
Instaed of scanf() use fgets() followed by sscanf().
Check return values of almost all functions with a prototype in <stdio.h>.
#include <stdio.h>
int main(void) {
int i;
char test[1024]; // I try to avoid identifiers starting with "str"
char tmp[10000]; // input buffer
// first line
if (fgets(tmp, sizeof tmp, stdin)) {
if (sscanf(tmp, "%d", &i) != 1) {
/* conversion error */;
}
} else {
/* input error */;
}
// second line: read directly into test
if (fgets(test, sizeof test, stdin)) {
size_t len = strlen(test);
if (test[len - 1] == '\n') test[--len] = 0; // remove trailing ENTER
// use i and test
printf("i is %d\n", i);
printf("test is \"%s\" (len: %d)\n", test, (int)len);
} else {
/* input error */;
}
return 0;
}

Resources