I'm learning C in ansi c and I have a question.
I what to stop reading from the console when the user prints enter.
Ideally the user will enter this
1 2 3 4 5
and right after typing 5 the program would output
1
2
3
4
5
I have this code:
#include <stdio.h>
#include <stdlib.h>
#define SIZE 5
int main()
{
int vector[SIZE] = {0}, number, counter = 0,i;
while (scanf("%d", &number) != EOF && counter < SIZE){
vector[counter] = number;
counter++;
}
for (i = 0; i < counter; i++){
printf("%d\n", vector[i]);
}
return 0;
}
Now, the program if I input 1 2 3 4 5 and then hit ENTER, it does not stop and if I type 1 2 3 4 5 6 it program stops.
Two things.
I believe the condition to scan until EOF it's not doing nothing. Second, program stops just because the counter < SIZE
So, how can I stop after user hits enter?
thanks
One option is to swap the order of your conditions:
while (counter < SIZE && scanf("%d", &number) != EOF){
As the conditions are evalulated left-to-right, your program waits for new input (via scanf(%d...)) before it evaluates counter < SIZE. This change would cause the program to immediately exit the while loop after the 5 is typed, BEFORE a user presses enter, as it evaluates counter < SIZE after the 5 is processed.
(If you choose this option, you will also want to print an endline character before you begin to repeat the numbers, as the user will not have pressed enter before it begins printing)
If you want the program to exit the loop when the user presses enter, you'll have to change your scanf. scanf(%d... reads a digit, and an endline character is not a digit.
Here's one approach off the top of my head (warning, untested):
int number;
char nl = 0;
while (counter < 5 && scanf("%d%c", &number, &nl) != EOF)
{
...
/**
* If we picked up a newline, exit the loop.
*/
if (nl == '\n')
break;
}
Try checking the last character of input to see if its a \n. ENTER does not constitute EOF, it generally is a \n character.
Related
Im trying to verify if the user enters the right value with the return values.
He should enter an even number between 0 and 100.
I think I got it right, but now my problem is, that the user has to enter the "enter" key twice to end the scanf function.
Do I have another possibility to avoid the user from doing so?
Heres the code I wrote:
#include <stdio.h>
#include <windows.h>
int main( void )
{
int ok, input = -1;
char c;
while(input < 1 || input > 100 || input%2 != 0) { //repeat loop if Input is not even or betwenn 0 and 100
printf("Enter an odd number between 1 und 100: ");
int ok = scanf("%d%c", &input, &c); //Input is correct if ok = 2 and c = 10'\n'
while((c = getchar()) != '\n' && c != EOF) {} //this loop empties the input buffer to avoid infinite loops if users enters a character
}
printf("You habe chosen the number %d ", input);
getchar();
return 0;
}
I believe you don't need the second while loop at all.
As suggested above, the scanf() is waiting for enter to be pressed once, and then getchar() is waiting after.
If you remove the second while loop, the code should run correctly and only require the enter key to be pressed once.
I hope this helps!
Here's a small portion of a practice I'm doing preventing erroneous inputs.
while(1) {
printf("Choose From 1 to 7 ");
if( scanf("%d", &nNum ) != 1) {
printf("Please only choose from the numbers 1-7.");
fgets(sErraticInputs, 100 , stdin);
} else if (nNum > 7 || nNum <= 0) {
printf("Please only choose from the numbers 1-7.");
} else {
break;
}
}
I was doing a good job, until I entered "6;p". It executed the 6 portion and ran correctly, but technically speaking it should have taken the whole thing as the input, and proceeded with the error message.
First of all I don't think the posted code can give the said result. The break statement will end the while(1) when 6 has been read so there will not be printed an error message.
If we assume that the break isn't part of your real code this is what happens:
When scanf is told to read an integer, it will continue reading from the input stream as long as the next character (together with the previous read characters) can be converted into an integer. As soon as the next character can not be used as part of an integer, scanf will stop and give you the result of what it has parsed so far.
In your case the input stream contains
6;p\n
So scanf will read the 6 and stop (i.e. return 6). The input stream now contains:
;p\n
Consequently this will be the input for your next scanf and cause the input error, you saw.
One way to solve this would be to flush stdin after all scanf - both on success and on failure:
nNum = 0;
while(nNum != 7) // Just as an example I use input 7 to terminate the loop
{
printf("Choose From 1 to 7 ");
if( scanf("%d", &nNum ) != 1 || nNum > 7 || nNum <= 0)
{
printf("Please only choose from the numbers 1-7.");
}
else
{
printf("Valid input %d\n", nNum);
// **************************** break;
}
fgets(sErraticInputs, 100 , stdin); // Always empty stdin
}
note: Using fgets with size 100 doesn't really ensure a complete flush... you should actually use a loop and continue until a '\n' is read.
With the change above input like 6;p will be taken as a valid input with value 6 and the ;p will be thrown away.
If that's not acceptable, you could drop the use of scanf and do the parsing yourself. There are several options, e.g. fgets or fgetc
The example below uses fgetc
#include <stdio.h>
#include <stdlib.h>
int get_next()
{
int in = fgetc(stdin);
if (in == EOF) exit(1); // Input error
return in;
}
void empty_stdin()
{
while(get_next() != '\n') {};
}
int main(void) {
int in;
int nNum = 0;
while(nNum != 7)
{
printf("Choose From 1 to 7 \n");
in = get_next();
if (in == '\n' || in <= '0' || in > '7') // First input must be 1..7
{
printf("Please only choose from the numbers 1-7.\n");
if (in != '\n') empty_stdin();
}
else
{
nNum = in - '0';
in = get_next();
if (in != '\n') // Second input must be \n
{
printf("Please only choose from the numbers 1-7.\n");
empty_stdin();
}
else
{
printf("Valid input: %d\n", nNum);
}
}
}
return 0;
}
This code will only accept a number (1..7) followed by a newline
Here's why the "whole thing" is not taken as the input. From the man pages:
The format string consists of a sequence of directives which describe
how to process the sequence
of input characters. If processing of a directive fails, no further input is read, and scanf()
returns. A "failure" can be either of the following: input failure, meaning that input characters
were unavailable, or matching failure, meaning that the input was inappropriate...
Here's the full text. Have a look at this as well.
One approach would be to read in the whole input using fgets and check whether the length of the input is greater than 1. For an input of length 1, check if the input is a number and so on...
I've looked at multiple solutions but none of them worked for me.
I'm asking the user to enter numbers in a loop, but if the user enters a specific number the loop should break.
This is what I've got so far.
#include <stdio.h>
#include <stdlib.h>
#define MAXNUMBERS 5
int getNumbers(int array[])
{
int i;
int n = 0;
printf("Enter max. %d numbers, enter empty line to end:\n", MAXNUMBERS);
for (i = 0; i < MAXNUMBERS; i++)
{
scanf("%d", &array[i]);
fflush(stdin);
n++;
if (array[i] == '5')
{
break;
}
}
return n;
}
int main()
{
int array[MAXNUMBERS];
int amount_numbers;
amount_numbers = getNumbers(array);
printf("Numbers entered: %d\n", amount_numbers);
printf("First three: %d %d %d", array[0], array[1], array[2]);
return 0;
}
Input:
1
5
4
3
2
Output:
Numbers entered: 5
First three: 1 5 4
If the user enters 5 the loop should break.
I'm using 5 as an example, I later want it to do with an empty line. But it doesn't even work with 5.
It just keeps prompting the user to enter another number after he entered 5.
The actual problem is '5' != 5 the former is the character 5 which is in fact it's ascii value, and the latter is the number 5, since you are reading integers, i.e. using the "%d" specifier in scanf() you should use 5, but it would be better if it was just a int variable, and you could initialize it to any number you like before the loop starts.
Your loop is wrong anyway because if the user enters a non-numeric value then your program will invoke undefined behavior. Besides you already invoke undefined behavior with fflush(stdin), so
Remove fflush(stdin)1
7.21.5.2 The fflush function
If stream points to an output stream or an update stream in which the most recent operation was not input, the fflush function causes any unwritten data for that stream to be delivered to the host environment to be written to the file; otherwise, the behavior is
undefined.
So the behavior is undefined for an input stream like stdin, or even if the most recent operation was input.
You must check that the value was read properly, and then check in the loop condition if it equals the value you want to stop the loop with, try this
int readNumber()
{
int value;
printf("input a number > ");
while (scanf("%d", &value) == 1)
{
int chr;
printf("\tinvalid input, try again...\n");
do { /* this, will do what you thought 'fflush' did */
chr = getchar();
} ((chr != EOF) && (chr != '\n'));
printf("input a number > ");
}
return value;
}
int getNumbers(int array[])
{
int i;
int stop = 5;
printf("Enter max. %d numbers, enter empty line to end:\n", MAXNUMBERS);
array[0] = 0;
for (i = 0 ; ((i < MAXNUMBERS) || (array[i] == stop)) ; i++)
array[i] = readNumber();
return i;
}
1This is a quote from the C11 draft 1570.
if (array[i] == '5')
You're checking whether array[i] is equal to the ASCII value of the character '5'.
Remove the '' to make it compare against the integer 5.
You are checking if an integer is equal to the character '5', which is then being cast to an ascii value of '5'.
Try using this:
if (array[i] == 5)
Disregard everything!
I should have written
if (array[i] == 5)
without the quotes!
I'm an idiot!
I sat 2 hours at this error...
I have this c program where I am inputing a number N followed by N more numbers. For example, I'll enter 100 followed by 100 more numbers. For some reason, after so many inputs the scanf function will stop working properly. It's as if it has stopped taking input and will just continue one with whatever value is in size.
The use case I came up with is 100 1 2 3 4 5 6 7 8 9 10... (repeated ten times). then after three or four times of that I'll type in 100 10 9 8 7 6 5 4 3 2 1... (repeated ten times) and then there will be an infinite loop of print statements.
int main(int argc, const char * argv[]) {
int histogram[10000];
int i;
while (1) {
int *rectPtr = histogram;
int size;
scanf("%d", &size);
if (!size) return 0;
for (i = 0; i < size; ++i) {
scanf("%d", rectPtr);
rectPtr++;
}
printf("%d", 1);
printf("\n");
}
return 0;
}
Distrust infinite loops.
In a series of comments, I said:
You're not testing the return value from scanf(), so you don't know whether it is working. The pair of printf() statements is odd; why not write printf("%d\n", 1); or even puts("1");?
Your code does not test or capture the return value from scanf(), so you do not know whether scanf() is reporting a problem. As a general rule, test the return value of input functions to make sure what you thought happened did in fact happen. You could also print out the values read just after you read them:
if (scanf("%d", rectPtr) != 1)
{
fprintf(stderr, "scanf() failed\n");
return 1;
}
printf("--> %d\n", *rectPtr);
rectPtr++;
Similarly when inputting size. Also consider if (size <= 0) return 0;. And using fgets() plus `sscanf() can make reporting errors easier.
j.will commented:
It is great to know if scanf fails, but I want to know why it fails and prevent it from failing. How do I do that?
I responded:
I understand you'd like to know. With scanf(), the best you can do after a failure is usually to read all the characters that follow up to a newline or EOF, and if you want to know what went wrong, then you print those characters too, because scanf() leaves the last character that it read in the input buffer ready for the next input operation.
void gobble(void)
{
printf("Error at: <<");
int c;
while ((c = getchar()) != EOF && c != '\n')
putchar(c);
puts(">>");
if (c == EOF)
puts("<<EOF>>");
}
The first character in the output is what caused the failure.
See also How to use sscanf() in loops?
Hacking your code to match this:
#include <stdio.h>
static void gobble(void)
{
printf("Error at: <<");
int c;
while ((c = getchar()) != EOF && c != '\n')
putchar(c);
puts(">>");
if (c == EOF)
puts("<<EOF>>");
}
int main(void)
{
enum { MAX_VALUES = 10000 };
int histogram[MAX_VALUES];
int size;
while (printf("Number of items: ") > 0 && scanf("%d", &size) == 1 &&
size > 0 && size <= MAX_VALUES)
{
int *rectPtr = histogram;
for (int i = 0; i < size; ++i)
{
if (scanf("%d", rectPtr) != 1)
{
gobble();
return 1;
}
rectPtr++;
}
printf("size %d items read\n", size);
}
return 0;
}
IMO, you need to check the return value of scanf() for proper operation. Please check the below code. I have added some modifications.
To exit from the program, you need to press CTRL+ D which will generate the EOF. Alternatively, upon entering some invalid input [like a char instead of int] wiil also cause the program to beak out of while() llop and terminate.
I have put the sequence to check first scanf(). All others need to be checked, too.
#include <stdio.h>
#include <stdlib.h>
int main(int argc, const char * argv[]) {
int histogram[10000] = {0};
int i;
int *rectPtr = histogram;
int size = 0;
int retval = 0;
printf("Enter the number of elements \n");
while ( (retval = scanf("%d", &size)) != EOF && (retval == 1)) {
rectPtr = histogram;
if (!size) return 0;
printf("Enter %d elements\n", size);
for (i = 0; i < size; ++i) {
scanf("%d", rectPtr); //check in a simmilar way to above
rectPtr++;
}
printf("%d\n", 1111111);
printf("Enter the number of elements: \n");
}
return 0;
}
The output of a sample run
[sourav#broadsword temp]$ ./a.out
Enter the number of elements: 2
Enter 2 elements
1
2
1111111
Enter the number of elements: 3
Enter 3 elements
1
2
3
1111111
Enter the number of elements: 9
Enter 9 elements
0
9
8
7
6
5
4
3
2
1111111
Enter the number of elements: r
[sourav#broadsword temp]$
histogram is declared to have size 10000. You say you do 100 1 2 3 ... repeated 10 times. If I correctly understand that uses 1000 slots in histogram.
If you repeat the test more than 10 times, you exhaust histogram and begin to write past the end of array causing undefined behaviour.
So you must either :
reset recPtr = histogram at each iteration
control recPtr - histogram + size <= sizeof(histogram) after reading size (IMHO better)
And as other said, you should always control input operations : anything can happen outside of your program ...
I have an assignment to "Write a C program that allows a user to enter up to 20 integers (it will stop accepting numbers based on a sentinel value or based on the 20-integer limit being reached). The program should then display the numbers in the reverse order of entry."
I decided to make my sentinel value "EOF" (or CTRL+D / CRTL+Z). My code has some really erratic behavior:
You have to press the EOF keys twice (which also creates a blank entry that is counted int the array.
The first digit from the first entry gets truncated.
Everything else seems to work OK but, this is clearly not the desired results. Below is my code. Can you explain what's wrong?
main() {
int i,iMax;
double dblMean;
int x[MAX];
printf("Please Enter Up to 20 Integers\n");
for (i=0; i<MAX; i++)
{
printf("%d.> ",i+1);
if (getchar() == EOF) /* Check for CTRL+D OR CTRL+Z */
break; /* Exit loop if EOF entered :: Must be pressed twice and eats first character of first entry */
scanf("%d",&x[i]);
iMax=i;
}
printf("\nYou entered %d numbers\n",iMax+1); /* Should be i+1 but EOF had to be entered twice */
printf("\nIndex Item Reverse Item\n");
for (i=0; i<=iMax; i++)
printf("%3d%4c%8d%9c%11d\n",i,' ',x[i],' ',x[iMax-i]);
return 0;
}
EDITED:
Here is my final code, Thank you all for your help:
#include <stdio.h>
#include <stdlib.h>
#define MAX 20
int main()
{
int i,iMax;
int x[MAX];
printf("Please Enter Up to 20 Integers\n");
for (i=0; i<MAX; i++)
{
printf("%d.> ",i+1);
if (scanf("%d",&x[i]) != 1) /* Checks for CTRL+D OR CTRL+Z */
break; /* EOF returns -1 and loop will be exited */
iMax=i; /* iMax must be set for second loop to exit properly */
/* Can sizeof x be used instead for second loop? */
}
printf("\nYou entered %d numbers\n",iMax+1); /* Displays number of items entered... Will be <= MAX*/
printf("\nIndex Item Reverse Item\n");
for (i=0; i<=iMax; i++) /* Why wont 'for (i=0; i<=sizeof x; i++)' work? */
printf("%3d%4c%8d%9c%11d\n",i,' ',x[i],' ',x[iMax-i]);
return 0;
}
The getchar() call reads (and effectively discards) the first digit because it is not EOF.
You don't test that scanf() worked; you should.
for (i = 0; i < MAX; i++)
{
if (scanf("%d", &x[i]) != 1)
break;
}
At this point, there are i integers in the array; you don't really need to set iMax in the loop. You could simply set it when the loop exits.
Also, in addition to checking scanf, here is what you need to know:-
EOF is not a character. The EOF is a macro that getchar() returns when it reaches the end of input or encounters some kind of error. The ^D is not "an EOF character". What's happening under linux when you hit ^D on a line by itself is that it closes the stream, and the getchar() call reaches the end of input and returns the EOF macro. If you type ^D somewhere in the middle of a line, the stream isn't closed, so getchar() returns values that it read and your loop doesn't exit.
see http://www.c-faq.com/stdio/getcharc.html for a nice description.