I'm a beginner in C, and I've got problem I can't figure out, and wasn't able to find a solution on other threads here.
I'm trying to read integers from a keyboard input/ txt file with the following code:
int grades[MAX_GRADES_LENGTH]={0}, histogram[HISTOGRAM_SIZE]={0};
int maxGradesHistogramBucket=0, median=0, gradesLength=0;
double avg=0.0;
int grade=0;
printf("Enter grades:\n");
while (scanf("%d",&grade) != EOF)
{
grades[gradesLength]=grade;
gradesLength=gradesLength+1;
}
I'm supposed to set these "grades" in the grades[] array and count the length of the array along the loop.
Somehow the loop is misbehaving, it seems that some inputs are ok with the loop, but for some inputs(most of them actually) the scanf() doesn't get the EOF, whether it's an actual end of file, or the ^D command in the terminal.
I've heard the scanf() is not the most reliable method to read data, but unfortunately I can't use anything else since we haven't learned any other method in class so we can only use scanf() on our homework.
I've tried to change the!= EOF with == 1 and its all the same.
for the input
100 0 90 10 80 20 70 30 60 40 50
for example it works fine.
but for the input:
0 50 100
the loop is infinite.
I'm using a macbook pro btw (if it matters).
Any ideas?
If you type a letter instead of a number, scanf() will return 0 (as in, "zero successfully converted numbers") and not EOF (as in, "there was no data left to read"). The correct test is to ensure that you got the expected number of values — in this case, 1:
while (scanf("%d", &grade) == 1)
If you need to know whether you got to EOF or got no result (but reading the rest of the line might clear the problem), then capture the return value from scanf():
int rc;
while ((rc = scanf("%d", &grade)) == 1)
{
}
if (rc != EOF)
…read the rest of the line, or at least the next character, before resuming the loop…
And, if you really want to, you could then write:
int rc;
while ((rc = scanf("%d", &grade)) != EOF)
{
if (rc == 1)
grades[gradesLength++] = grade;
else
{
printf("Discarding junk: ");
int c;
while ((c = getchar()) != EOF && c != '\n')
putchar(c);
putchar('\n');
if (c == EOF)
break;
}
}
The code in the else clause could plausibly be put into a function. It might also report the messages to standard error rather than standard output. It is courteous to let the user know what it was that you objected to. You could stop before newline with a different test (&& !isdigit(c) && c != '+' && c != '-', using isdigit() from <ctypes.h>). However, the user doesn't have a chance to re-edit the stuff they put after the letters, so you may be going to misinterpret their input. It is probably better just to throw away the rest of the line of input and let them start over again.
As chux noted, after reading a character that could be part of an integer, that character needs to be put back into the input stream. Therefore, if I were going to analyze the rest of the line and restart scanning at the first data that could actually be part of an integer, I'd consider using something like:
#include <ctype.h>
static inline int could_be_integer(int c)
{
return isdigit(c) || c == '+' || c == '-');
}
and then the else clause might be:
else
{
printf("Discarding junk: ");
int c;
while ((c = getchar()) != EOF && c != '\n' && !could_be_integer(c))
putchar(c);
putchar('\n');
if (could_be_integer(c))
ungetc(c, stdin);
else if (c == EOF)
break;
}
This gets messy, as you can see. Sometimes (as Weather Vane noted in a comment), it is easier to read a line with fgets() and then use sscanf() in a loop (see How to use sscanf() in a loop?). Be wary of suggestions about Using fflush(stdin); it isn't automatically wrong everywhere, but it won't work on a MacBook Pro under normal circumstances.
On the whole, simply ignoring the rest of the line of input is usually a better interface decision.
It works for me.
I enclosed your snippet thus:
#include <stdio.h>
#include <errno.h>
#define MAX_GRADES_LENGTH 20
#define HISTOGRAM_SIZE 20
main()
{
int grades[MAX_GRADES_LENGTH]={0}, histogram[HISTOGRAM_SIZE]={0};
int maxGradesHistogramBucket=0, median=0, gradesLength=0;
double avg=0.0;
int grade=0;
int i;
printf("Enter grades:\n");
while (scanf("%d",&grade) != EOF)
{
grades[gradesLength]=grade;
gradesLength=gradesLength+1;
}
if (errno)
perror("grade");
for (i = 0; i < gradesLength; ++i) {
printf("%d\n", grades[i]);
}
}
and ran it:
$ a.out
Enter grades:
100 0 90 10 80 20 70 30 60 40 50
100
0
90
10
80
20
70
30
60
40
50
$ a.out
Enter grades:
0 50 100
0
50
100
$
Perhaps you are looking in the wrong place. Maybe the bug is in your output routine?
Personally, if had to do this, given some ambiquity over what scanf returns when, and without rewriting it, then this small change is probably more reliable:
int i, r;
printf("Enter grades:\n");
while ((r = scanf("%d",&grade)) > 0)
Related
Beginner here.
In the ANSI C textbook by K&R, page 20, they ask: How would you test the word count program?
I have copied exactly from the text book, using the CodeBlocks IDE, console application. I have seen many great input tests online, but my question is even dumber. How do I actually input something? Nothing happens when I press enter. Do I have this problem because I am using an IDE and therefore not learning how to run C programs properly?
Thanks in advance. I added a picture to show you what I mean
Here is the code:
#include <stdio.h>
#define IN 1 /* inside a word */
#define OUT 0 /* outside a word */
/* counts lines, words and characters as input */
main()
{
int c, nl, nw, nc, state;
state = OUT;
/* set these three constants to 0: */
nl = nw = nc = 0;
while ((c = getchar()) != EOF){
++nc;
if (c == '\n')
++nl;
/* || == OR (&& == AND)
evaluation of the following line
will stop as soon as the truth or
falsehood is known, so the order matters */
if (c == ' ' || c == '\n' == c == '\t')
state = OUT;
else if (state == OUT){
state = IN;
++nw;
}
}
printf("%d %d %d\n", nl, nw, nc);
}
On MacOS:
gcc kr_wc.c -o kr_wc
./kr_wc < example_text.txt
Example output:
40 260 1397
Where example_text.txt is this file:
1.5.4 Word Counting
The fourth in our series of useful programs counts lines, words, and
characters, with the loose definition that a word is any sequence of
characters that does not contain a blank, tab or newline. This is a
bare-bones version of the UNIX program wc.
#include <stdio.h>
#define IN 1 /* inside a word */
#define OUT 0 /* outside a word */
/* count lines, words, and characters in input */
int main() {
int c, nl, nw, nc, state;
state = OUT;
nl = nw = nc = 0;
while ((c = getchar()) != EOF) {
++nc;
if (c == '\n')
++nl;
if (c == ' ' || c == '\n' || c == '\t')
state = OUT;
else if (state == OUT) {
state = IN;
++nw;
}
}
printf("%d %d %d\n", nl, nw, nc);
}
Every time the program encounters the first character of a word, it
counts one more word. The variable state records whether the program
is currently in a word or not; initially it is "not in a word", which
is assigned the value OUT. We prefer the symbolic constants IN and
OUT to the literal values 1 and 0 because they make the program more
readable. In a program as tiny as this, it makes little difference,
but in larger programs, the increase in clarity is well worth the
modest extra effort to write it this way from the beginning. You'll
also find that it's easier to make extensive changes in programs
where magic numbers appear only as symbolic constants.
The program to count words in K&R 2nd edition is made to run on an environment in which you signal the end of input somehow. Normally, as they used UNIX all the time, they used Ctrl-D sequence (which is valid if you run the program in Linux or any Unix-like operating system) This has been so since the early beginning of the UNIX system.
Windows signals the end of input in a console application by the input of a Ctrl-Z (probably followed by a keyboard return key)
If you redirect the input from a file (as when you say a.out <my_input_file.txt) you'll get the number of words at the end, when there's no more input in the file.
You are running the program in an IDE, which is something that normally hides you where standard input and standard output go, or how to signal the window you show how to say that there's not more input to the program.
For the program to get to it's end, you have first to know how to get to an end on input.
The examples in K&R omit a return type of main, which is not valid in modern C, so add int before main():
#include <stdio.h>
#define IN 1 /* inside a word */
#define OUT 0 /* outside a word */
/* counts lines, words and characters as input */
int main()
{
int c, nl, nw, nc, state;
state = OUT;
/* set these three constants to 0: */
nl = nw = nc = 0;
while ((c = getchar()) != EOF){
++nc;
if (c == '\n')
++nl;
/* || == OR (&& == AND)
evaluation of the following line
will stop as soon as the truth or
falsehood is known, so the order matters */
if (c == ' ' || c == '\n' == c == '\t')
state = OUT;
else if (state == OUT){
state = IN;
++nw;
}
}
printf("%d %d %d\n", nl, nw, nc);
}
"How do I actually input something? Nothing happens when I press enter."
If you got problems with your IDE just run it online.
"How would you test the word count program"?
To cite the authors of a bundle of K&R solutions with an answer to that particular question here:
It sounds like they are really trying to get the programmers to learn how to do a unit test. I would submit the following:
input file contains zero words
input file contains 1 enormous word without any newlines
input file contains all white space without newlines
input file contains 66000 newlines
input file contains word/{huge sequence of whitespace of different >kinds}/word
input file contains 66000 single letter words, 66 to the line
input file contains 66000 words without any newlines
input file is /usr/dict contents (or equivalent)
input file is full collection of moby words
input file is binary (e.g. its own executable)
input file is /dev/null (or equivalent)
66000 is chosen to check for integral overflow on small integer machines.
Creating a simple C function that prevents the user from entering anything other than a number between 1 and 9. Any other input including letters symbols and any number that is either less than 1 and greater than 9 should not be accepted.
So far its pretty straight forward; however, the part where the code is supposed to check that the entered character is not a symbol or letter is not working the way I wanted it to work.
int validateUserInput(){
printf("%s\n", "Please enter a number from 1 to 9: ");
char value = getchar();
int numValue = value;
char temp;
int digitCounter = 0;
while((temp = getchar()) != '\n'){
digitCounter++;
}
//if there is more than 1 digit.
if(digitCounter>0){
printf("%s\n","Input too long!");
return validateUserInput();
}
// if the char entered is not between 1 and 9
// this part is giving me a hard time.
else if(numValue < 49 || numValue > 57){
printf("%s\n", "Imput is not within the valid parameters");
return validateUserInput();
}
return numValue;
}
There are a couple of additional points to look at when taking input with your input routine. What if the user needs to cancel input? (e.g. presses ctrl+d, or ctrl+z on windows) As you have it written, there is no way to cancel the loop. (while that works for your purpose of forcing only a single input of 1-9, it leaves no way to cancel) If you trap EOF, you provide both, a way to cancel, and a way to indicate cancellation back in the calling function (e.g. by checking the return against EOF)
While recursion has its place, be aware that each recursion is itself a separate function call that requires a separate stack, and all the other trappings of a function call. There are many times when that can be avoided with a simple goto statement. glibc makes regular use of goto (e.g., check the source of qsort, getdelim, etc..) In your case, a single goto label can completely eliminate the need for recursion. For example, you could do something similar to the following while meeting all of your criteria:
int validateuserinput()
{
int c, extra, tmp;
getinput:; /* a simple goto can avoid recursion completely */
extra = 0;
printf ("Please enter a number from 1 to 9: ");
/* prompt/loop while chars not between 1 and 9 */
for (c = getchar(); (c < '1' || '9' < c ); c = getchar()) {
if (c == '\n') { /* no digits in input */
fprintf (stderr, "error: invalid input.\n");
goto getinput;
}
else if (c == EOF) { /* trap EOF */
fprintf (stderr, "\nerror: input canceled.\n");
return c;
}
}
/* empty input buffer -- increment extra count */
for (tmp = getchar(); tmp != '\n' && tmp != EOF; tmp = getchar())
extra++;
if (extra) { /* if extra chars -- input too long */
fprintf (stderr, "error: input too long.\n");
goto getinput;
}
return c - '0'; /* return integer value instead of ASCII value */
}
(just a style note, C generally avoids the use of camelCase variable names in favor of lower-case, that's just a generality, it's entirely up to you)
You can check the function (and respond to a cancellation) with the following short bit of code:
#include <stdio.h>
int validateuserinput();
int main (void) {
int n;
if ((n = validateuserinput()) != EOF)
printf ("\n valid input : %d\n", n);
return 0;
}
Example Use/Output
Testing accepting input of only 1-9:
$ ./bin/inputhandler
Please enter a number from 1 to 9: Hello World!
error: invalid input.
Please enter a number from 1 to 9: ?
error: invalid input.
Please enter a number from 1 to 9: 0
error: invalid input.
Please enter a number from 1 to 9: 23
error: input too long.
Please enter a number from 1 to 9: 6
valid input : 6
Testing input cancellation (e.g. ctrl+d, or ctrl+z on windows)
$ ./bin/inputhandler
Please enter a number from 1 to 9:
error: input canceled.
While there is nothing wrong with using recursion, it always helps to ask "Does this need to be a recursive function to begin with?" Sometimes the answer will be yes, but often there are simple ways to avoid the additional overhead. (note: the overhead with a few recursions is minimal so it isn't a big consideration in your case, but if you unwittingly call a recursive function that spins a million times, it rapidly becomes a concern)
Look over all answers, and if you have any questions, just let us know.
You only set numValue once - for the first getchar() - looks like you expect it to be set after every getchar().
while((temp = getchar()) != '\n'){
digitCounter++;
}
could be
while((temp = getchar()) != '\n'){
if (temp < 49 || temp > 57 || ++digitCounter>0)
return validateUserInput();
}
This way makes it simple I think
I have a very simple C example program that does a crude count of characters words and spaces from input. The program compiles without error but when tested the program doesn't return any of the int variables via the print function. I am using VS2012 for coding and compiling. Stepping into the code shows that the values are being calculated correctly. Is there something wrong with my code or the compiler?
#include <stdio.h>
#define IN 1
#define OUT 0
/* count digits, white space, others */
main()
{
int c, nl, nw, nc, state;
state = OUT;
nl = nw = nc = 0;
while ((c = getchar()) != EOF){
++nc;
if(c == '\n'){
++nl;
}
if (c == ' ' || c == '\n' || c == '\t'){
state = OUT;
} else if (state == OUT){
state = IN;
++nw;
}
}
printf("%d %d %d\n", nl, nw, nc);
}
It works if you run it with stdin coming from a file a.exe < test.txt. It works if you run it with stdin coming from the console on Linux. It does not work if you run it with stdin coming from the console on Windows.
It must be some sort of Windows console oddity.
In order to see the actual output, start the debugger with Ctrl-F5. This will keep the console window open.
See this answer for more information.
Your loop is permanent. Meaning it doesn't end.. so it doesn't reach the call of printf.
EOF is indicator that clearly doesn't work like that here. You must break the loop on a specific key. Like enter for instance, which character decimal representation is 13 for carriage return. 10 for NL.
I've written a small C program where I wanted to display the numeric ASCII value that corresponds to certain key presses.
My code follows. The problem is, after running the program, it accepts input, but doesn't do anything else. It doesn't even reach the first printf statement. I can't figure out what the issue is - is there a problem with mixing the getchar() function with the printf() function in the same program?
#include <stdio.h>
int main() {
const int numKeys = 256;
int keys[numKeys];
int i;
for (i = 0; i < numKeys; i++) {
keys[i] = 0;
}
printf("\n Start pressing some keys!\n\n");
int c;
while ((c = getchar()) != EOF) {
printf(" CAPTURED: %d\n", c);
keys[c]++;
}
printf("\n\n ** RESULTS ** \n\n");
for (i = 0; i < numKeys; i++) {
if (keys[i] != 0) {
printf(" Key with value %d was called %d times.", i, keys[i]);
}
}
}
I should clarify that I have a Windows XP Pro machine, with Cygwin installed. I use Cygwin for my development space, so I wonder if there is something different when running this type of program in that environment.
I found the problem. I think you want to use
while ((c = getchar()) != EOF && c != '\n')
Instead if you want to have it print the results after the person hits enter/return.
problem 1 : getting to printf(" CAPTURED: %d\n", c); without having to press the Enter key
solution : is by using getche() in while loop.
problem 2 : getting to 'printf("\n\n ** RESULTS ** \n\n");' or essentially breaking while loop?
solution : you cannot. you will never get EOF as long as you read from keyboard.
workaround : close stdin or use a escape character other than EOF.
EDIT : workaround2 :
->use getchar() itself. but to print those entered char u need to press Enter key. now on windows ctrl+z gives EOF but this should be the **FIRST** input on the line after you press Enter key. well this is not a good solution.
if you want a "Press key display times pressed scenario. there is just no simple way(AFAIK)"
I believe that the first printf statement gets executed, but due to buffering is not displayed on the screen immediately. Use fflush(stdout) to send the contents of the buffer to the screen. Ie:
printf("\n Start pressing some keys!\n\n");
fflush(stdout);
I am working on a program to filter a list of craigslist results; I want to find a relatively cheap room for rent. The finished program will remove lines that have a price over $600, and create a new file, but for now I am removing every line with a $ character, and printing to the terminal.
The program works fine when run on its own source, but when I run it on an html page of craigslist results saved from Firefox, it prints until the closing html bracket and throws a stack smashing detected warning and a backtrace. I am learning C from K&R so if this code looks antiquated that's why.
# include <stdio.h>
# define MAXLINE 300
main()
{
char line[MAXLINE];
int c;//current character
int p = 0;//position in the line
int flag = 0;//there is a dollar sign
while ((c = getchar()) != EOF){
line[p++] = c;
if (c == '$'){
flag = 1;
}
if (c == '\n'){
if(flag == 0){//there was no $, print the line
int i;
for(i=0;i<p;i++){
putchar(line[i]);
line[i] = '\0';
}
}
p = 0;
flag = 0;
}
}
}
I imagine the problem is just that the HTML contains at least one line that is more than MAXLINE characters long. You don't check anywhere whether you're about to exceed the size of the array; if you do, you would indeed smash the stack. Your while loop could check whether p was less than MAXLINE, print a message if not, and stop. You couldn't do anything else without fairly significant changes to your program.