Using fgets and strtol to get a single integer - c

I am learning C and I am trying to use fgets() and strtol() to accept user input and just take the first character. I am going to make a menu that will allow a user to select options 1-3 and 4 to exit. I want each option to only be selected if '1', '2', '3', or '4' are selected. I don't want 'asdfasdf' to work. I also don't want '11212' to select the first option since it starts with a 1. I created this code so far while I started testing and for some reason this loops over the question and supplies a 0 to the input.
#include <stdio.h>
#include <stdlib.h>
int main() {
char a[2];
long b;
while(1) {
printf("Enter a number: ");
fgets(a, 2, stdin);
b = strtol(a, NULL, 10);
printf("b: %d\n", b);
}
return 0;
}
output
Enter a number: 3
b: 3
Enter a number: b: 0
Enter a number:
It should be:
Enter a number: 3
b: 3
Enter a number: 9
b: 9

You need to have enough room for the '\n' to be read or else it will be left in the input buffer and the next iteration it will be read immediately and thus make fgets() return with an empty string and hence strtol() returns 0.
Read fgets()'s documentation, it reads until a '\n' or untill the buffer is full. So the first time, it stops because it has no more room to store characters, and then the second time it still has to read '\n' and it stops immediately.
A possible solution is to increase the buffer size, so that the '\n' is read and stored in it.
Another solution, is to read all remaining characters after fgets().
The second solution could be cleanly implemented by reading one character at a time instead, since you are just interested in the first character you can discard anything else
#include <stdio.h>
#include <stdlib.h>
int main()
{
int chr;
while (1) {
// Read the next character in the input buffer
chr = fgetc(stdin);
// Check if the value is in range
if ((chr >= '0') && (chr <= '9')) {
int value;
// Compute the corresponding integer
value = chr - '0';
fprintf(stdout, "value: %d\n", value);
} else {
fprintf(stderr, "unexpected character: %c\n", chr);
}
// Remove remaining characters from the
// input buffer.
while (((chr = fgetc(stdin)) != '\n') && (chr != EOF))
;
}
return 0;
}

Related

Issue while creating a simple calculator in C

I am creating a very basic calculator in C but the output is not coming as desired.
#include<stdio.h>
int main(int argc, char const *argv[])
{
/* code */
char ch;
int a,b,p=0;
scanf("%d",&a);
while(1)
{
ch=getchar();
if(ch==';')
{
p=2;
break;
}
scanf("%d",&b);
if(ch=='+')
{
a+=b;
}
if(ch=='-')
{
a-=b;
}
if(ch=='*')
{
a*=b;
}
if(ch=='/' && b!=0)
{
a/=b;
}
if(ch=='/' && b==0)
{
printf("INVALID INPUT\n");
p=2;
break;
}
}
if(p!=0)
printf("%d",a);
return 0;
}
The Output is always coming as the initial value which has been assigned to "a".
Output which is coming-
4
+
5
;
4
Expected output -
4
+
5
;
9
Can you please help me with this issue of why the expression is not getting evaluated correctly?
The line
scanf("%d",&a);
will consume the first number from the input stream, but it will leave the newline character on the input stream.
Therefore, when you later call
ch=getchar();
it will read that newline character. It will not read the + character.
If you want to read the + character, then you can change that line to the following:
scanf( " %c", &ch );
This line will first discard all whitespace characters and will match the first non-whitespace character on the input stream.
Afterwards, your program will have the desired output:
4
+
5
;
9
An alternative solution would be to discard the rest of the line after every call to scanf that uses the %d format specifier. That way, calling getchar immediately afterwards should read the + character as intended.
You can discard the remainder of an input line using the following code:
//discard remainder of input line
{
int c;
do
{
c = getchar();
} while ( c != EOF && c != '\n' );
}
Here is a more compact way of writing the same thing:
//discard remainder of input line
for ( int c; (c=getchar()) != EOF && c != '\n'; )
;
The only problem is in scanning the inputs.
As a tab or a new line must seperate the value supplied to scanf.
In short, just add \n at the end of scanf.
scanf("%d\n", &a);
scanf("%d\n", &b);
this should do it.

How to accept ";' semicolon, as input, and not execute the next line of codes?

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...

How to scan in values and ignore the characters

I am a newbie to C and I was looking over some questions where I pondered upon a question where we need to scan in values using the users input. Example
1 2 3 45 6 7. So Automatically we scan these values into a 2D array.
One thing that troubles me is what If the user inputs
1 2 3 2 3 Josh, how can we ignore Josh and only scan in the values into the array.
I looked at using getchar and use a flag variable but I am unable to figure out the conundrum of differentiating between the integer and character.
/* This is something that I tried */
#include <stdio.h>
int main(int argc, char *argv[]) {
int a;
int b;
int A[10];
while (((a = getchar()) != '\n') && (b = 0)) {
if (!(a >= "A" && a <= "Z")) {
scanf("%d", A[b]);
}
b++;
}
}
}
I think one good method for achieving what you want is using scanf with the format "%s", which will read everything as a string, effectively splitting the input according to white spaces. From the manual:
s
Matches a sequence of non-white-space characters; the next
pointer must be a pointer to character array that is long
enough to hold the input sequence and the terminating null
byte ('\0'), which is added automatically. The input string
stops at white space or at the maximum field width, whichever
occurs first.
To convert the string to integer, you can use atoi. From the manual:
The atoi() function converts the initial portion of the string
pointed to by nptr to int.
So, if it converts the initial portion of the string into an integer, we can use that to identify what is a number and what's not.
You can build a simple "word detector" for atoi.
Using the function isalpha from ctype.h you can do:
int isword(char *buffer)
{
return isalpha(*buffer);
}
And rewriting your reading program you have:
#include <stdio.h>
#include <ctype.h>
int isword(char *buffer)
{
return isalpha(*buffer);
}
int main(void)
{
char input[200];
int num;
while (1) {
scanf("%s", input);
if (!strcmp(input, "exit")) break;
if (isword(input)) continue;
num = atoi(input);
printf("Got number: %d\n", num);
}
return 0;
}
You should keep in mind that the name isword is fallacious. This function will not detect if buffer is, in fact, a word. It only tests the first character and if that is a character it returns true. The reason for this is the way our base function itoa works. It will return zero if the first character of the buffer is not a number - and that's not what you want. So, if you have other needs, you can use this function as a base.
That's also the reason I wrote a separate function and not:
if (!isalpha(input[0]))
num = itoa(input);
else
continue;
The output (with your input):
$ ./draft
1 2 3 2 3 Josh
Got number: 1
Got number: 2
Got number: 3
Got number: 2
Got number: 3
exit
$
About assigments and &&
while (((a = getchar()) != '\n') && (b = 0))
As I said in a comment, this loop will never work because you're making a logical conjunction(AND) with an assignment that will always return zero. That means the loop condition will always evaluate to false.
In C, assignments return the value assigned. So, if you do
int a = (b = 10);
a will have now hold the value 10. In the same way, when you do
something && (b = 0)
You're effectively doing
something && 0
Which will always evaluate to false (if you remember the AND truth table):
p q p && q
---------------
0 0 0
0 1 0
1 0 0
1 1 1
Your code is completely wrong. I suggest to delete it.
You could use scanf with %d to read in numbers. If it returns 0, there is some invalid input. So, scan and discard a %s and repeat this process:
int num = -1;
while(num != 0)
{
printf("Enter a number, enter 0 to exit:");
if(scanf("%d", &num) == 0) /* If scanf failed */
{
printf("Invalid input found!");
scanf("%*s"); /* Get rid of the invalid input (a word) */
}
}

Scanning values in C till hit a new-line char, '\n'

How can I scanf() the integer values I enter into an array until I hit enter.
I believe I can use getchar() != '\n'.
but how do I loop through the line ?
Suppose my input is 20 21 2 12 2. I want an array that has all those inputs.
What given functions could I use in order to scan them all in.
You are trying to read integers as characters so once read you need to convert it to integers.
Read the line to a buffer using fgets() then parse the input buffer to get integers.
Store the integers to the array.
The code looks like
char buf[300];
int a[5],i=0;
fgets(buf,sizeof(buf),stdin);
char *p = strtok(buf," ");
while(p != NULL)
{
char *endptr;
a[i] = strtol(p,&endptr,10);
if ((*endptr != '\0') && (isspace(*endptr) == 0))
printf("warning: invalid value detected\n");
else
i++;
p = strtok(NULL," ");
}
You can use the alternative strtol() instead of atoi() to convert string to integer.
PS: Your buf should be large enough to hold the whole line. fgets() read till newline character.
If you use getchar() you obtain digits one by one, so you need
to store them first in the buffer, and when white space comes,
you convert those digits into a number, and store it into array.
here is the explanation of the code I made for you.
1st if statement : if obtained character is a digit, store it in buf
2nd if statement : if obtained character is a white space or EOL and at least 1 digit is stored in buf, convert digits into number and store it in array a.
3rd if statement : if obtained character is not a digit or a white space or a EOL, warns users.
4th if statement : if obtained character is a EOL, end loop.
The code below works fine.
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int main(){
#define BUFSIZE 50
#define ARRAYSIZE 5
int i,k,a[ARRAYSIZE];
char c,buf[BUFSIZE];
for(i=0,k=0;(i<BUFSIZE)&&(k<ARRAYSIZE);){
c=getchar();
if(isdigit(c)){
buf[i++] = c;
}else if((i>0) && (c==' ' || c=='\n')){
buf[i] = '\0';
a[k++] = atoi(buf);
i=0;
}else if(!(c==' ' || c=='\n')){
printf("warning : invalid value %c is detected\n",c);
i=0;
}
if(c=='\n'){
break;
}
}
printf("input :");
for(i=0;i<ARRAYSIZE;i++){
printf("%d, ",a[i]);
}
printf("\n");
}

How to limit input length with scanf

In this program I have taken a dimensional character array of size[3][4],
as long as I enter a 3 characters for each row it will work well.
For example: if I enter abc abd abd I get the same output but if i enter more letters in the first or second or 3rd row I get an error.
How should I check for null character in 2 dimensional?
# include <stdio.h>
#include <conio.h>
# include <ctype.h>
void main()
{
int i=0;
char name[3][4];
printf("\n enter the names \n");
for(i=0;i<3;i++)
{
scanf( "%s",name[i]);
}
printf( "you entered these names\n");
for(i=0;i<3;i++)
{
printf( "%s\n",name[i]);
}
getch();
}
As pointed out by #SouravGhosh, you can limit your scanf with "%3s", but the problem is still there if you don't flush stdin on each iteration.
You can do this:
printf("\n enter the names \n");
for(i = 0; i < 3; i++) {
int c;
scanf("%3s", name[i]);
while ((c = fgetc(stdin)) != '\n' && c != EOF); /* Flush stdin */
}
How should I chk for null character in 2 dimensional ... [something has eaten the rest part, I guess]
You don't need to, at least not in current context.
The problem is in your approach of allocating memory and putting input into it. Your code has
char name[3][4];
if you enter more that three chars, you'll be overwriting the boundary of allocated memory [considering the space of \0]. You've to limit your scanf() using
scanf("%3s",name[i]);
Note:
change void main() to int main(). add a return 0 at the end.
always check the return value of scanf() to ensure proper input.
EDIT:
As for the logical part, you need to eat up the remainings of the input words to start scanning from the beginning of the next word.
Check the below code [Under Linux, so removed conio.h and getch()]
# include <stdio.h>
# include <ctype.h>
int main()
{
int i=0; char name[3][4];
int c = 0;
printf("\n enter the names \n");
for(i=0;i < 3;i++)
{
scanf( "%3s",name[i]);
while(1) // loop to eat up the rest of unwanted input
{ // upto a ' ' or `\n` or `EOF`, whichever is earlier
c = getchar();
if (c == ' ' || c == '\n' || c == EOF) break;
}
}
printf( "you entered these names\n");
for(i=0;i<3;i++)
{
printf( "%s\n",name[i]);
}
return 0;
}
(Cringing after reading the answers to date.)
First, state the problem clearly. You want to read a line from stdin, and extract three short whitespace separated strings. The stored strings are NUL terminated and at most three characters (excluding the NUL).
#include <stdio.h>
void main(int, char**) {
char name[3][4];
printf("\n enter the names \n");
{
// Read tbe line of input text.
char line[80];
if (0 == fgets(line, sizeof(line), stdin)) {
printf("Nothing read!\n");
return 1;
}
int n_line = strlen(line);
if ('\n' != line[n_line - 1]) {
printf("Input too long!\n");
return 2;
}
// Parse out the three values.
int v = sscanf(line, "%3s %3s %3s", name[0], name[1], name[2]);
if (3 != v) {
printf("Too few values!\n");
return 3;
}
}
// We now have the three values, with errors checked.
printf("you entered these names\n%s\n%s\n%s\n",
name[0], name[1], name[2]
);
return 0;
}
you might consider something on the order of scanf( "%3s%*s",name[i]);
which should, if I recall correctly, take the first three characters (up to a whitespace) into name, and then ignore anything else up to the next white space. This will cover your long entries and it does not care what the white space is.
This is not a perfect answer as it will probably eat the middle entry of A B C if single or double character entries are mode. strtok, will separate a line into useful bits and you can then take substrings of the bits into your name[] fields.
Perhaps figuring out the entire requirement before writing code would be the first step in the process.

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