So thats my programm:
This program scans and prints numbers, but I want it to print "Mistake" if a character is entered. It produces an infinite loop if I put something like an "a" in. Why?
#include <stdio.h>
int main(void){
printf("Enter a number: \n");
int a;
while(scanf("%d",&a)!=EOF){
if(46<'a'<58){
}
else{
printf("Mistake.");
return -1;
}
printf("%d\n",a);
}
}
First of all, you cannot chain the relational operators in C. You need to change
if(46<'a'<58){
to
if ((46 < 'a') && ( 'a' < 58 )) {
to make it work.
That said, a char value is not a match for %d format specifier. You need to either
use %c to scan the input as char and check the ASCII value
use %d to scan the int input and check for the return value of scanf() for success. Also, in case scanf() fails for a non-int value, you need to clean up the input buffer before you loop again to avoid the infinite looping.
Read input as a string with fgets(), then test it as needed.
Using scanf("%d", ...) returns 0, 1, or EOF when it fails, reads an int or end-of-file is detected. Code could test that return value, but robust code simple does not use scanf().
char buf[100];
while(fgets(buf, sizeof buf, stdin) != NULL) {
int a;
if (sscanf(buf, "%d",&a) == 1) {
printf("%d\n",a);
} else {
puts("Mistake.");
}
}
if(46<'a'<58) does not work. it first compares 46<'a' which is 1:true and then compares 1 < 58 which is also true. #124...
46 < 'a' < 58 is always true, because 46 < 'a' evaluates to 1, and then 1 < 58 evaluates to 1. Also I suppose you want a, rather than 'a', which is actually a literal. Additionally, I think your code has some logical problems.
Here is the refined code:
#include <stdio.h>
int main(void){
puts("Enter a number:");
char a;
while(scanf("%c", &a) != EOF) {
if(('0' <= a && a <= '9') || a == '\n'){
putchar(a);
}
else
{
puts("Mistake.");
return -1;
}
}
}
Related
In the following code example from K&R's book, if I replace putchar(c) with printf("%c", c) the code works the same. But if I replace it with printf("%d", c) it gives gibberish output.
#include <stdio.h>
int main() {
int c;
c = getchar();
while (c != EOF) {
putchar(c);
c = getchar();
}
}
From here, I learned that the getchar() converts the stdin to an 8-bit character whose value ranges from 0 to 255.
Now I want to print the value of c using putchar(c) in one line and printf("%d", c) in another line. So I wrote the following code:
#include <stdio.h>
int main() {
int c, b;
c = getchar();
b = c;
while (c != EOF && c != 10) {
printf("%c",c);
c = getchar();
}
printf("\n");
while (b != EOF && b != 10) {
printf("%d\t",b);
b = getchar();
}
}
I used the condition c != 10 as the newline character is read as 10 by getchar(). I expected the code to work as
$ ./a.out
783
783
55 56 51
but the program terminates as
$ ./a.out
783
783
55
I understand that getchar() takes input from stdin and the variable b is not stdin. So how should I copy the variable c to b so that my program works as I expect it to?
The problem is that your code does not (and cannot, as it stands) 'remember' the inputs you gave in the first loop. So, after you have finished that loop, your second loop is wanting to read in the characters for b (after it has output the first value, which is remembered from the earlier b = c line).
So, after outputting 55 (the integer value of the character 7), it is waiting for further input.
Probably the easiest way to get the output that you're looking for is to have an array of input characters. Then, you can output the %c values as you read them (as before), then re-run the outputs using the %d format in a subsequent for loop.
Here is a demonstration that does what I think you're after:
#include <stdio.h>
#define MAXINS 20 // Set to the maximum number of input characters you want to allow
int main()
{
int c[MAXINS];
int i = 0, n = 0;
c[0] = getchar();
while (i < MAXINS && c[i] != EOF && c[i] != 10) {
printf("%c", c[i]);
c[++i] = getchar();
++n;
}
printf("\n");
for (i = 0; i < n; ++i) {
printf("%d\t", (int)(c[i]));
}
return 0;
}
Feel free to ask for further clarification and/or explanation.
EDIT: On the point in the your first paragraph, "But if I replace it with printf("%d", c) it gives gibberish output." Well, when I try the following code and give 783 and then hit return (which generates a newline) I get the expected 55565110 as the output:
int main()
{
int c;
c = getchar();
while (c != EOF) {
printf("%d", c);
c = getchar();
}
return 0;
}
This may look like gibberish, but it's just the same output as you 'expect' in your later code, but without the spaces and with the addition of the 10 for the newline.
You need to have every character stored, because once you read a char from stdin, it is not present in stdin anymore.
Since you want the newline character in the end as a part of the input, you should use fgets to take the input.
Say you are taking an input that could have a maximum of 100 characters.
#include <stdio.h>
int main(void) {
char c[100]; // A char array
fgets(c,100,stdin);
int x=0;
while (c[x] != 10 && c[x] != EOF)
printf("%c",c[x++]);
printf("\n");
x = 0;
while (c[x] != 10 && c[x] != EOF) // You can simply compare it with the newline character too.
printf("%d ",c[x++]);
printf("\n");
return 0;
}
There are many ways to do this. You can also read stdin character-by-character ans store it in an array. However, since you need to display the ASCII values of the characters in another line after displaying the characters themselves, you will have to store them in an array.
You are copying only the first input, to copy the whole string you need to store each input in a buffer and check if the string doesn't overflow that buffer on each iteration:
int main(void)
{
enum {size = 256};
char buffer[size];
size_t count = 0;
int c;
while ((c = getchar()) && (c != '\n') && (c != EOF))
{
printf("%c", c);
if (count < size)
{
buffer[count++] = (char)c;
}
}
printf("\n");
for (size_t iter = 0; iter < count; iter++)
{
printf("%d\t", buffer[iter]);
}
printf("\n");
}
If you don't want to limit the buffer to an arbitrary size then you need to change your approach to use dynamic memory (realloc or a linked list)
I'm having trouble finding out how to set up a loop where i enter input and then
stop the input by pressing 'e' or 'E'. The input entered is integers but needs to be stopped with a character. That is where i get lost. I have seen a bunch of information about using ascii conversions but i dont know how efficient that would be. This code is broken but it is as far as i could get. Any information would be helpful.
int main(void)
{
char num;
int sub;
while (sub != 'e' || sub != 'E') {
scanf("%d", &num);
sub = #
printf("%d", num);
}
return 0;
}
Simple.
#include <stdio.h>
#include <ctype.h>
int main(void) {
char c = getchar();
int num;
while (c != 'e' || c != 'E') {
if (isdigit(c))
num = c - '0';
c = getchar();
}
return 0;
}
But you don't have to use an ascii character as a way to stop input. You can use EOF which is -1. It is Ctrl-D on UNIX systems and Ctrl-Z on Windows.
int c;
while ((c = getchar()) != EOF)
A direct way to distinguish between an input of int, 'e' and , 'E' is to read a line of user input with fgets() and then parse it.
#define LINE_SZ 80
char buf[LINE_SZ];
while (fgets(buf, sizeof buf, stdin) && buf[0] != 'e' && buf[0] != 'E') {
if (sscanf(buf, "%d", &num) != 1) {
Handle_other_non_int_input();
}
sub = #
printf("%d", num);
}
As noted in the comments, (sub != 'e' || sub != 'E') is always true. If sub can never be e and E at the same time.
Note that sub is an int and not an integer pointer (int *).
The line sub = # assigns sub with num's address.
And the value of sub is used in the control expression of the while loop before it is initialised. sub has garbage value at that point which is indeterminate. You have to initalise it with some value before using it.
Do
int num, rv;
while( 1 )
{
rv=scanf("%d", &num);
if(rv==0)
{
if( (num=getchar())=='e' || num=='E' )
{
break;
}
else
{
while(getchar()!='\n');
continue;
}
}
printf("\n%d", num);
}
A value is read into num by scanf() whose return value is stored in rv.
scanf() returns the number of successful assignments which in this case should be 1 if an integer value was read into num since %d is the format specifier.
If rv is 1, it is a number and is printed. Otherwise it could be a character which won't read by the scanf() and would remain unconsumed in the input buffer. The first byte of this data is read by the getchar() and if this is e or E, the loop is exited but otherwise the input buffer is cleared till a \n is encountered and the next iteration of the loop is done without going into the part where the printing takes place.
What is wrong with this ? Also, I have to use scanf(). It is supposed to read any integers and sum them, the loop is to stop when 0 is entered..
main (void){
int a;
int r=0;
while(scanf(" %d",&a)){
r=r+a;
}
printf("the sum is %d\n",r);
return 0;
}
Quoting from man
These functions return the number of input items assigned. This
can be
fewer than provided for, or even zero, in the event of a matching fail-
ure. Zero indicates that, although there was input available, no conver-
sions were assigned; typically this is due to an invalid input character,
such as an alphabetic character for a `%d' conversion.
The value EOF is
returned if an input failure occurs before any conversion such as an end-
of-file occurs. If an error or end-of-file occurs after conversion has
begun, the number of conversions which were successfully completed is
returned.
So, that pretty much explains what is returned by scanf().
You can solve the problem by adding ( 1 == scanf("%d", &a) && a != 0 ) as the condition in your while loop like
int main (void)
{
int a;
int r=0;
while( 1 == scanf("%d", &a) && a != 0 )
{
r=r+a;
}
printf("the sum is %d\n",r);
return 0;
}
Also note that you have to specify the type of main as int main().
I would also like to add that the loop will end when you enter a character like 'c' ( or a string ) and it will show the sum of all the numbers you entered before entering the character.
scanf() doesn't return what it has written to the variable. It returns the total number of items successfully filled.
EDIT:
You would be much better off using fgets() to read from stdin and then using sscanf() to get the integer, which you can check against 0.
#define BUFF_SIZE 1024
int main (void)
{
int a;
int r = 0;
char buffer[BUFF_SIZE] = {0};
while(1) {
fgets(buffer, sizeof buffer, stdin);
sscanf(buffer, "%d", &a);
if(!a)
break;
r = r + a;
}
printf("the sum is %d\n", r);
return 0;
}
i need help with short Code in C. I must read floats on input line seperated with space and input is ended with float 0 or EOF.
How to do this if i dont know how many numbers or in input, or how it works and ask to EOF if i am reading just numbers and not chars?
Thanks for any response.
example of input in one line:
12 11 10 45 50 12 EOF
12 10 11 45 0
int main(void)
{
float num;
float sum = 0;
do{
scanf("%f", num);
sum += num;
} while(EOF || num == 0);
return 0;
}
From the man page of scanf -
scanf returns the number of items successfully matched and assigned
which can be fewer than provided for, or even zero in the event of an
early matching failure. The value EOF is returned if the end of input
is reached before either the first successful conversion or a matching
failure occurs.
This means that scanf will return EOF only when it encounters EOF as the first input when it is called because EOF must be preceded with a newline '\n' else it won't work (depending on the OS). You must also account for the matching failure scanf may encounter.
#include <stdio.h>
int main(void) {
float num;
float sum = 0;
int val;
while((val = scanf("%f", &num)) != EOF && val == 1) {
sum += num;
}
if(val == 0) {
printf("matching failure. input is not a float.\n");
}
else {
printf("end of input.\n");
}
return 0;
}
From scanf reference:
On success, the function returns the number of items of the argument
list successfully filled. This count can match the expected number of
items or be less (even zero) due to a matching failure, a reading
error, or the reach of the end-of-file.
If a reading error happens or the end-of-file is reached while
reading, the proper indicator is set (feof or ferror). And, if either
happens before any data could be successfully read, EOF is returned.
If an encoding error happens interpreting wide characters, the
function sets errno to EILSEQ.
So, you may rewrite your do-while loop to something like
int retval;
while((retval = scanf("%f", &num)) != EOF && retval > 0 && num != 0) {
sum += num;
}
if(retval == 0) {
printf("input read error.\n");
}
to match your constraints.
Also note you need to prefix your variable with & when passing it to scanf(), since the function expects a pointer to deal with (you need to pass variable address).
EDIT:
see this topic concerning EOF problems in Windows
You can re write your code like this
int main(void)
{
float num;
float sum = 0;
do
{
scanf("%f", &num);
sum += num;
} while((!feof(stdin)) && (num != 0));
printf("%f", sum);
return 0;
}
Here feof indicates end of input stream.
The following may be a slightly more robust way to do this:
#include <stdio.h>
#include <string.h>
int main(void) {
int sum=0;
int num;
char *p;
char buf[1000];
fgets(buf, 1000, stdin);
p = strtok(buf," ");
while(p!=NULL) {
if(sscanf(p, "%d", &num) == 1) sum+=num;
p = strtok(NULL, " ");
}
printf("the sum is %d\n", sum);
}
Test:
> testme
1 2 3 4 0
the sum is 10
> testme
1 2 3 4 ^D
the sum is 10
Note - you have to enter ctrl-D twice to get the desired effect when you are at the end of a line.
you can get your doubt clear by reading "C programming a modern approach by K N King"
This book provides proper clarification on this topic
Test the result of scanf() for 0, 1 or EOF.
Test the value scanned for 0.0.
int main(void) {
float num;
float sum = 0;
int cnt;
while ((cnt = scanf("%f", &num)) == 1) {
if (num == 0.0) break;
sum += num;
}
// cnt should be EOF, 0 or 1
if (cnt == 0) {
printf("Input is not a number\n");
}
else {
printf("Sum %f\n", sum);
}
return 0;
}
Although, in general, scanf() returns values EOF, 0, 1, ... "number of format specifiers", a value of 0 occurs rarely. Example input is "+".
I'm writing a program in C that is suppose to ask the user for a number.
The number has to be greater than zero and cannot have letters before or after the number. (ie: 400 is valid but abc or 400abc or abc400 is not). I can make my program invalidate everything besides 400abc. How would I make it invalidate an input if it starts valid then turns invalid? (I'm about 2 months into an intro to c class so my knowledge is very limited.)
#include<stdio.h>
int check(void);
void clear_input(void);
main()
{
int num;
printf("Please enter a number: ");
num = check();
printf("In Main %d\n", num);
}
int check(void){
int c;
scanf("%d", &c);
while (c < 0){
clear_input();
printf("Invalid, please enter an integer: ");
scanf("%d", &c);
}
return c;
}
void clear_input(void){
char junk;
do{
scanf("%c", &junk);
}while (junk != '\n');
}
You can also check whether ascii value of each char scanned from user input should lie in range 48-57, It will only then be integer value.
strtol can be used to do it, but it takes some extra work.
After running this code:
char *endptr;
int n = strtol(num_text, &endptr, 10);
n will contain the number. But you still have to check that:
1. *endptr=='\0' - this means strtol didn't stop in the middle. In 400abc, endptr will point to abc. You may want to allow trailing whitespace (in this case, check that endptr points to an all-whitespace string.
2. num_text isn't empty. In this case, strtol will return 0, but an empty string isn't a valid number.
Read the input as a line, using fgets.
Check if all characters are numeric.
If not, it's invalid. If yes, use sscanf to get the line into an int.
Check if the int is in the range; you're done.
Scanf with %d will treat the "400abc" as 400, all the trailing characters would be ignored, so there is nothing to worry about.
If you definitely want to treat "400abc" as an invalid input, then maybe you shouldn't use %d in scanf, use %s instead?
One way is to read the whole line as a string and check by yourself if it contains any non-digits.
The other way is reading the integer and then looking into the input using fgetc() to see if the next character after the detected input is valid. Or you could even use the same scanf() for this:
char delim;
if(scanf("%d%c", &c, &delim) == 2 && !isspace(delim))
// the input is invalid
You can read the number in a character array and validate it by checking if all the characters lie in the ascii range 48 to 57 ( '0' to '9' ) .If so your no. is valid otherwise you can safely regard it as invalid input.Here the the demonstration :
#include<stdio.h>
#include<string.h>
int conv( char * word )
{
int ans=0;
int res=0;
for(int i=0;i<strlen(word);i++)
if(word[i]>='0' && word[i]<='9')
ans=(ans*10) + (word[i] - '0');
else
res=-999;
if(res==-999)
return res;
else
return ans;
}
int main()
{
char a[10];
gets(a);
int b=conv(a);
if(b==-999)
printf("Invalid Entry.\n");
else
printf("Success.No is %d.\n",b);
return 0;
}
You can adjust for negatives as well by checking the first character in the word array to be '-' and adjusting the sign accordingly.
This is C99, so compile with -std=c99
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <stdbool.h>
bool getNum(int *n) {
char c, s[10];
if (!scanf("%9s", s))
return false;
for (int i=0; c=s[i]; i++)
if (!isdigit(c))
return false;
*n = atoi(s);
return true;
}
int main() {
int n;
if (getNum(&n))
printf("you entered %d\n", n);
else
printf("you did not enter a number\n");
}
The following is your check function rewritten to fix your problem, so try this:
int check(void){
int n;
char c;
while (EOF==scanf("%d%c", &n,&c) || n < 0 || !isspace(c)){
clear_input();
printf("Invalid, please enter an integer: ");
}
return n;
}