Why is this while loop not exiting ?
I want to input a series of number from users and then use them one by one for further processing. I do not have the size of numbers list. Can someone please tell me how to do this ?
#include<stdlib.h>
#include<stdio.h>
int main()
{
int x;
scanf("%d",&x);
while(x!='\n')
{
printf("%d",x);
scanf("%d",&x);
}
return 0;
}
Example:
Input:
3 5 6
Output:
3 5 6
Actual Output:
3 5 6
but loop doesn't exit
The %d conversion specifier never stores the \n into the buffer. It reads up to the \n in the buffer and leaves it there. So, this is essentially an infinite loop.
You may need to use a format specifier like %c which actually reads and stores the \n.
Although, if I may suggest, try to make use of getchar(), which reads the next character from the standard input and returns the value as an int. However, for an input of digits, you need to parse and store the integer and/or float value accordingly, but it will certainly provide you more flexibility.
What about this?
while(scanf("%d", &x) != EOF) {
printf("%d",x);
if (getchar() == '\n') break;
}
Why not use argv and strtol ?
#include <stdlib.h>
#include <stdio.h>
int main(int argc, char **argv) {
int i = 0, x;
for(i = 1; i < argc; i++) {
x = strtol(argv[i],NULL,10);
printf("%d\n",x);
}
return 0;
}
while(scanf("%d",&x) != EOF) {
printf("%d",x);
}
...would do it. Don't ignore scanf returns. The reason your code is looping is you're really comparing x against decimal value of 10 ('\n'). Also, please remember that you asked scanf to give you integers and that's what it gives you, it will not give you newlines, even if it could.
Edited, so that I added comparison to EOF specifically.
You need to remember that scanf scans a stream, in this case the input stream, it will not return until that stream is done.
You should really use strings and sscanf() for what you want to do.
Related
I'm pretty new in C, I used to work in Python, and I'm trying to see if something that I read is integer number. If not, to read until I manage to entry a number.
I did some research and I found out that the function scanf actually returns 1 if the read is done suitably, and 0 otherwise.
So, I have written this code, and I don't understand why this is an infinite loop, writing "Give an integer" on the console
#include <stdio.h>
int main() {
int a;
int b = 1;
do {
printf("Give an integer\n");
b = scanf("%d", &a);
} while (b == 0);
}
The problem with scanf() is that it stops reading when the first white space is encountered for most specifiers like "%d", so it's left in the input and that's why reading again would cause a problem if you don't discard such white space because it will then return immediately the next time you call it. There is a mandatory white space that is introduced when you press Enter or Return, the '\n' new line character (or line feed).
If you want to read an integer and make sure you did you can try like this
#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
int
main(void)
{
long int value; // You can adapt this to use `int'
// but be careful with overflow
char line[100];
while (fgets(line, sizeof(line), stdin) != NULL) {
char *endptr;
value = strtol(line, &endptr, 10);
if ((isspace(*endptr) != 0) && (endptr != line))
break;
}
fprintf(stdout, "%ld\n", value);
return 0;
}
read strtol()'s manual to understand this code
You could as suggested in the comments, remove the white space characters from the input, but IMHO that is harder and you would still have other problems with scanf(), like handing empty input which is not straight forward.
I don t understand why this is an infinite loop, writing "Give an intiger" on the console
The problem is that scanf() does not consume data that it cannot match against the specified format. It leaves such characters unread in the input stream. Therefore, if you try reading again from the same stream with the same format, without consuming at least one character by some other means, then you can be certain that the input will again not be matched. And again. And again.
To avoid your infinite loop, you need to consume at least one character of the non-matching input after each matching failure. There are many ways you could do that; here's a fairly simple one:
#include <stdio.h>
int main() {
int a;
do {
printf("Give an intiger\n");
if (scanf("%d", &a)) {
// breaks from the loop on a successful match or an error
break;
}
// consume the remainder of one line of input without storing it
if (scanf("%*[^\n]") == EOF) {
break;
}
} while (1);
}
That consumes the whole remainder of the line on which the non-matching input is encountered, which will yield less surprising interactive behavior for some inputs than many alternatives would.
If you've a penchant for writing terse code, or if you don't like to break out of the middle of a loop, then you might write the same thing like this:
#include <stdio.h>
int main() {
int a;
do {
printf("Give an intiger\n");
} while ((scanf("%d", &a) == 0) && (scanf("%*[^\n]") != EOF));
}
Because the && operator short circuits, the second scanf() call will be executed only if the first returns zero, and the loop will exit after the first iteration wherein either the first scanf() call returns nonzero or the second returns EOF (indicating an error).
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#define N 30
int main()
{ char c, y, input[N];
int X, i=0, j;
printf("Give displacement\n");
scanf("%d",&X);
printf("Give chars\n");
while(((c=getchar()) !=EOF) && (i<N-1)){
input[i]=c;
i++;
}
input[i]='\0';
j=0;
for(j=0; j<=i-1; j++){
if (isalpha(input[j])){
if (isupper(input[j]))
y=(input[j]-'A'+X)%26+'A';
else
y=(input[j]-'a'+X)%26+'A';
putchar(y);
}
}
return 0;
}
hey all. well, this code doesnt seems to works as it should. It skips 2 positions at the table instead of one. Which makes the program unusable since i need a table of 30 positions. I think that the problem is in the while loop, but i really cant find it. Any help would be apprecieted.
Thanks in advance.
After the call to scanf, there's a newline left in the input buffer. That newline becomes the first character read by getchar.
To get the newline out of the buffer, add a separate call to getchar right after the scanf call:
scanf("%d",&X);
getchar();
That will give you the additional character you're missing.
Also, as mentioned in the comments, c should be defined as an int instead of a char, because getchar returns an int. Otherwise, the test for EOF will always be false.
if you want 30 characters you have to define N as 31
[0 - 29] + '\0'
In your first while loop i believe that the second test is wrong:
Use it like this:
while(((c=getchar()) !=EOF) && (i<=N-1)){
input[i]=c;
i++;
}
This question already has answers here:
Why is scanf() causing infinite loop in this code?
(16 answers)
Closed 9 years ago.
In line 5 I read an integer and isint is getting 1 if it reads an integer or 0 if it's not an integer. If isint is 0 I have a loop asking user to give an integer and I read until the user gives an integer. I try this code giving a character instead of an integer but I have an infinite loop. The program just doesn't wait to give a new input. What's wrong with my code?
#include <stdio.h>
int main(void) {
int arg1;
//int arg2;
int attacknum = 1;
int isint = 1;
//printf("Insert argument attacks and press 0 when you have done this.\n");
printf("Attack %d\n", attacknum);
attacknum++;
printf("Give attacking argument:");
isint = scanf("%d", &arg1); //line 5
while(isint == 0){
printf("You did not enter a number. Please enter an argument's number\n");
isint = scanf("%d", &arg1);
printf("is int is %d\n", isint);
}
return 0;
}
As others have mentioned, if scanf can't parse the input, it leaves it unscanned.
Generally scanf is a poor choice for interactive input because of this kind of behavior, and because it doesn't match the line-at-a-time interface experienced by the user.
You are better off reading one line into a buffer using fgets. Then parse that line using sscanf. If you don't like the input, throw the whole line away and read another one.
Something like this:
#include <stdio.h>
int main(void)
{
char line[256];
int arg1;
int isint;
while (1) {
printf("Give attacking argument:");
fgets(line, sizeof line, stdin);
isint = sscanf(line, "%d",&arg1);
if (isint) break;
printf("You did not enter a number.Please enter an argument's number\n");
}
printf("Thanks for entering %d\n", arg1);
return 0;
}
(For production code you'll want to handle long lines, check return codes, also check for trailing garbage after the number, etc.)
Actually, an even better approach would be to not use scanf if you just want to read an integer, and instead use strtol. That gives you a handy pointer to the character just after the number, and you can check that it's whitespace or nul.
When scanf is confronted with a non-digit it will not consume any input and return that zero integers were read. The non-digit will stay in the input for the next call to scanf that will behave the same as the first call, etc.
In answer to your question below. You could use fgetc to parse at least one character, but this will give the error messages for every character already typed. Typically I think you want to skip until a newline. To this end you could use fgets as suggested by poolie. Or you could add the following after scanf.
int ch;
if (isint == 0)
while ((ch = fgetc(stdin)) != EOF && ch != '\n')
{
/* Skip characters */
}
P.S: In your case it is probably better to put it just before the first printf in the loop.
I started learning C programming and in this program I am trying to get user input and then a line at a time and decide if it contains non-int characters. I've been trying this method:
scanf("%d", &n);
if (isalpha(n))
{
i = -1;
}
I googled a bit and learned the function isalpha is good way to do it. However, I'm getting a segmentation fault every time I test the fragment above with non-int characters (letters for example). Any suggestion would be appreciated.
The %d format specifier forces scanf() to only accept strings of digits. Given anything else, it will fail and leave n unfilled (and assuming you didn't initialize n before, it will be filled with garbage).
The crux of the problem is that isalpha() expects a value between 0 and 255, and has an assertion to enforce it. At least on my VC++ compiler, it causes a crash with an access violation when given an invalid value (in non-debug mode).
To solve this you just have to switch to a %c format specifier. Converting n to a char would also be advisable as that makes your intent of reading a single character clearer.
EDIT: Given your clarifications in the comments, you can leave everything as is and simply check the return value of scanf() instead of going the isalpha() route. It returns the number of values read successfully, so when it encounters a non-integer or end of file, it will return 0. E.g.:
int main() {
int n;
while (scanf("%d", &n)) {
printf("Got int: %d\n", n);
}
}
I have no idea why you're getting a seg-fault. I'd have to see more of your program.
But using "%d" for scanf will only accept integer values and you'll get "0" for n that isn't an integer and therefore isalpha(n) will always be false and i will never be set to -1.
Perhaps you aren't initializing i and therefore it is never set. If you are referencing it later, that's probably the source of your seg-fault.
Use scanf("%c", &n), like this:
int main(char** argc, int argv) {
char n = 0;
int i = 0;
scanf("%c", &n);
if (isalpha(n)) {
i = -1;
}
printf("you typed %c, i=%d", n, i);
}
Make sure you have a character buffer to store the value in. Scan it as a string, and then use isalpha():
char buffer[32];
sscanf("%32s", buffer);
// loop and check characters...
if(isalpha(buffer[i])) ....
Note the use of %32s, this is to prevent buffer overflows (32 == size of buffer)
Given that n is an integer, we can diagnose that you are reading a value into n which is not in the range 0..255 plus EOF (normally -1), so that the code for isalpha(n) is doing something like:
(_magic_array[n]&WEIRD_BITMASK)
and the value of n is causing it to access memory out of control, hence the segmentation fault.
Since scanf():
Returns the number of successful conversions, and
Stops when there is a non-integer character (not a digit or white space or sign) in the input stream,
you can use:
#include <stdio.h>
int main(void)
{
char n = 0;
while (scanf("%c", &n) == 1)
printf("you typed %d\n", n);
return 0;
}
I am trying to print the ASCII values of 3 char-type characters. When I input the first char it doesn't print the value of the char. After the first char it starts to give the value.
#include <stdio.h>
int main() {
char ch;
int t;
while(t < 3){
scanf("%c\n", &ch);
printf("%c - %d\n", ch,ch);
t++;
}
}
http://i54.tinypic.com/2mdqb7d.png
Variable t is not automatically initialized to 0 by compiler. So You need to initialize t with 0. If printf doesn't print immediately it means the data is buffered. If you want to see immediatley you may consider flushing stdout right after printf.
I saw this several times, and don't know the root cause, but solution that works is:
scanf("\n%c", &ch);
It probably has something to do with buffered end of line character.