This is a super basic question... I am relearning C (haven't used it for more than 5 years). I can't get this code to work. I am trying to scan a user input (ascii character) as an integer, and show the ascii code for the entered character.
#include <stdio.h>
int main(int argc, char *argv[]) {
int character;
printf("Welcome to ASCII:\n");
do {
scanf("%d",&character);
printf("ascii: %d\n",character);
} while(character != 999);
printf("Done.\n");
return 0;
}
It just shows 0 for every input...
" I am trying to scan a user input (ascii character) as an integer, and show the ascii code for the entered character"
What you should do is exact opposite. You should read a character and display it as an integer, i.e.:
char c;
scanf("%c", &c); // <-- read character
printf("%d", c); // <-- display its integral value
input: a, output: 97
Also note that while(character != 999) isn't very lucky choice for a terminating condition of your loop. Checking the return value of scanf to determine whether the reading of character was successful might be more reasonable here:
while (scanf("%c", &character)) {
printf("ascii: %d\n", character);
}
try this:
#include <stdio.h>
int main(int argc, char *argv[]) {
char character;
printf("Welcome to ASCII:\n");
do {
scanf("%c",&character);
getchar(); // to get rid of enter after input
printf("ascii: %d\n",character);
} while(character != 999);
printf("Done.\n");
return 0;
}
output:
d
ascii: 100
s
ascii: 115
You are trying to read an integer(scanf("%d", &...)) and it's normal this operation to fail and the value to be 0 - a default value for int variable. Change the "%d" to "%c" and it should work.
change to :
printf("ascii: %c\n",character);
However , What your condition speciftied 999?
Related
I know there are many questions on the same topic of scanf until EOF is reached, but here's a particular case I haven't seen. Suppose I want to make a C program where the user enters a single character, and the program prints back the character and the number of times the user has entered a character until they press CTRL+D (EOF)
This is what I have:
#include <stdio.h>
int main(){
char thing;
int i=0;
while(scanf("%c", &thing) != EOF){
printf("time:%d, char:%c\n",i,thing);
i++;
}
return 0;
}
However, the output is not as expected. It's the following:
f
time:0, char:f
time:1, char:
p
time:2, char:p
time:3, char:
m
time:4, char:m
time:5, char:
I'm not too sure why i is being incremented again, and why printf gets executed again. Perhaps I'm missing something.
Try
#include <stdio.h>
int main(){
char thing;
int i=0;
while(scanf("%c", &thing) != EOF){
if (thing!='\n') {
printf("time:%d, char:%c\n",i,thing);
i++;
}
}
return 0;
}
#user2965071
char ch;
scanf("%c",&ch);
With such a snippet one reads any ASCII character from the stream including new line, return, tab, or escape. Thus, inside the loop I would test the symbol read with one of the ctype-functions.
Something like this:
#include <stdio.h>
#include <ctype.h>
int main(){
char thing;
int i=0;
while(1 == scanf("%c", &thing)){
if (isalnum(thing)) {
printf("time:%d, char:%c\n",i,thing);
i++;
}
}
return 0;
}
As for me, I think it's not a good idea to check scanf for returning EOF. I would rather check for the number of good read arguments.
I am making a text based game, and i am having a big problem with input. Here is a small example of my problem code.
#include <stdio.h>
#include <stdlib.h>
char c;
int main(int argc, char *argv[]){
system("clear");
while(1){
printf("\nInput a character.\n");
c = getchar();
printf("\nYour input: %c\n", c);
sleep(1);
system("clear");
}
return 0;
}
So, if you compile/run this, and type in 'abc', it will just take each one, and send it through the loop. What I need it to do is only take the very first character that someone types in, no matter how many they do type in.
And, PS: I have tried it this way, and it does the same thing:
#include <stdio.h>
#include <stdlib.h>
char c[2];
int main(int argc, char *argv[]){
system("clear");
while(1){
printf("\nInput a character.\n");
scanf("%1s", c);
printf("\nYour input: %c\n", c[0]);
sleep(1);
system("clear");
}
return 0;
}
EDIT: It also adds a space to what ever you type in, I assume it is a \0, but im not sure. Thanks!
When you use scanf, enter a string and hit the ENTER key, a string and a character are placed in the input buffer, they are namely: the entered string and the newline character. The string or character by character gets consumed by the scanf but the newline remains in the input buffer, unless you consume that too.
getchar(), on the other hand will not wait for ENTER key, it would read character by character, then your logic.
I think you can add 1 more line to read all the characters that come after the first one until there is a newline character (i.e. the user presses Enter):
while (getchar() != '\n');
Adding to your example, it would be like this:
#include <stdio.h>
#include <stdlib.h>
char c;
int main(int argc, char *argv[]){
system("clear");
while(1){
printf("\nInput a character.\n");
c = getchar();
printf("\nYour input: %c\n", c);
sleep(1);
system("clear");
while (getchar() != '\n');
}
return 0;
}
Use getch() which does not wait for a newline.
What i think you look for is something like this code, to save very first character, you can also check if c == '\n' to continue your operation, but i dont know what you want after saving very first character:
int i,c;
char save;
for ( i = 0;(c=getchar())!= EOF ; i++)
{
if ( i == 0)
save = c;
}
You can use fgets(), and extract its first character, like...
char ch[2], c;
fgets(ch, 2, stdin);
c = ch[0];
I created a program to make a diamond out of *'s. I am looking for a way to check if the type of input is an integer in the C language. If the input is not an integer I would like it to print a message.
This is what I have thus far:
if(scanf("%i", &n) != 1)
printf("must enter integer");
However it does not display the message if it's not an integer. Any help/guidance with this issue would be greatly appreciated!
you can scan your input in a string then check its characters one by one, this example displays result :
0 if it's not digit
1 if it is digit
you can play with it to make your desired output
char n[10];
int i=0;
scanf("%s", n);
while(n[i] != '\0')
{
printf("%d", isdigit(n[i]));
i++;
}
Example:
#include <stdio.h>
#include <string.h>
main()
{
char n[10];
int i=0, flag=1;
scanf("%s", n);
while(n[i] != '\0'){
flag = isdigit(n[i]);
if (!flag) break;
i++;
}
if(flag)
{
i=atoi(n);
printf("%d", i);
}
else
{
printf("it's not integer");
}
}
Use fgets() followed by strtol() or sscanf(..."%d"...).
Robust code needs to handle IO and parsing issues. IMO, these are best done separately.
char buf[50];
fgets(buf, sizeof buf, stdin);
int n;
int end = 0; // use to note end of scanning and catch trailing junk
if (sscanf(buf, "%d %n", &n, &end) != 1 || buf[end] != '\0') {
printf("must enter integer");
}
else {
good_input(n);
}
Note:
strtol() is a better approach, but a few more steps are needed. Example
Additional error checks include testing the result of fgets() and insuring the range of n is reasonable for the code.
Note:
Avoid mixing fgets() and scanf() in the same code.
{ I said scanf() here and not sscanf(). }
Recommend not to use scanf() at all.
strtol
The returned endPtr will point past the last character used in the conversion.
Though this does require using something like fgets to retrieve the input string.
Personal preference is that scanf is for machine generated input not human generated.
Try adding
fflush(stdout);
after the printf. Alternatively, have the printf output a string ending in \n.
Assuming this has been done, the code you've posted actually would display the message if and only if an integer was not entered. You don't need to replace this line with fgets or anything.
If it really seems to be not working as you expect, the problem must be elsewhere. For example, perhaps there are characters left in the buffer from input prior to this line. Please post a complete program that shows the problem, along with the input you gave.
Try:
#include <stdio.h>
#define MAX_LEN 64
int main(void)
{ bool act = true;
char input_string[MAX_LEN]; /* character array to store the string */
int i;
printf("Enter a string:\n");
fgets(input_string,sizeof(input_string),stdin); /* read the string */
/* print the string by printing each element of the array */
for(i=0; input_string[i] != 10; i++) // \0 = 10 = new line feed
{ //the number in each digits can be only 0-9.[ASCII 48-57]
if (input_string[i] >= 48 and input_string[i] <= 57)
continue;
else //must include newline feed
{ act = false; //0
break;
}
}
if (act == false)
printf("\nTHIS IS NOT INTEGER!");
else
printf("\nTHIS IS INTEGER");
return 0;
}
[===>] First we received input using fgets.Then it's will start pulling each digits out from input(starting from digits 0) to check whether it's number 0-9 or not[ASCII 48-57],if it successful looping and non is characters -- boolean variable 'act' still remain true.Thus returning it's integer.
I have program that asks to enter a string (mystring) and a char (ch). Then it deletes all entered chars (ch) from the string (mystring). For example "abcabc" and char 'a' then the result shoud be "bcbc".
-When I use scanf the program works nicely if the string does not have spaces. If I enter "abc abc abc" It reads and processes only the first 3 letters (until space).
Then I was advised to use gets(mystr); because it can read all the stirng. But when I use gets the result is the same as the input string and nothing happens.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 100
int main(int argc, char *argv[])
{
char mystr[N] ,result[N];
char ch;
int i,k;
k=0;
printf("enter string \n");
//gets(mystr);///////////////////////////
//scanf("%s",&mystr);///////////////////
printf("enter char \n");
scanf("%c",&ch);
scanf("%c",&ch);
for ( i = 0; i <= strlen(mystr); i++ )
{
if (mystr[i] != ch)
{
result[k]=mystr[i];
k++;
}
}
puts(result);
system("pause");
return 0;
}
scanf("%c",&ch);
scanf("%c",&ch);
That second scanf is your problem. It's picking up the new-line character that you enter after the letter you want to remove (and overwrites the previous value of ch).
Get rid of it.
Please note, as the man page says:
Never use gets(). Because it is impossible to tell without knowing the data in advance how many
characters gets() will read, and because gets() will continue to store characters past the end of
the buffer, it is extremely dangerous to use. It has been used to break computer security. Use
fgets() instead.
hmm - not sure what the problem is - use getstr, but not scanf for the string, and it works for me in visual studio
int main(int argc, char *argv[])
{
char mystr[N] ,result[N];
char ch;
int i,k;
k=0;
printf("enter string \n");
gets(mystr);///////////////////////////
//scanf("%s",&mystr);///////////////////
printf("enter char \n");
scanf("%c",&ch);
// scanf("%c",&ch);
for ( i = 0; i <= strlen(mystr); i++ )
{
if (mystr[i] != ch)
{
result[k]=mystr[i];
k++;
}
}
puts(result);
system("pause");
return 0;
}
Use this one:
char temp[2];
scanf("%1s",temp);
ch = temp[0];
and use gets
scanf when used with chars has some problems (it gets the "old" new line). Here we "cheat" a little and we use scanf to get a string that can have up to one character. A string of 1 character clearly needs a second character for the terminator, so an array of 2 characters.
Be aware that using a scanf for the character to search, you won't be able to insert the space character.
Note that gets is an "evil" function. You can easily do buffer overruns using it (it doesn't check that the buffer is big enough). The "right" way to do it is normally: fgets(mystr, N, stdin); (the "file" variant of gets has a maximum number of characters that can be read and will append a \0 at the end). Note that if you insert 150 characters in a fgets, 99 will go to your string (because you gave 100 of max size), 1x \0 will be appended and the other characters will remain in the buffer "ready" for the next scanf/gets/fgets... (to test it, reduce the buffer to a smaller value, like 5 characters, and do some tests)
You can use fgets() as suggested by xanatos with a small hack, so you can reliably handle return characters. Just change the '\n' to '\0' in the string obtained using fgets.
And in your program, you forgot to terminate the new string with a '\0'.
So here's the code you're looking for.
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#define N 100
int main(int argc,char **argv){
char string[N],str1[N];
char ch;
int i,k = 0;
fgets(string,N,stdin);
string[strlen(string)-1] = '\0';
scanf("%c",&ch);
printf("\n%s , %c",string,ch);
for (i=0;i<=strlen(string);i++)
if(string[i] != ch)
str1[k++] = string[i];
str1[k] = '\0';
printf("\n%s , %s\n",string,str1);
return 0;
}
I was trying out a simple program in C for validating user data. The program is supposed to identify whether a user entered character is a number, alphabet or a special character.
Somehow , the code identifies every kind of input character as a number. I have appended the code below, I'd be grateful if someone could kindly point out where I'm going wrong ?
//Program to take input from the user and determine whether it is character, number, or a special character
#include<stdio.h>
#include<conio.h>
#include<string.h>
char ch;
int main()
{
clrscr();
printf("Enter a character \n");
scanf("%c \n",ch);
if ((ch>='A'&& ch<='Z')||(ch>='a'&& ch<='z') )
{
printf("The character entered is an alphabet \n" );
}
else if ((ch>=0)&&(ch<=9))
{
printf("Character entered is an number \n");
}
else
{
printf("Character entered is a special character");
}
return 0;
}
scanf accepts a pointer as the argument for %c. In other words,
scanf("%c \n",ch);
should be written as:
scanf("%c\n",&ch);
Without the reference operator (&), scanf receives the value of ch. In this case, the value is garbage, because ch is unset.* Referencing ch gives scanf a pointer to ch, not ch itself, so scanf can modify the value of ch by dereferencing the pointer (using the dereference operator, *).
There's also the issue with digit checking that Himadri mentioned.
* This is actually undefined behaviour.
Oh, Arun very silly mistake.
In your second condition in else if you have to right 0 and 9 in single quotation mark.
So, your code will be -
if ((ch>='A'&& ch<='Z')||(ch>='a'&& ch<='z') )
{
printf("The character entered is an alphabet \n" );
}
else if ((ch>='0')&&(ch<='9'))
{
printf("Character entered is an number \n");
}
else
{
printf("Character entered is a special character");
}
May be this is the only mistake. Now, it should work.
A few comments on style:
conio.h and clrscr() are non-standard.
Global variables are bad (char ch). Declaring them non-static is also bad.
Always check the return value of scanf. This will help you catch input format errors.
In this case, as we need to just a single character, getchar is more appropriate.
This is how I would've written this program:
#include <stdio.h>
#include <ctype.h>
int main()
{
int ch; /* We use an int because it lets us check for EOF */
printf("Enter a character: ");
fflush(stdout); /* Remember to flush the output stream */
ch = getchar();
if (ch == EOF)
{
printf("end-of-file or input-error\n");
return 1;
}
if (isalpha(ch))
printf("The character entered is an alphabet\n" );
else if (isdigit(ch))
printf("Character entered is an number\n");
else
printf("Character entered is a special character\n");
return 0;
}