The following code uses getchar() to accept a line of input.
#include <stdio.h>
#include <stdlib.h>
int main()
{
char *rawString = (char *)malloc(200*sizeof(char));
char *rawStringInitial = rawString;
char c;
c=getchar();
while(c!='\n')
{
*rawString=c;
rawString++;
c=getchar();
}
*rawString='\0';
printf("\n[%s]\n",rawStringInitial);
return(0);
}
While typing, if I press backspace, shouldn't it also be received by getchar() & stored in the rawString-pointed location? However the output simply shows the final string without any special characters. Could someone explain why?
Standard input is (usually) buffered; non-printing characters like backspace are handled by the terminal server, and library functions like getchar() will never see them.
If you need to read raw keystrokes, then you will need to use something outside of the C standard library.
#include<stdio.h>
#include<conio.h>
#include<string.h>
void get_string(char *string);
void main(){
char *stringVar;
clrscr();
printf("Enter String : ");
get_string(stringVar);
printf("String Enter : %s",stringVar);
getch();
}
void get_string(char *string){
char press;int i=0;
do{
press=getch();
if(press!=8){
printf("%c",press);
string[i]=press;
i++;
}
else if(i>0){printf("\b%c\b",0);sting[i]=NULL;i--;}
}while(press!13);
}
This is Will Work.
Related
I am getting this error whenever I run my code in visual Studio:
#include <stdio.h>
#include <ctype.h>
int main() {
char username[10];
printf("Enter Username: ");
scanf_s("%[^\n]", &username);
while (isupper(username)) {
if (username == '-') {
printf("Username cannot contain UpperCase Letters");
}
}
}
Error Image
I don't think you can pass whole array to isupper. Also if you don't want to return anything instead of int main() use void main() or just return 0 in the end or when you want to end after your program executed successfully. As for using scan_s or scanf or getline or whatever I won't say anything because its a different matter and your syntax of scanf_s is certainly wrong.
Also following code will not check for any buffer overflow (not a good practice, you will see even though we gave size 20 char array, this code will work even for larger input which is certainly not a good thing). So you can either limit the size of input or better to read an entire line via fgets() (or getline() if available) and parse the string yourself.
#include <stdio.h>
#include <ctype.h>
#include <string.h>
int main(){
char username[20];
printf("Enter Username: ");
// scanf("%[^\n]", username); <--- Instead of this
scanf_s("%20c", username, 20); // <----Try Using this
int i=0;
while (i<strlen(username)) {
if (isupper(username[i])) {
printf("Username cannot contain UpperCase Letters\n");
return 0;
}
i++;
}
return 0;
}
My first guess would be that your while is an endless loop, try to do it like this:
int i;
for(i=0; i<strlen(username);i++){
if(isupper(username[i])){
printf("Username cannot contain UpperCase Letters");
}
}
I wrote a simple program to read a string.
void main()
{
char *str; /*didn't allocate memory*/
scanf(" %s",str);
printf("%s",str);
}
But it is causing a segmentation fault. Whereas the next one isn't.
void main()
{
char *str;
scanf(" %c",str);
printf("%c\n",str);
}
Would someone mind to clarify how actually this works?
You string isn't allocated. which mean you are writing somewhere you didn't ask for.
however what you can do is:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[])
{
char c;
while (scanf("%c",&c) && c != '\n')
printf("%c",c);
printf("\n");
return 0;
}
It will read every characters you send in input until you press return.
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];
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(void)
{
char p,q;
printf("Hello enter char: ");
p=getchar();
printf("the char is: %c\n",p);
printf("Hello enter char: ");
q=getchar();
printf("the char is: %c\n",q);
return 0;
}
(WHY IS MY OUTPUT for the second printf and scanf not waiting for me to input a char before exiting the program?.....what i mean is u know where it says q=getchar();??? shouldnt it wait for to input a char before exiting the program? but for some reason the program just exits when it goes to the next line...
when pressing enter,a character '\n' is inputing.So your getchar() was used before you enter the second character.I think you want the code below:
#include <stdio.h>
#include <stdlib.h>
int main(void) {
char p,q;
printf("Hello enter char: ");
p=getchar();
printf("the char is: %c\n",p);
int c;
while((c = getchar()) != '\n' && c != EOF && c != ' ') ;
printf("Hello enter char: ");
q=getchar();
printf("the char is: %c\n",q);
return 0;
}
You can also use a getch() instead of getchar() to avoid pressing enter key.
#include <stdio.h>
#include <conio.h>
int main(void)
{
char p,q;
printf("Hello enter char: ");
p=getch();
printf("the char is: %c\n",p);
printf("Hello enter char: ");
q=getch();
printf("the char is: %c\n",q);
return 0;
}
When encountering invalid user inputs, use getchar() to read char, and other similar instances where there are undesired characters stuck at input stream(like in your case it was a newline) I define a constant named FLUSH
#define FLUSH while(getchar() != '\n')
to solve the problem. What this statement does is that it reads a character and then throws it away. Now if you try to place it after one of your getchars i.e.
p=getchar();
printf("the char is: %c\n",p);
FLUSH;
it will read the newline then stops because the condition within the while statement no longer holds.
Note: Using getchar() for prompts leaves a '\n' in the input stream you will find this troublesome once you make another prompt and have not eradicated that '\n'.