That's two backspaces, with a space between them. I don't think you can print a delete and get anything sensible. I can't see your computer from here, please post your code.
I meant I would print "\b \b", but I just want to give the user the option of deleting characters with backspace or delete.
So I was messing around with implementing backspace and also reading it into into an array today and think I've got it working fairly well. Correct me if I'm wrong, but getchar() turns the input into a unique integer right? Well I want to display the password that was entered just to check, but if everything has been converted into integers, how do I display it?
And I'll narrow my earlier question changing the colors into something that is answerable. Basically I want to do something along these lines:
Code:
while ((ch = getchar()) != '\n')
{
if(ch == '\b') /*Here is where I want to give the option of using delete*/
{
printf("\b \b");
i--;
}
else
{
password[i] = ch;
if((ch == '1') /*I'm not sure if this is done correcting. I want to print a red "1" if a 1 was typed*/
{
printf("\033[1;3;31;49m1\033[0m"); /*Output a red '1'*/
}
else((ch == '2') /*Same issue*/
{
printf("\033[1;3;34;49m2\033[0m"); /*Output a blue '2'*/
}
else((ch == '3') /*Same issue*/
{
printf("\033[1;3;32;49m3\033[0m"); /*Output a green '3'*/
}
i++;
}
}
I guess this is kinda similar to my first question, this time comparing something that has been turned into an integer to an input. I commented on the lines I am referring to.
[QUOTE=bigdrock44;302481777]I meant I would print "\b \b", but I just want to give the user the option of deleting characters with backspace or delete.
So I was messing around with implementing backspace and also reading it into into an array today and think I've got it working fairly well. Correct me if I'm wrong, but getchar() turns the input into a unique integer right? Well I want to display the password that was entered just to check, but if everything has been converted into integers, how do I display it?
And I'll narrow my earlier question changing the colors into something that is answerable. Basically I want to do something along these lines:
Code:
while ((ch = getchar()) != '\n')
{
if(ch == '\b') /*Here is where I want to give the option of using delete*/
{
printf("\b \b");
i--;
}
What this does is print one backspace, which moves the cursor back one. Then it prints a space, which prints a blank overtop of the spot it moved to(which doesn't happen just from printing a backspace). then it backspaces again.
But that probably won't work since you're using printf, which is buffered! It'll wait until you feed it a \n to print anything. Try fprintf(stderr, "\b \b"); instead.
Code:
else
{
password[i] = ch;
if((ch == '1') /*I'm not sure if this is done correcting. I want to print a red "1" if a 1 was typed*/
{
printf("\033[1;3;31;49m1\033[0m"); /*Output a red '1'*/
}
That code works absolutely fine on my system... (again, though, it's buffered -- fprintf(stderr, "...); instead) but what I'd do is:
I got it to do what I was basically asking. Here's the issue, though. I want it to output the "*" or the colored numbers, whatever the situation may be, but I want to store the actual keystroke. For example, here's my code:
Code:
void read_unbuffered(void)
{
int ch;
int password[4];
int i;
setbuf(stdin, NULL);
echo_off();
i = 0;
while ((ch = getchar()) != '\n')
{
if(ch == '\b' && i != 0)
{
fprintf(stderr, "\b\b \b\b"); /*Don't mind this. As you can see below
i--; after each entry it puts a space (for clarity), so I double backspace
and double space to make up for this*/
}
else if((ch >= '1') && (ch <= '6') && i < 4)
{
password[i] = ch;
if(ch == '1')
{
fprintf(stderr, "\033[1;3;31;49m%c \033[0m", ch);
}
else if(ch == '2')
{
fprintf(stderr, "\033[1;3;34;49m%c \033[0m", ch);
}
else if(ch == '3')
{
fprintf(stderr, "\033[1;3;33;49m%c \033[0m", ch);
}
else if(ch == '4')
{
fprintf(stderr, "\033[1;3;35;49m%c \033[0m", ch);
}
else if(ch == '5')
{
fprintf(stderr, "\033[1;3;32;49m%c \033[0m", ch);
}
else if(ch == '6')
{
fprintf(stderr, "\033[1;3;39;49m%c \033[0m", ch);
}
i++;
}
}
printf("\n");
for(i = 0; i < 4; i++) /*To see what is stored in the array*/
{
printf("%d ", password[i]);
}
echo_on();
}
Say I type in "1234". This is what happens on the screen when I type that:
Code:
1234 /*This is good. This is what I want. To showed colored numbers as I type.*/
49 50 51 52 /*These are the values stored in the array*/
So the code works well in that as I type in 1234, it is actively changing the color as I type. Since the code is using getchar(), however, it is storing the values 49, 50, 51, and 52 in the array. I want 1, 2, 3, and 4 to be stored in the array, because basically I want this function to return this array, and if it is returning an array of 49 50 51 52 it will not make sense in my program. Does this make sense?
void read_unbuffered(void)
{
int ch;
int password[4];
int i;
static const int color[]={0, 31, 34, 33, 35, 32, 39};
setbuf(stdin, NULL);
echo_off();
i = 0;
while ((ch = getchar()) != '\n')
{
if(ch == '\b' && i != 0)
{
fprintf(stderr, "\b\b \b\b"); /*Don't mind this. As you can see below
i--; after each entry it puts a space (for clarity), so I double backspace
and double space to make up for this*/
}
else if((ch >= '1') && (ch <= '6') && i < 4)
{
// 1 is not '1'. An ASCII 1, like you type in, is 49 in decimal.
// An ASCII 9 is 57.
// ASCII has them all in a row, so '5' - '0' == 5, etc.
password[i] = ch-'0';
// Use an array so you don't need umpteen different print statements
fprintf(stderr, "\033[1;3;%d;49m%c \033[0m", color[ch-'0'], ch);
i++;
}
}
printf("\n");
for(i = 0; i < 4; i++) /*To see what is stored in the array*/
{
printf("%d ", password[i]);
}
echo_on();
}
Last edited by Corona688; 12-21-2010 at 09:12 PM..
Reason: whoops, not all printf is the same...
Oh okay, yeah I noticed that it was along those lines but I didn't know if it would be wrong to do that.
So now I have everything working correctly, I just want the function to return an array, not be void. But as I'm thinking about it it's not possible to return an array right? I must return a pointer? I'm not really sure how this works but is it along these lines?
Code:
int password[4];
int* read_unbuffered(void)
{
/*code*/
return password;
}
int main()
{
password = read_unbuffered();
return 0;
}
Last edited by bigdrock44; 12-21-2010 at 10:29 PM..
The difference between an array and a pointer is nil as far as your code's concerned. They both work with the [] operator the exact same way.
Instead of using a global variable, which works but is ugly and unsafe, you should pass a pointer into the function for it to use. This also leaves room for a proper return value to check if an error happened.
Code:
int read_unbuffered(int *password)
{
...
if(success)
return(1);
else
return(0);
}
int main(void)
{
int arr[4];
if(read_unbuffered(arr))
{
fprintf(stderr, "read pass\n");
}
else
{
fprintf(stderr, "couldn't read pass\n");
}
return(0);
}
Well I'm actually not using it as a typical password that grants access to something. I'm making a Mastermind game, maybe you've heard of it. Basically, one player types a code and the other player has to figure out what it was based on some given hints. The whole thing is done, I just had to cover up the code with *'s so that the other player wouldn't be able to see on the screen what it was.
So I need to somehow return the array from the read_unbuffered() function so that I can continue with my program which is currently comparing what the Codebreaker input to that of the Codemaker. If it's not possible, I can just copy the whole functions code into what I already have, but for clarity and legibility I was trying to just utilize it as a separate function.
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