for clarity and legibility I was trying to just utilize it as a separate function.
But it is a seperate function. I don't understand.
If you don't care about thread safety, you could use a global variable (or a static, local variable) and return an "int *". But this isn't reccomended because it's not threadsafe. When you look for examples of how it's done in system libraries, you see all sorts of things like sprintf(), fgets() and the like which all take buffers you give them. And the ones which don't, like gets(), are much maligned.
Like you said, it is a separate function, but I need to utilize it so that it returns the array that the user typed in. I don't want a simple true or false to be returned because that is not what I'm doing. I know I use the word "password" but it's more so a code. I'll try to give a basic idea of what my program does. I'll just pseudo code it.
Code:
int main()
{
/*Prompt player one for secret code*/
codemaker[4] = read_unbuffered(); /*Here is where I want to use read_unbuffered to somehow return the array.*/
/*Comper codemaker's secret code to the codebreaker's guess*/
for (i = 0; i < 4; i++)
{
codemaker[i] = codebreaker[i]; /*This is way simplified and wouldn't even work, but just to give a vague idea of what's going on.
}
printKey(); /*Prints a key so that the codebreaker can see what's right and wrong about his guess*/
return 0;
}
As you can see, I use codemaker[] throughout the program and I need it to be set to the array that is created in read_unbuffered.
Let me make sure I understand everything correctly first. When you have variables within a function, they are local to that function correct? It will not have any affect on the variables within the main, right? So in my read_unbuffered() I'm using password[] and in my main I'm using codeMaker[]. These have no bearing on each other right?
So if I want to alter codeMaker (located in main()) with the function read_unbuffered I should not use a pointer?
Here's an excerpt from my main:
Code:
/*Determine if code is valid*/
valid = 0;
while(valid != 1)
{
printf("Player 2, enter a secret code using values 1-6: ");
getcode();
/*Determine if any of the inputed coordinates are invalid*/
for(i = 0; i < 4; i++)
{
/*If invalid, warn user and re-prompt*/
if(codeMaker[i] < 1 || codeMaker[i] > 6) /*This codeMaker is defined in the main*/
{
printf("Please enter values from 1 to 6!\n\n");
i = 4;
}
/*If valid AND last coordinate, continue on*/
else if(i == 3)
{
valid = 1;
}
}
}
And, just to refresh, here is the function:
Code:
void getcode()
{
int ch;
int i;
int codeMaker[4];
setbuf(stdin, NULL);
echo_off();
i = 0;
while ((ch = getchar()) != '\n')
{
if(ch == '\b' && i != 0)
{
fprintf(stderr, "\b\b \b\b");
i--;
codeMaker[i] = 0;
}
else if((ch >= '1') && (ch <= '6') && i < 4)
{
codeMaker[i] = ch - '0';
fprintf(stderr, "* ");
i++;
}
}
printf("\n");
echo_on();
}
As you can see, I've changed the name of read_unbuffered() to getcode(). Also in getcode() I changed the array from password[] to codeMaker[] in attempt to make this work.
So, since the codeMaker[] in getcode() is local to getcode(), it is not affecting the codeMaker[] in main(), right? Which means, when I check for validity in main() it is checking a different codeMaker[] that it needs to be checking.
So if I want to alter codeMaker (located in main()) with the function read_unbuffered I should not use a pointer?
That's the precise opposite of what I just told you. You're confusing yourself with too much doublethink.
Pass a pointer to codeMaker into read_unbuffered. Just like the way dozens of stdio functions take a pointer to a buffer, to let them read or write to that buffer. That tells it what memory it should modify, so it can modify the contents of that array directly without passing anything back (because returning an array is nonsensical, and returning a pointer doesn't modify its contents, so both are useless in this context). As long as the memory's valid, a pointer to memory doesn't care about variable scope.
Quote:
So, since the codeMaker[] in getcode() is local to getcode(), it is not affecting the codeMaker[] in main(), right?
Correct. It bears no resemblance to the example I showed you, however, which does modify variables outside itself. I'll flesh it out a little more.
Code:
void read_unbuffered(int *password)
{
int ch;
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();
}
int password_global[4];
int main(void)
{
int password_local[4];
// giving a pointer means no need to return anything -- the memory is modified directly.
read_unbuffered(password_local); // modifies password_local
read_unbuffered(password_global); // modifies password_global
}
Last edited by Corona688; 12-23-2010 at 01:28 PM..
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