passing arg makes pointer to integer without a cast

From the code below i get the following warning message from the display_char function.

warning: passing arg 1 of 'valid char' makes pointer to integer without a cast

Code:

---

void display_char(char c, char position)
{
        if(position > 5)
        {
                return;
        }
        if (valid_char(c) == 0);
        {
                return;
        }

        TextBuffer[position] = c; 
}

---

Code:

---

int valid_char(char *check)
{
        char i,j, k;
        char validChar[] = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z',
                            'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z',
                            '0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '*', '+', '-', '.', ' ', '_', '\0'};
        char validCharSize = 0;
        char inputCharSize = 0;
        char valid = 0;
       
        for(i = 0; validChar[i] != '\0'; i++)
        {
                validCharSize++;       
        }
       
        for(i = 0; check[i] != '\0'; i++)
        {
                inputCharSize++;       
        }
       
        for(j = 0; j <= inputCharSize; j++)
        {
                for(k = 0; k < validCharSize; k++)
                {
                        if(check[j] == validChar[k])
                        {
                                valid++;
                        }
                }
        }
       
        if(valid == inputCharSize)
        {
                return 1;
        }
        else
        {
                return 0;
        }
}

---

I don't really get what the warning message is saying and so it was no surprise to be when my attepmts to fix this failed.
Any help would be greatly appreciated.

Thanks
-Nick

The message is pretty strait forward.

char c in display_char() is a char. valid_char() takes a char pointer. So your character is being implisidly typecast as a char pointer.

Hi

thanks for the reply

I'm not at my the computer with the code on it right now so i can't test my guess.

I assume the problem is solved by passing the pointer *c instead of the char itself to valid_char.

correct?

If all you are doing is testing a single character, just pass a character and don't bother passing a pointer.


Quzah.

&c actually, but that still won't solve your problem because valid_char() expects a null terminating string.

You could either create a 1 char string in display_char(), or probably better change/make a new function for valid_char() that takes a single character.