The execution of the code always goes into the else statement - c

Question

Some very strange things happen in my source code.
The following function works well and it prints 'y' when the password is correct and prints 'n' when it is incorrect. But if i add some UART1_Write and Delay functions to the else statement the bug comes out and even if the password is "zxc" (correct) it ALWAYS enters the else statement.
I'm using MikroC PRO for PIC v6.0.0, the robot system is made of PIC18F452 and RN-42 bluetooth module connected to it. I am testing with a laptop with a bluetooth and TeraTerm.
For more info: http://instagram.com/p/pLnU9eDL8z/#
Here it is the well working routine:
void authenticate() {
char *input = "";
char *password = "zxc\0";
unsigned char ready = 0;
while (connected && !ready) {
if (UART1_Data_Ready()) {
UART1_Read_Text(input, "|", 17);
strcat(input, "\0");
if (strcmp(input, password) == 0) {
UART1_Write('y');
ready = 1;
} else {
UART1_Write('n');
ready = 1;
}
}
}
}
This version of the routine ALWAYS goes in the ELSE statement of the strcmp(input, password) == 0 part:
void authenticate() {
char *input = "";
char *password = "zxc\0";
unsigned char ready = 0;
while (connected && !ready) {
if (UART1_Data_Ready()) {
UART1_Read_Text(input, "|", 17);
strcat(input, "\0");
if (strcmp(input, password) == 0) {
UART1_Write('y');
ready = 1;
} else {
UART1_Write('n');
Delay_ms(100);
UART1_Write('$');
Delay_ms(100);
UART1_Write('$');
Delay_ms(100);
UART1_Write('$');
Delay_ms(100);
UART1_Write('K');
Delay_ms(100);
UART1_Write(',');
Delay_ms(100);
UART1_Write('-');
Delay_ms(100);
UART1_Write('-');
Delay_ms(100);
UART1_Write('-');
Delay_ms(100);
UART1_Write('\n');
ready = 1;
}
}
}
}
It is important to send all these addition symbols in order to get RN-42 into command mode and disconnect the user if the password is wrong.
Please help me solve the problem. Any ideas appreciated!

Answer 1

As others have pointed out in the comments section a major issue with your code is that you are trying to store the UART data to memory that does not belong to you.
When you declare char *input = "";, you haven't actually allocated any space except for a single byte that stores '\0'. Then, when you use UART1_Read_Text(), you tell that function you may have up to 17 characters that will be read before finding the delimiter - all of which should be stored at the location pointed to by input.
The description of that library function can be found here. Also, based on the library description it looks like UART1_Read_Text() already adds the null-termination to the UART data. I base this assumption off the description of UARTx_Write_Text and the example that they provide on their website. However, I would recommend that you verify that is indeed the case.
Also, your initialization of password is redundant and char *password = "zxc\0" should be changed to char *password = "zxc". When you declare a string literal using double quotation marks it is automatically null-terminated. This excerpt is from "C in a Nutshell":
A string literal consists of a sequence of characters (and/or escape sequences) enclosed in double quotation marks... A string literal is a static array of char that contains character codes followed by a string terminator, the null character \0... The empty string "" occupies exactly one byte in memory, which holds the terminating null character.
Based on the above, I would go about it a little more like this:
#define MAX_NUM_UART_RX_CHARACTERS 17
void authenticate()
{
char input[MAX_NUM_UART_RX_CHARACTERS + 1];
char *password = "zxc";
unsigned char ready = 0;
while (connected && !ready)
{
if (UART1_Data_Ready())
{
UART1_Read_Text(input, "|", MAX_NUM_UART_RX_CHARACTERS);
if (strcmp(input, password) == 0)
{
UART1_Write('y');
ready = 1;
}
else
{
UART1_Write('n');
ready = 1;
}
}
}
}