syntax error
if the stop bits does not appear when it is supposed to, the UART considers the entire word to be garbled and will report a framing error regardless of the whether the data was received correctly or not, the UART automatically discards the start,parity and stop bits.
type1 error is more dangerous
It depends on whether it is the Type I Error or the Type II Error that is increased.
(100/400) * 12 = 3
The compiler is software that allows any high-level language to be translated into the machine language, compilers are language specific, i.e. C language has its own compiler which will translate the code written in C language to the machine language, compiler for Java will do the same task for the Java code
syntax error
C++ programs won't compile if they contain compiler errors. The compiler will tell you precisely where the error is, and the type of error, unless the error is in a macro. The compiler cannot see macro definitions because they are inline expanded prior to compilation.
we need compiler to run the programme. compiler will check the error in the problem. compiler can check the error with the help of run time compiler. they can be static or dynamically.
Simply, the error that your compiler catches is called compiler error. Doesn't matter if you are in Eclipse, NetBeans or Intellij IDEA, all these IDE's are very smart. They can detect if you made any any mistakes in your program. Common compiler errors are- -When you try and access a variable that is out of scope -When you forget to give a semicolon. -When you will try to use a reserved keyword like - return. -Any syntax error that your compiler might find suspicious, will cause compiler to throw an error.
When a program contains a compiler error, the compiler will detect it, preventing the program from compiling. Compiler errors must be fixed before a program will compile successfully.
I would need to see the code where the error occurs. But I would guess the error might well be before the place where the compiler complains that there is an error - simply because the previous instruction was not complete.
It will get a compiler error in Java.
Either you get a compiler error. Or you get a compiler warning, and the variable won't be stored in register.
You failed to find (or understand) the compiler's error messages. Ask your teacher for help.
Debugging
Not without casting. A char is a 16 bit type, whereas a byte is an 8 bit type. Therefore the compiler cannot guarantee that the 16 bit value will fit into the 8 bit value without overflowing. If you attempt to stick a char into a byte, you will get a compiler error. To override this, you can cast the char value to a byte during assignment. However, you might get some unexpected results. A few examples below: char a = 'A'; byte b = a; //compiler error char a = 'A'; byte b = (byte)a; //valid, no error. b=65 char a = 172; byte b = (byte)a; //valid, no error, but b=-84 because of overflow.