CODEWITHSUNDEEP
ccompiler-errorssyntax-error
Garrick
Garrick

Reputation: 689

What kind of error is there in C code snippet?

I am examining errors in different C programs and differentiating between them.

Currently, I am confused what type of error is there in this code.

#include <stdio.h>

int main(void) {
    in/*hello*/z; 
    double/*world*/y; 
    printf("hello world"); 
    return 0;
}

When i run this program, i get compilation error as :

prog.c: In function 'main':
prog.c:4:5: error: unknown type name 'in'
     in/*this is an example*/z; 
     ^
prog.c:5:30: warning: unused variable 'y' [-Wunused-variable]
     double/*is it an error?*/y; 
                              ^
prog.c:4:29: warning: unused variable 'z' [-Wunused-variable]
     in/*this is an example*/z; 
                         ^

I know that warning will not prevent from compiling, but there is an error

error: unknown type name 'in'

So, Is this syntax or semantic error ?

Upvotes: 0

Views: 570

Answers (3)

sg7
sg7

Reputation: 6298

I think it is typical typo:

in/*hello*/z;

and you wanted int type for your variable z:

int /*hello*/ z;

Upvotes: 0

Gerald Shyti
Gerald Shyti

Reputation: 51

There is obviously a syntax error. First of all we have to clarify a thing: you have to understand clearly what's the difference between syntax and semantics. Syntax are the "grammar rules" of a programming language, when the compiler does not compile, you've done syntax errors. When you do a semantics error you have done a code that compiles successfully, but, when executed does things you do not want to. C is a strong-typed language: this means that you have to declare variables before using them. You have done a couple of errors, but don't worry, let's analyze them together. First error: you used a type of variable not possible: and the compiler simply showed it up. For the "in" type i'm assuming that you meant int, but when you code you have constantly to ask yourself: "Is this variable useful now?" The answer to this question is: "No", because you just want to make an output. So the correct implementation of this simple procedure is:

http://groups.engin.umd.umich.edu/CIS/course.des/cis400/c/hworld.html

Hope that this helps.

Bye.

Gerald

Upvotes: 1

DUman
DUman

Reputation: 2590

The code does not compile, so this is a syntax error.

A syntax error is when the code is not valid at all, and cannot be compiled. Such as trying to declare a variable of a type that does not exist (as in your snippet), or performing an invalid assignment, or many other things.

5 = a;

That is a syntax error because you cannot assign a value to a number.

int if;

That is a syntax error because if is a keyword that cannot be a variable name.

A semantic error is code that is valid and compiles but does not do what you would like. Consider the following simple function to square a number:

int square (int value)
{
     return value ^ 2;
}

That function does not square the number passed to it but rather performs a bitwise XOR. Yet, in some other languages, ^ 2 would be the syntax to square a number, so this is a semantic error that could conceivably exist in C code.

Upvotes: 0

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