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ccompiler-constructionlalr

Why is a multiplication without an LHS valid in C?

Why is this piece of C code valid?

int main() {
    int x = 0, y = 0;

    x * y; // See this?!
}

The most ubiquitous examples that explain the need for back-tracking in parser often shows this. And then goes on to say that depending on what x is, the meaning of * will change. For some time, I believed it was just a way of explaining until I compiled it and found to my surprise, it compiles and works!

Why is a multiplication without an LHS valid in C and what is it used for?

Edit This confusion is because, the same piece of code won't compile in Java. It seems Java compiler tries to prevent operations without LHS.

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Solution

  • This is referred to as an expression statement.

    From section 6.8.3 of the C standard:

    1
    expression-statement:

    expressionopt ;

    2 The expression in an expression statement is evaluated as a void expression for its side effects.

    In your case, the expression is a multiplication. This statement has no side effect, so it has no effect at all.

    If on the other hand you had this:

    i++;

    That's an expression that does have a side effect.

    This is also considered an expression:

    a = b * c;

    In this case, the side effect is the assignment operator.