How would C parse a = b---c?

Question

If C has to compile the following code

int a = 5, b = 3, c = 7;

a = b---c;

Would it have to parse it as a = b-- - c, a = b - --c, or is it undefined behavior?

Answer 1

The compiler considers the longest sequence of characters to determine a token.

So this statement

a = b---c;

is equivalent to

a = b-- - c;

That is in the expression of the right side of the assignment there is the postfix decrement operator -- followed by the additive operator -.

From the C Standard (6.4 Lexical elements)

4 If the input stream has been parsed into preprocessing tokens up to a given character, the next preprocessing token is the longest sequence of characters that could constitute a preprocessing token. ...

Thus as result the values of the variables after executing this statement will be

a = -4, b = 2, c = 7

Also pay attention to that a valid sub-expression may be enclosed in parentheses. You may write for example

a = ( b-- )-c;

but you may not write

a = ( b- )--c;

On the other hand, you may also write

a = b-( --c );

In this case the values of the variables after executing the statement will be

a = -3, b = 3, c = 6

Answer 2

It is parsed as : a = b-- - c, and b-- means post-decrement, i.e. decrement after taking the value of b to compute b - c

run this code and watch the output :

#include <stdio.h>

int main()
{
    int a = 5, b = 3, c = 7;

    a = b---c;

    printf ("a is : %d\n", a);
    printf ("b is : %d\n", b);
    printf ("c is : %d\n", c);
}

output :

a is : -4
b is : 2
c is : 7