3 ms·
Exactly. Some programs are inherently ambiguous. In such cases, we cannot do much. - Example 1: `void f(){x * y;}` It's impossible to know whether that is a m
by unusualcool 9y ago
Exactly. Some programs are inherently ambiguous. In such cases, we cannot do much.
- Example 1: `void f(){x * y;}`
It's impossible to know whether that is a multiplication or pointer declaration.
- Example 2: `void f(){x y = 0;}`
We can only know that x is a scalar type (in C terminology), but it could be an int, int * , int * * , etc.