为什么会编译？ [C ++]

Question

I tried to look for an answer online but couldn't quite find it.

Today I saw these lines of code:

    int main(){
    int n = 7;
    while(n /= 10);
    }

It doesn't make much sense, but the question was only 'will it compile?'. To which I answered no, but I was wrong.

My question is, why? Why does

    n /= 10

behave like a bool (or an int) here?

Answer 1

An assignment (including a compound assignment like /= ) is an expression, which yields the value that was assigned ¹ .

So, you can do something like: x = y = z = 0; , which assigns 0 to z , takes the result of that (also 0) and assigns it to y , and takes the result of that (still 0) and assigns it to x .

From there, it's making use of the implicit conversion from int to bool , in which 0 converts to false , and any non-zero value converts to true .

---

^{1. Note: that's what happens for built-in types. By convention, when/if you overload operator= for a class, you have it return *this; , so it works the same way, as a user would/will expect--but that part's not mandatory--you can overload your operator= to return a different value or an entirely different type--but this is almost always a bad idea and should usually be avoided.}

Answer 2

What you have here for your while loop is as follows:

while ( expression );

If the expression is true or non 0 the loop will continue; otherwise if it evaluates to false or 0 it will terminate. So looking back at your original:

int n = 7;
while ( n /= 10 );

This then becomes:

while ( n = 7 / 10 ); 

Here the full expression is n = 7 / 10 ; This should result in 0 due to truncation of integer arithmetic. The value by implicit conversion from int to bool becomes false . As the yielded result is 0 .

There is nothing here preventing this from compiling. As this is no different than having:

while ( false );

However with assignment and arithmetic operations; this may not always be the case, but in your case it is. Consider this next example: This will still compile but the loop will not terminate:

int n = 5;
while( n + n );

This will then become:

while( 5 + 5 );
...
while( 10 );
...
while( true );

Which will still compile but the loop will continue infinitely.

Answer 3

Just like += and -= work, *= and /= work.

In fact, there are also &= and |= .

These all evaluate to the new value that has been assigned.

And, as you know, you don't have to put a boolean in a while / for / if condition; you only need to put something there that can be converted to a boolean.

For example, if (42) , or for (char* ptr = begin; ptr; ++ptr) , or while (n /= 10) .

Answer 4

C++ will convert the n /= 10 to bool. Integers = 0 converted to bool evaluates to false. Integers != 0 converted to bool evaluates to true. That while will be evaluated as while(false) .