了解C编程中的逻辑表达式

Question

I am learning C on my own and following a book with exercises that do not have answers. I am having trouble understanding why the below expression evaluates to "true" or "1" when running it in my compiler. I understand precedence, associativity, and how logical operators work, but the a += trips me up. I don't know how that relates to true and false. I appreciate any help or explanation as to how this evaluates to true.

 int a = 1, b = 2, c = 3;

 a += !b && c == ! 5;

Answer 1

a += !b && c == ! 5

parses as

a += ((!b) && (c == (! 5)))

We can evaluate each sub-expression in turn:

• b is 2 .
• !b is 0 (because ! turns all non-zero values into zero, and zero into one).
• 0 && ... does not evaluate its right-hand side but returns 0 immediately.
• a += 0 is equivalent to a = a + 0 , which doesn't change the value of a , but returns its new value (same as the old value in this case), 1 .
• 1 is not zero, so it's true.

In case this is what's tripping you up: The assignment operators are, well, operators. The have an effect (assigning a value to a variable), but they also have a result. For all assignment operators the result is the value being assigned:

int n = 2;
printf("%d\n", n += 3);

outputs 5 and also sets n to 5 .

Just for completeness, c == (!5) would have evaluated as follows:

• !5 is 0 .
• c == 0 is 0 (because c is 3 and == returns 1 for true and 0 for false).

Answer 2

Expression a += !b && c == ! 5; a += !b && c == ! 5; is equivalent to a = a + (!b && c == ! 5); . Hope that helps, and I hope that you never face such an expression in practice.