Because the expression num1++ evaluates to num1. You may want to do: ++num1 which evaluates to num1 + 1. Note however that both expressions increment num1 by one. Evaluating num1 in the next statement evalutes to the incremented value. In…
Why does this code print 10 times “Hello world.”? No, this will be an infinite loop because the following statement resets the value of k to 1: k=+1; Also, k=k++ will not change the value of k because it is…
You practically already answered your question: Because declaration does not assign a value. Therefore your second sample never makes sense. The first sample consists of two separate statements each of which could make sense in a certain context. Therefore it…
The way that the ++ “postfix” increment operator works involves first performing a conversion to number type. The value of that conversion is always the return value from the operation, even if that value is NaN. solved Why NaN is…
For the original code: x++ = y++ This line will never be evaluated because it is not legal C and will not be compiled. The result of x++ is not an lvalue and is not permitted on the left side…
Does this help? a = 5; i=++a + ++a + a++; => i=6 + 7 + 7; (a=8) a = 5; i=a++ + ++a + ++a; => i=5 + 7 + 8; (a=8) The main point is that ++a increments…
In the case of for (i=start; i<end; i++) vs for (i=start; i<end; ++i) their meanings are completely identical, because the value of the expressions i++ and ++i are not used. (They are evaluated for side effects only.) Any compiler which…
This is not legal C++ code. The statement a++ = b; is not legal. Intuitively, you can only place something on the left-hand side of an assignment expression if it represents an object, not a value. For example, we can’t…
How are the pre and post increment / decrement operators are evaluated in C++ when they happen to occur repeatedly in a single cout statement? [duplicate] solved How are the pre and post increment / decrement operators are evaluated in…