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 short In C, why doesn’t num1++ increment in the printf()? num1++ does increment num1 but … Read more
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 processed something like int temp = k; k++; k = temp; You can try the … Read more
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 compiles. 1 solved Why can’t variables be declared and modified at once? (C++)
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 returned by a post-increment operator? [closed]
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 of an assignment. For the updated code: float x,y; *x++ = *y++ This is not … Read more
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 the value and immediately returns it. a++ also increments the value (in the background) but … Read more
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 produces different code for the two is pathologically bad and should be chucked in the … Read more
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 write x + y = z; because x + y is a value, not an … Read more
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 C++ when they happen to occur repeatedly in a single cout statement? [duplicate]