pre-increment

(Note: I am not asking about the definitions of pre-increment vs. post-increment, or how they are used in C/C++. Therefore, I do not think this is a duplicate question.) Developers of C (Dennis Ritchie et al) created increment and decrement operators for very good reasons. What I don't understand is why they decided to create the distinction of pre- vs post- increments/decrements? My sense is that these operators were far more useful when C was being developed than today. Most C/C++ programmers use one or the other, and programmers from other languages find the distinction today bizarre and

This question already has an answer here: Undefined behavior and sequence points 5 answers #include "stdafx.h" #include <iostream> #include <conio.h> using namespace std; int main() { int n = 5; cout<< n++ <<" "<< ++n << " "<< n++; _getch(); return 0; } When I run this program on Visual Studio the output is 7 8 5. I think it is compiler dependent. ( Correct me if I am wrong) But shouldn't it be either 7 7 5 or 5 7 7 ? The sequence, in which the various n++ / ++n are executed, is undefined by the C standard and might change over time or depending on target machine and/or optimization options.

This question already has an answer here: Java: Prefix/postfix of increment/decrement operators? 10 answers I wondering to know why this snippet of code give output 112 How this last digit 2 was creating? public static void main(String[] args) { int i = 0; System.out.print(++i); System.out.print(i++); System.out.print(i); Why does this happen? Your snippet it's translated as int i = 0; i = i + 1; // 1 System.out.print(i); // 1 System.out.print(i); // 1 i = i + 1; // 2 System.out.print(i); // 2 That's why the final result it's a 2. ++i it's incrementing the variable before being called by the

I have never seen the usecase for pre-increment and post-increment in actual code. The only place i see them most often are puzzles. My opinion is, it introduces more confusion rather than being useful. is there any real use case scenario for this can't this can be done by using += y = x++ y = x x += 1 It's just a shorter way of writing the same thing and it's only confusing to those who don't deeply understand C (a) . The same argument could be made for replacing: for (i = 0; i < 10; i++) printf ("%d\n", i); with: i = 0; while (i < 10) { printf ("%d\n", i); i = i + 1; } since any for can also

This question already has an answer here: Not able to understand error condition wrt lvalues 3 answers I have been trying to understand pointer concepts by writing simple code, and I got an error problem, and it seems like I couldn't solve it or understand it. #include <stdio.h> int *foo(void); int main(void) { printf("%d\n", *foo()); return 0; } int *foo(void) { static int num = 1; ++num; return &(++num); } Here is the error message. error: lvalue required as unary ‘&’ operand return &(++num); Function 'foo()' returns a pointer to int, and main is supposed to be print the returned int by