increment

My interest is in the difference between for and while loops. I know that the post-increment value is used and then incremented and the operation returns a constant pre-increment. while (true) { //... i++; int j = i; } Here, will j contain the old i or the post-incremented i at the end of the loop? zennehoy Since the statement i++ ends at the ; in your example, it makes no difference whether you use pre- or post-increment. The difference arises when you utilize the result: int j = i++; // i will contain i_old + 1, j will contain the i_old. Vs: int j = ++i; // i and j will both contain i_old +

I want to create a non-thread-safe chunk of code for experimentation, and those are the functions that 2 threads are going to call. c = 0 def increment(): c += 1 def decrement(): c -= 1 Is this code thread safe? If not, may I understand why it is not thread safe, and what kind of statements usually lead to non-thread-safe operations. If it is thread-safe, how can I make it explicitly non-thread-safe? Single opcodes are thread-safe because of the GIL but nothing else: import time class something(object): def __init__(self,c): self.c=c def inc(self): new = self.c+1 # if the thread is interrupted

This question already has an answer here: Why are these constructs using pre and post-increment undefined behavior? 14 answers int b=0,a=1;b= ++a + ++a; what is the value of b? what is the calculation for it? [duplicate] 2 answers I'm new to C language so plz sum1 help me out. A C code written int i=3; printf("%d",++i + ++i); Complier gvs O/P =9. How? Thanx in advance The results are undefined. You're modifying a variable more than once in an expression (or sequence point to be more accurate). Modifying a variable more than once between sequence points is undefined, so don't do it. It might be

I am learning C language and quite confused the differences between ++*ptr and *ptr++ . For example: int x = 19; int *ptr = &x; I know ++*ptr and *ptr++ produce different results but I am not sure why is that? templatetypedef These statements produce different results because of the way in which the operators bind. In particular, the prefix ++ operator has the same precedence as * , and they associate right-to-left. Thus ++*ptr is parsed as ++(*ptr) meaning "increment the value pointed at by ptr ,". On the other hand, the postfix ++ operator has higher precedence than the dereferrence operator

I was browsing Google Code when I chanced upon this project called JSpeed - optimization for Javascript. I noticed one of the optimization was to change i++ to ++i in for loop statements. Before Optimization for (i=0;i<1;i++) {} for (var i = 0, j = 0; i < 1000000; i++, j++) { if (i == 4) { var tmp = i / 2; } if ((i % 2) == 0) { var tmp = i / 2; i++; } } var arr = new Array(1000000); for (i = 0; i < arr.length; i++) {} After optimization for(var i=0;i<1;++i){} for(var i=0,j=0;i<1000000;++i,++j){if(i==4){var tmp=i>>1;} if((i&1)==0){var tmp=i>>1;i++;}} var arr=new Array(1000000);for(var i=0,arr

Why is ++i is l-value and i++ not? Johannes Schaub - litb Well as another answerer pointed out already the reason why ++i is an lvalue is to pass it to a reference. int v = 0; int const & rcv = ++v; // would work if ++v is an rvalue too int & rv = ++v; // would not work if ++v is an rvalue The reason for the second rule is to allow to initialize a reference using a literal, when the reference is a reference to const: void taking_refc(int const& v); taking_refc(10); // valid, 10 is an rvalue though! Why do we introduce an rvalue at all you may ask. Well, these terms come up when building the