Pointer A predincrements and pointer B postincrements in cycle. Why aren't they equival? What's wrong with pointer B? | Sololearn: Learn to code for FREE!
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Pointer A predincrements and pointer B postincrements in cycle. Why aren't they equival? What's wrong with pointer B?

//Part of code with cycle: do { --*I; ++*A; *B++; cout << " A: " << *A << ", B: " << *B <<endl; } while (*I); //Full code: https://code.sololearn.com/cRyHc43YY9ca/?ref=app

20th Jun 2019, 12:40 PM
–ü–į—Ä–į–ľ–Ņ–į–ľ–Ņ–į–ľ
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11 Respostas
+ 2
"But result are very different. When A = 2, B can be 0 or 35677 or something else, and next execution A = 3, B =0." The reason for that becomes clear when you come to grip with the fact of the uncertainty of implementation (compiler) reaction against presented undefined behavior which is caused by blindly moving pointers ahead into the regions (next 4 bytes in each iteration) not being assigned to the process. In this situation, everything is possible. The program might crash, or a segmentation fault might be triggered, or the program would happily generate incorrect results. Now you can probably figure out how those numbers come up "randomly", Example: If B is a pointer declares as `int *B = new int;` and initialized as `*B = 0` at address `0x1720c20` (before iteration), then for 3 iterations we'll have the following changes each time the execution flow reaches to `*B++`, Itr #1: the value of B = undefined, address = 0x1720c24 Itr #2: the value of B = undefined, address = 0x1720c28 Itr #3: the value of B = undefined, address = 0x1720c2c Also, please note that each dynamically allocated storage with `new` keyword must be accompanied by its respective `delete` keyword in order to fully handed the utilized memory to the operating system, otherwise the program causes memory leakage.
20th Jun 2019, 2:45 PM
To Seek Glory in Battle is Glorious
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+ 1
++*B;
20th Jun 2019, 1:24 PM
Nico Ruder
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+ 1
Nico Ruder ~ swim ~ Now I got it. Thank for your help!
20th Jun 2019, 2:01 PM
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+ 1
no problem
20th Jun 2019, 2:06 PM
Nico Ruder
Nico Ruder - avatar
+ 1
"delete I, A, B" Although the above construct is valid, yet if you have a look at the way the comma operator[1] treats both sides of itself and most importantly the precedence of comma and `delete` operators [2], you come to realize that what you are doing is actually deallocating "only" the pointer `I`. That is, (delete I) , (A, B) First, the left-hand-side (LHS) of `, ` (inside the left parenthesis) being evaluated and the result (if any) is discarded. Then, after checking for any side effect of LHS the RHS of `, ` get through the same process. So, you want to make sure to separately delete each as delete I; delete A; delete B; [1] https://en.cppreference.com/w/c/language/operator_other#Comma_operator [2] https://en.cppreference.com/w/cpp/language/operator_precedence
21st Jun 2019, 1:14 PM
To Seek Glory in Battle is Glorious
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0
the pluses before B i would say
20th Jun 2019, 1:21 PM
Nico Ruder
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0
because ++var increaments first and than goes on and var++ increments first after the next execution
20th Jun 2019, 1:28 PM
Nico Ruder
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0
because ++var increaments first and than goes on and var++ increments first after the next execution
20th Jun 2019, 1:29 PM
Nico Ruder
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0
But result are very different. When A = 2, B can be 0 or 35677 or something else, and next execution A = 3, B =0.
20th Jun 2019, 1:39 PM
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~ swim ~ Just tryed it with int array, works generally correctly.
20th Jun 2019, 2:20 PM
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To Seek Glory in Battle is Glorious Thank you for the detailed answer, I fixed the mistake, and have another question: is next part of code correct? "delete I, A, B; I = A = B = 0;"
21st Jun 2019, 12:18 PM
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