This question already has answers here:
bool operator ++ and --
(4 answers)
Closed 2 years ago.
why incase of boolean, overflow doesn't occur in circular fashion. eg say a=126 when you reach 128 and you increment it, a goes to -127 if range is -127 to 128. similarly for boolean it is 0 to 1 so it should move around 0101010101 and so on. please clarify
using namespace std;
int main()
{
bool a;
for (a = 1; a <= 5; a++)
cout << a;
return 0;
}
From cppreference
If the operand of the pre-increment operator is of type bool, it is set to true
If the operand of the post-increment operator is of type bool, it is set to true
Note that this behaviour was removed in C++17 and your code won't compile with newer standards (probably because it was confusing).
Related
This question already has answers here:
Undefined behavior and sequence points
(5 answers)
Is the behaviour of i = i++ really undefined?
(9 answers)
Closed 3 years ago.
I am facing problems in understanding the increment orders in C++.
I am aware that increments are unary operators thus they come after parenthesis from right to left.
My question is when do we increase the number?
Here is a simple code:
#include <iostream>
using namespace std;
int main()
{
int a1;
int a(12),b(3);
a1=7+10%3-5;
b=a/b++;
cout<<a1<<"\t"<<b<<endl;
return 0;
}
Here I get a=3 that's right but b=5 , I think it is 3 because we start from the right and increase by 1 then 12/4 gives 3.
Note that the C++ grammar implies that the associativity of the postfix increment is from left to right, and that of the prefix increment is right to left.
The behaviour of b = a / b++; is actually undefined. This is because = is not a sequencing point, so there are simultaneous reads and writes on b.
(The same applies to C.)
It's a variant on i = i++;: for more on that see Is the behaviour of i = i++ really undefined?
This question already has answers here:
What happens when I assign a number larger than INT_MAX to an int?
(3 answers)
Closed 4 years ago.
On Linux x86_64:
size_t some_constant = std::numeric_limits<size_t>::max();
int my_int = some_constant;
size_t my_size_t = my_int;
cout << (my_size_t == some_constant) << endl;
this prints 1
Is this an example of UB?
Not undefined, just implementation defined.
[conv.integral]
3 If the destination type is signed, the value is unchanged if it can be
represented in the destination type; otherwise, the value is
implementation-defined.
So whatever your implementation of C++ does, it must document it somehow. Which also makes it far more likely that it won't break suddenly, because implementers usually like to keep documented behavior unchanged.
This question already has answers here:
How do I detect unsigned integer overflow?
(31 answers)
Closed 6 years ago.
Edit (IMPORTANT): This question was more specific than the question this was marked as a duplicate against. This was asking how to do it with a boolean function. But now I know that it just doesn't work like that. This question shows that a + b = c can only be overflow checked if you write if(c - b == a) and that there is no operator independent way of checking.
Once the overflow happened, you cannot detect it
-lorro
I have been looking for a way to detect integer overflow in C++ using an if statement.
Possible pseudo code:
#include <iostream>
#include <...>
using namespace std;
bool isOverflow(...);
int main()
{
int a = INT_MAX + 1;
if (isOverflow(...))
{
cout << "Overflow" << endl;
}
else
{
cout << "No Overflow" << endl;
}
return 0;
}
bool isOverflow
{
...
}
OK to be honest this pseudo code preference may not work, but i've seen this question asked many times and have not found any useful answers. It may require unsigned or unsigned long long, although I'm not necessarily encouraging the use of those.
EDIT:
I would like to use it with a multiplication sentence with 3 numbers:
a * a * b
I am aware that there is a pow function in <math.h> but that's off topic.
I am also aware that if I want an accurate int result from pow I would use:
int(pow(base, index) + 0.5)
It depends on what sort of operation you are using and what are the types of the operands.
e.g. if you want to detect an overflow after addition, and both operands are unsigned integers, then an overflow will have occurred if the result is less than the sum of both operands.
bool overflow;
if (a+b < a)
overflow = true;
else
overflow = false;
For signed integers, you can refer to the excellent post here
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
In C arrays why is this true? a[5] == 5[a]
How is it possible that this is valid C++?
void main()
{
int x = 1["WTF?"];
}
On VC++10 this compiles and in debug mode the value of x is 84 after the statement.
What's going on?
Array subscript operator is commutative. It's equivalent to int x = "WTF?"[1]; Here, "WTF?" is an array of 5 chars (it includes null terminator), and [1] gives us the second char, which is 'T' - implicitly converted to int it gives value 84.
Offtopic: The code snippet isn't valid C++, actually - main must return int.
You can read more in-depth discussion here: In C arrays why is this true? a[5] == 5[a]
int x = 1["WTF?"];
equals to
int x = "WTF?"[1];
84 is "T" ascii code
The reason why this works is that when the built-in operator [] is applied to a pointer and an int, a[b] is equivalent to *(a+b). Which (addition being commutative) is equivalent to *(b+a), which, by definition of [], is equivalent to b[a].
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Best way to detect integer overflow in C/C++
how do we check if any arithmetic operation like addition, multiplication or subtraction could result in an overflow?
Check the size of the operands first, and use std::numeric_limits. For example, for addition:
#include <limits>
unsigned int a, b; // from somewhere
unsigned int diff = std::numeric_limits<unsigned int>::max() - a;
if (diff < b) { /* error, cannot add a + b */ }
You cannot generally and reliably detect arithmetic errors after the fact, so you have to do all the checking before.
You can easily template this approach to make it work with any numeric type.