In the following piece of code the output is 1 but I'm not sure why. Has i been implicitly changed to a boolean? How can an integer value be set to an expression like this?
#include <iostream>
using namespace std;
int main (int argc, const char *argv[])
{
int a = 1, b = 1, c = 1, i = 1;
i = b < a < c;
cout << i;
return 0;
}
Has i been implicitly changed to a boolean?
No. Rather, a boolean expression has been implicitly converted to int, then stored in i, as commented by #WhozCraig.
How can an integer value be set to an expression like this?
Because it follows the C++ Standard.
When you do:
i = b < a < c;
since the first < and the second < have the same operator precedence (since they are the same operator), so:
Operators that have the same precedence are bound to their arguments
in the direction of their associativity. For example, the expression a
= b = c is parsed as a = (b = c), and not as (a = b) = c because of right-to-left associativity of assignment, but a + b - c is parsed (a > + b) - c and not a + (b - c) because of left-to-right associativity of addition and subtraction.
as mentioned in C++ Operator Precedence, which means that this expression of yours will be parsed as if it was written like this:
i = (b < a) < c;
which will evaluate the condition inside the parentheses, which is False.
Now bool to int conversion is implicit:
ยง4.7/4 from the C++ Standard says (Integral Conversion)
If the source type is bool, the value false is converted to zero and the value true is converted to one.
which means that false will be converted to 0, and then we basically end up doing:
i = 0 < c;
which evaluates to true, since c is equal to 1. Now i is of type int, which means that another, second implicit conversion from boolean to integer takes place, eventually assigning 1 to i.
I tried to get the result of -1 modulo 1000000007 using the % operator of C++
and fmod function.
The output is -1, but -1 modulo 1000000007==1000000006.
What have I done wrong?
Plainly said, you took the wrong operator.
C++ and C % is not modulo, but remainder.
assert(a / b * b + a % b == a); // for integral types
If a is non-negative, modulo and remainder are the same.
Otherwise the return-value is negative, just add b.
template<class T>
inline constexpr auto
modulo(T a, T b) -> decltype(a%b) {
auto r = a % b;
if (r < 0) r += b;
return r;
}
Or (also) for C:
#define modulo(a, b) (a % b < 0 ? a % b + b : a % b)
For completeness: Before C++11, a / b could always round down instead of always to 0, though C++03 already had a note that the next standard would probably mandate rounding to 0.
See Wikipedia on modulo:
Modulo is the remainder of euclidiean division, and always in range 0 <= modulo < divisor
And on remainder:
In mathematics, the remainder is the amount "left over" after performing some computation.
Modulo function that handles negative divisors as well as positive divisors:
template<class T>
inline constexpr auto
modulo(T a, T b) -> decltype(a%b) {
auto r = a % b;
if((b > 0 && r < 0) || (b < 0 && r > 0))
r += b;
return r;
}
I ran into a behavior which I didn't expect using bitwise operations on unsigned ints. I'll cut right to my example.
unsigned int a = 0;
unsigned int b = 0;
std::printf("a & b: %u\n", a & b);
std::printf("a == b: %i\n", a == b);
std::printf("a & b == a: %i\n", a & b == a);
The above code produces the following output:
a & b: 0
a == b: 1
a & b == a: 0
The last line is what confuses me. Shouldn't a & b == a evaluate to true, since a & b == (unsigned int)0 and a == (unsigned int)0?
You're getting this behavior because you didn't realize == comes before & in the C operator precedence table. In fact, a good compiler will warn you straight away about your code:
t.cpp:10:35: warning: & has lower precedence than ==; == will be evaluated first [-Wparentheses]
std::printf("a & b == a: %i\n", a & b == a);
^~~~~~~~
t.cpp:10:35: note: place parentheses around the '==' expression to silence this warning
std::printf("a & b == a: %i\n", a & b == a);
^
( )
t.cpp:10:35: note: place parentheses around the & expression to evaluate it first
std::printf("a & b == a: %i\n", a & b == a);
^
( )
Make sure your warnings are turned on, like g++ -Wall -Wextra -Werror.
You should write:
(a & b) == a
Now you'll get 1 since a & b will be evaluated first:
(a & b) = 0, 0 == 0 is 1.
In your case, a & b == a is evaluated as a & (b == a), b == a is 1, and a & 1 is 0.
Due to =='s precedence over &, a & b == a gets evaluated as a & (b == a) (and not as (a & b) == a as you appear to have expected).
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Undefined Behavior and Sequence Points
Working out the below code by hand:
#include <stdio.h>
int func (int a, int b) {
static int c = 1;
return a + b * (c *= -1);
}
int main () {
int a = 2, b = 3;
int c = func(a, b);
a *= a++;
b *= ++b;
printf("%d %d %d %d\n", a, b, c, func(a, b));
}
I calculate the variables in printf() to be as follows:
a = 5, b = 16, c = -1, func(a, b) = -11
however my compiler tells me the last value is in fact 21.
Output:
a = 5, b = 16, c = -1, func(a, b) = 21n
I'd calculate my value as (16*-1) + 5
Can anyone tell me where I have gone wrong?
a *= a++;
b *= ++b;
both statements are undefined behavior in C. They are violating C sequence points rules.
a *= a++;
is equivalent to:
a = a * a++;
and modifying an object twice between the previous and the next sequence point is undefined behavior in C (C99, 6.5p2).
Simple question - In c++, what's the neatest way of getting which of two numbers (u0 and u1) is the smallest positive number? (that's still efficient)
Every way I try it involves big if statements or complicated conditional statements.
Thanks,
Dan
Here's a simple example:
bool lowestPositive(int a, int b, int& result)
{
//checking code
result = b;
return true;
}
lowestPositive(5, 6, result);
If the values are represented in twos complement, then
result = ((unsigned )a < (unsigned )b) ? a : b;
will work since negative values in twos complement are larger, when treated as unsigned, than positive values. As with Jeff's answer, this assumes at least one of the values is positive.
return result >= 0;
I prefer clarity over compactness:
bool lowestPositive( int a, int b, int& result )
{
if (a > 0 && a <= b) // a is positive and smaller than or equal to b
result = a;
else if (b > 0) // b is positive and either smaller than a or a is negative
result = b;
else
result = a; // at least b is negative, we might not have an answer
return result > 0; // zero is not positive
}
Might get me modded down, but just for kicks, here is the result without any comparisons, because comparisons are for whimps. :-)
bool lowestPositive(int u, int v, int& result)
{
result = (u + v - abs(u - v))/2;
return (bool) result - (u + v + abs(u - v)) / 2;
}
Note: Fails if (u + v) > max_int. At least one number must be positive for the return code to be correct. Also kudos to polythinker's solution :)
unsigned int mask = 1 << 31;
unsigned int m = mask;
while ((a & m) == (b & m)) {
m >>= 1;
}
result = (a & m) ? b : a;
return ! ((a & mask) && (b & mask));
EDIT: Thought this is not so interesting so I deleted it. But on the second thought, just leave it here for fun :) This can be considered as a dump version of Doug's answer :)
Here's a fast solution in C using bit twiddling to find min(x, y). It is a modified version of #Doug Currie's answer and inspired by the answer to the Find the Minimum Positive Value question:
bool lowestPositive(int a, int b, int* pout)
{
/* exclude zero, make a negative number to be larger any positive number */
unsigned x = (a - 1), y = (b - 1);
/* min(x, y) + 1 */
*pout = y + ((x - y) & -(x < y)) + 1;
return *pout > 0;
}
Example:
/** gcc -std=c99 *.c && a */
#include <assert.h>
#include <limits.h>
#include <stdio.h>
#include <stdbool.h>
void T(int a, int b)
{
int result = 0;
printf("%d %d ", a, b);
if (lowestPositive(a, b, &result))
printf(": %d\n", result);
else
printf(" are not positive\n");
}
int main(int argc, char *argv[])
{
T(5, 6);
T(6, 5);
T(6, -1);
T(-1, -2);
T(INT_MIN, INT_MAX);
T(INT_MIN, INT_MIN);
T(INT_MAX, INT_MIN);
T(0, -1);
T(0, INT_MIN);
T(-1, 0);
T(INT_MIN, 0);
T(INT_MAX, 0);
T(0, INT_MAX);
T(0, 0);
return 0;
}
Output:
5 6 : 5
6 5 : 5
6 -1 : 6
-1 -2 are not positive
-2147483648 2147483647 : 2147483647
-2147483648 -2147483648 are not positive
2147483647 -2147483648 : 2147483647
0 -1 are not positive
0 -2147483648 are not positive
-1 0 are not positive
-2147483648 0 are not positive
2147483647 0 : 2147483647
0 2147483647 : 2147483647
0 0 are not positive
This will handle all possible inputs as you request.
bool lowestPositive(int a, int b, int& result)
{
if ( a < 0 and b < 0 )
return false
result = std::min<unsigned int>( a, b );
return true;
}
That being said, the signature you supply allows sneaky bugs to appear, as it is easy to ignore the return value of this function or not even remember that there is a return value that has to be checked to know if the result is correct.
You may prefer one of these alternatives that makes it harder to overlook that a success result has to be checked:
boost::optional<int> lowestPositive(int a, int b)
{
boost::optional<int> result;
if ( a >= 0 or b >= 0 )
result = std::min<unsigned int>( a, b );
return result;
}
or
void lowestPositive(int a, int b, int& result, bool &success)
{
success = ( a >= 0 or b >= 0 )
if ( success )
result = std::min<unsigned int>( a, b );
}
tons of the answers here are ignoring the fact that zero isn't positive :)
with tricky casting and tern:
bool leastPositive(int a, int b, int& result) {
result = ((unsigned) a < (unsigned) b) ? a : b;
return result > 0;
}
less cute:
bool leastPositive(int a, int b, int& result) {
if(a > 0 && b > 0)
result = a < b ? a : b;
else
result = a > b ? a : b:
return result > 0;
}
I suggest you refactor the function into simpler functions. Furthermore, this allows your compiler to better enforce expected input data.
unsigned int minUnsigned( unsigned int a, unsigned int b )
{
return ( a < b ) ? a : b;
}
bool lowestPositive( int a, int b, int& result )
{
if ( a < 0 && b < 0 ) // SO comments refer to the previous version that had || here
{
return false;
}
result = minUnsigned( (unsigned)a, (unsigned)b ); // negative signed integers become large unsigned values
return true;
}
This works on all three signed-integer representations allowed by ISO C:
two's complement, one's complement, and even sign/magnitude. All we care about is that any positive signed integer (MSB cleared) compares below anything with the MSB set.
This actually compiles to really nice code with clang for x86, as you can see on the Godbolt Compiler Explorer. gcc 5.3 unfortunately does a much worse job.
Hack using "magic constant" -1:
enum
{
INVALID_POSITIVE = -1
};
int lowestPositive(int a, int b)
{
return (a>=0 ? ( b>=0 ? (b > a ? a : b ) : INVALID_POSITIVE ) : INVALID_POSITIVE );
}
This makes no assumptions about the numbers being positive.
Pseudocode because I have no compiler on hand:
////0 if both negative, 1 if u0 positive, 2 if u1 positive, 3 if both positive
switch((u0 > 0 ? 1 : 0) + (u1 > 0 ? 2 : 0)) {
case 0:
return false; //Note that this leaves the result value undef.
case 1:
result = u0;
return true;
case 2:
result = u1;
return true;
case 3:
result = (u0 < u1 ? u0 : u1);
return true;
default: //undefined and probably impossible condition
return false;
}
This is compact without a lot of if statements, but relies on the ternary " ? : " operator, which is just a compact if, then, else statement. "(true ? "yes" : "no")" returns "yes", "(false ? "yes" : "no") returns "no".
In a normal switch statement after every case you should have a break;, to exit the switch. In this case we have a return statement, so we're exiting the entire function.
With all due respect, your problem may be that the English phrase used to describe the problem really does hide some complexity (or at least some unresolved questions). In my experience, this is a common source of bugs and/or unfulfilled expectations in the "real world" as well. Here are some of the issues I observed:
Some programmers use a naming
convention in which a leading u
implies unsigned, but you didn't
state explicitly whether your
"numbers" are unsigned or signed
(or, for that matter, whether they
are even supposed to be integral!)
I suspect that all of us who read it
assumed that if one argument is
positive and the other is not, then
the (only) positive argument value
is the correct response, but that is
not explicitly stated.
The description also doesn't define
the required behavior if both values
are non-positive.
Finally, some of the responses
offered prior to this post seem to
imply that the responder thought
(mistakenly) that 0 is positive! A
more specific requirements statement
might help prevent any
misunderstanding (or make it clear
that the issue of zero hadn't been
thought out completely when the
requirement was written).
I'm not trying to be overly critical; I'm just suggesting that a more precisely-written requirement will probably help, and will probably also make it clear whether some of the complexity you're concerned about in the implementation is really implicit in the nature of the problem.
Three lines with the use (abuse?) of the ternary operator
int *smallest_positive(int *u1, int *u2) {
if (*u1 < 0) return *u2 >= 0 ? u2 : NULL;
if (*u2 < 0) return u1;
return *u1 < *u2 ? u1 : u2;
}
Don't know about efficiency or what to do if both u1 and u2 are negative. I opted to return NULL (which has to be checked in the caller); a return of a pointer to a static -1 might be more useful.
Edited to reflect the changes in the original question :)
bool smallest_positive(int u1, int u2, int& result) {
if (u1 < 0) {
if (u2 < 0) return false; /* result unchanged */
result = u2;
} else {
if (u2 < 0) result = u1;
else result = u1 < u2 ? u1 : u2;
}
return true;
}
uint lowestPos(uint a, uint b) { return (a < b ? a : b); }
You are looking for the smallest positive, it is be wise to accept positive values only in that case. You don't have to catch the negative values problem in your function, you should solve it at an earlier point in the caller function. For the same reason I left the boolean oit.
A precondition is that they are not equal, you would use it like this in that way:
if (a == b)
cout << "equal";
else
{
uint lowest = lowestPos(a, b);
cout << (lowest == a ? "a is lowest" : "b is lowest");
}
You can introduce const when you want to prevent changes or references if you want to change the result. Under normal conditions the computer will optimize and even inline the function.
No cleverness, reasonable clarity, works for ints and floats:
template<class T>
inline
bool LowestPositive( const T a, const T b, T* result ) {
const bool b_is_pos = b > 0;
if( a > 0 && ( !b_is_pos || a < b ) ) {
*result = a;
return true;
}
if( b_is_pos ) {
*result = b;
return true;
}
return false;
}
Note that 0 (zero) is not a positive number.
OP asks for dealing with numbers (I interpret this as ints and floats).
Only dereference result pointer if there is a positive result (performance)
Only test a and b for positiveness once (performance -- not sure if such a test is expensive?)
Note also that the accepted answer (by tvanfosson) is wrong. It fails if a is positive and b is negative (saying that "neither is positive"). (This is the only reason I add a separate answer -- I don't have reputation enough to add comments.)
My idea is based on using min and max. And categorized the result into three cases, where
min <= 0 and max <= 0
min <= 0 and max > 0
min > 0 and max > 0
The best thing is that it's not look too complicated.
Code:
bool lowestPositive(int a, int b, int& result)
{
int min = (a < b) ? a : b;
int max = (a > b) ? a : b;
bool smin = min > 0;
bool smax = max > 0;
if(!smax) return false;
if(smin) result = min;
else result = max;
return true;
}
After my first post was rejected, allow me to suggest that you are prematurely optimizing the problem and you shouldn't worry about having lots of if statements. The code you're writing naturally requires multiple 'if' statements, and whether they are expressed with the ternary if operator (A ? B : C) or classic if blocks, the execution time is the same, the compiler is going to optimize almost all of the code posted into very nearly the same logic.
Concern yourself with the readability and reliability of your code rather than trying to outwit your future self or anyone else who reads the code. Every solution posted is O(1) from what I can tell, that is, every single solution will contribute insignificantly to the performance of your code.
I would like to suggest that this post be tagged "premature optimization," the poster is not looking for elegant code.