why does below program gives the output that b is greater than a? Even though b contains -2.
void main()
{
unsigned int a=12;
int b=-2;
if(a>b)
printf("a is greater");
else
printf("b is greater");
getch();
}
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Can you clarify your question? Why do you not expect 12 > -2?michaelrccurtis– michaelrccurtis2015-09-30 13:12:46 +00:00Commented Sep 30, 2015 at 13:12
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Um, math agrees with your results.Jonathon Reinhart– Jonathon Reinhart2015-09-30 13:13:24 +00:00Commented Sep 30, 2015 at 13:13
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Take a look on Usual Arithmetic ConversionMissu– Missu2015-09-30 13:16:32 +00:00Commented Sep 30, 2015 at 13:16
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The conversion goes the opposite way: signed -> unsigned.molbdnilo– molbdnilo2015-09-30 13:24:19 +00:00Commented Sep 30, 2015 at 13:24
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possible duplicate of Signed versus Unsigned IntegersJohn Bollinger– John Bollinger2015-09-30 13:54:43 +00:00Commented Sep 30, 2015 at 13:54
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4 Answers
First, to quote the C11 standard for relational operators, chapter 6.5.8
If both of the operands have arithmetic type, the usual arithmetic conversions are performed.
Now, following the description in chapter 6.3.1.8, Usual arithmetic conversions, if you try to perform arithmetic operation between a signed and an unsigned integer (type), the signed one will get promoted to unsigned type (higher rank) and then the operation will take place.
So, here, for the comparison, the value of b is getting converted to unsigned type and you're getting the wrong output there.
To quote the relevant part, from the same chapter
[...] Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.
You can also check on the usual arithmetic promotion rule here
That said, void main() should be int main(int argc, char* argv[]), or, at least, int main(void).
5 Comments
Psytho
Comparing is not an arithmetic operation.
Sourav Ghosh
@Alex.S
If both of the operands have arithmetic type, the usual arithmetic conversions are performed.
chux
@Alex.S How is comparing not an arithmetic operation?
Psytho
@chux Because comparison is not an arithmetic operation. en.wikipedia.org/wiki/Arithmetic
chux
@Alex.S At the risk of a duck/rabbit season exchange, a comparison like
a>b is done by subtraction which is certain arithmetic.To perform the comparison, both operands are first converted to the same type. In this case, int b is converted to the higher ranking unsigned. For the comparison values, it is then 12 > (-2 + (UINTMAX + 1))? which is false.
To compare in the usual mathematical sense:
unsigned int a;
int b;
if ((b < 0) || (a > b)) printf("a is greater");
Comments
It's one of the type promotion rules: if one argument is an int and the other an unsigned int then the int is promoted to an unsigned int, adding UINT_MAX + 1 if necessary.
This happens prior to the comparison.
Comments
if(a>b)
In this a is unsigned int and b is signed int , so due to implicit conversion , b will be converted to unsigned int type and b will have a large value (no need to say, it will be greater than a).
Therefore ,you get unexpected result.
According to C99- 6.3.1.8 Usual arithmetic conversions
[...]
3.Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.
