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# Strange behaviour of long long and unsigned int

 P: n/a Hi everybody, please, can someone explain me this behaviour. I have the following piece of code: long long ll; unsigned int i = 2; ll = -1 * i; printf("%lld\n", ll); Why this prints 4294967294 instead of -2? Thanks in advance Luke Mar 8 '06 #1
9 Replies

 P: n/a Just type cast -1 to long long. ll = (long long)-1 * i; Thats it. Mar 8 '06 #2

 P: n/a luke wrote: Hi everybody, please, can someone explain me this behaviour. I have the following piece of code: long long ll; unsigned int i = 2; ll = -1 * i; printf("%lld\n", ll); Why this prints 4294967294 instead of -2? http://c-faq.com/expr/intoverflow1.html Mar 8 '06 #3

 P: n/a "luke" writes: I have the following piece of code: long long ll; unsigned int i = 2; ll = -1 * i; printf("%lld\n", ll); Why this prints 4294967294 instead of -2? In ll = -1 * i; the expression -1 is of type int, and i is of type unsigned int, so the -1 is converted to unsigned int before the multiplication. Converting -1 to unsigned int yields UINT_MAX, which happens to be 4294967295 in your implementation. Multiplying 4294967295 by 2 wraps around (because it's unsigned arithmetic), yielding 4294967294. -- Keith Thompson (The_Other_Keith) ks***@mib.org San Diego Supercomputer Center <*> We must do something. This is something. Therefore, we must do this. Mar 8 '06 #4

 P: n/a Thanks. Where can I find the rules for arithmetic conversions? I found this rules (taken from an ANSI C manual): "First, if the corresponding real type of either operand is long double, the other operand is converted, without change of type domain, to a type whose corresponding real type is long double. Otherwise, if the corresponding real type of either operand is double, the other operand is converted, without change of type domain, to a type whose corresponding real type is double. Otherwise, if the corresponding real type of either operand is float, the other operand is converted, without change of type domain, to a type whose corresponding real type is float. Otherwise, the integer promotions are performed on both operands. Then the following rules are applied to the promoted operands: If both operands have the same type, then no further conversion is needed. Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank. 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. Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type. Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type." Are this rules good or maybe can you suggest anything better? Thanks Keith Thompson wrote: "luke" writes: I have the following piece of code: long long ll; unsigned int i = 2; ll = -1 * i; printf("%lld\n", ll); Why this prints 4294967294 instead of -2? In ll = -1 * i; the expression -1 is of type int, and i is of type unsigned int, so the -1 is converted to unsigned int before the multiplication. Converting -1 to unsigned int yields UINT_MAX, which happens to be 4294967295 in your implementation. Multiplying 4294967295 by 2 wraps around (because it's unsigned arithmetic), yielding 4294967294. Mar 8 '06 #5

 P: n/a luke wrote: Thanks. Where can I find the rules for arithmetic conversions? The ISO (and ANSI) C Standard. I found this rules (taken from an ANSI C manual): "First, if the corresponding real type of either operand is long double, the other .... Are this rules good or maybe can you suggest anything better? They are correct and useful, assuming you understand what is meant by integer rank. -- Thad Mar 9 '06 #6

 P: n/a Mmmh, I think rank is what is returned by sizeof() but I'm not sure about this. Can you explain me what rank really means? Thanks Luke Thad Smith wrote: luke wrote: Thanks. Where can I find the rules for arithmetic conversions? The ISO (and ANSI) C Standard. I found this rules (taken from an ANSI C manual): "First, if the corresponding real type of either operand is long double, the other ... Are this rules good or maybe can you suggest anything better? They are correct and useful, assuming you understand what is meant by integer rank. -- Thad Mar 9 '06 #7

 P: n/a In another ANSI C manual I found the following rules, which are different from the previous ones. Which one are good then?? 1. If either operand is long double, the other operand is converted to long double. 2. If either operand is double, the other operand is converted to double. 3. If either operand is float, the other operand is converted to float. 4. Integral promotions are performed on both operands, and then the rules listed below are followed. These rules are a strict extension of the ANSI "Usual Arithmetic Conversions" rule (Section 3.2.1.5). This extension ensures that integral expressions will involve long long only if one of the operands is of type long long. For ANSI conforming compilation, the integral promotion rule is as defined in Section 3.2.1.1 of the Standard. For non-ANSI compilation, the unsigned preserving promotion rule is used. 1. If either operand is unsigned long long, the other operand is converted to unsigned long long, 2. otherwise, if one operand is long long, the other operand is converted to long long, 3. otherwise, if either operand is unsigned long int, the other operand is converted to unsigned long int, 4. otherwise, if one operand is long int, and the other is unsigned int, and long int can represent all the values of an unsigned int, then the unsigned int is converted to a long int. (If one operand is long int, and the other is unsigned int, and long int can NOT represent all the values of an unsigned int, then both operands are converted to unsigned long int.) 5. If either operand is long int, the other operand is converted to long int. 6. If either operand is unsigned int, the other operand is converted to unsigned int. 7. Otherwise, both operands have type int. Regards Thad Smith wrote: Are this rules good or maybe can you suggest anything better? They are correct and useful, assuming you understand what is meant by integer rank. -- Thad Mar 9 '06 #8

 P: n/a "luke" wrote: [ Please do not top-post. Please do snip. Thank you. ] Mmmh, I think rank is what is returned by sizeof() but I'm not sure about this. No, but it's a related concept. sizeof returns the actual storage size any type or object takes in bytes of memory; rank is a (relative, and non-measurable) indicator of the ideal mathematical size of an integer type. Can you explain me what rank really means? It is defined precisely in paragraph 6.3.1.1 of the Standard, but basically, it's an indicator - not even a number - of how "large" a type is, in the sense of what it can and must be able to contain. For example: - a signed type and its corresponding unsigned type have the same rank - no signed integer types have the same rank, even if they look the same in all respects - a signed type which can hold larger numbers has higher rank than one which can hold smaller numbers - even if, say, INT_MAX == LONG_MAX, the rank of long is higher than that of int, and similar for the other default types - _Bool has the lowest rank of all and so on. The significance of this rank is that when integer types need to be compared to one another, types with lower rank are generally converted to types with higher rank. Richard Mar 9 '06 #9

 P: n/a Groovy hepcat luke was jivin' on 9 Mar 2006 01:49:02 -0800 in comp.lang.c. Re: Strange behaviour of long long and unsigned int's a cool scene! Dig it! In another ANSI C manual I found the following rules, which aredifferent from the previous ones. No they're not. They're the same. They're just stated differently. They amount to the same thing. Which one are good then?? Both. But the set of rules you've quoted here (which I've snipped) lacks a rule for when both operands have the same type. But otherwise it's fine. -- Dig the even newer still, yet more improved, sig! http://alphalink.com.au/~phaywood/ "Ain't I'm a dog?" - Ronny Self, Ain't I'm a Dog, written by G. Sherry & W. Walker. I know it's not "technically correct" English; but since when was rock & roll "technically correct"? Mar 12 '06 #10

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