[HH] c++ strings?

[Jerry Feldman]

Mark,
Initially what you say is true, but the C and C++ standards did diverge.
You have got to be careful in C++ to use iostream and not C stdio, but
if you do use C stdio, you should know what you are doing. Remember that
the classes cin, cout, cerr are classes, and the '<<' and '>>' are
overloaded operators that depend on the data type they operate on. In
printf, et. al. you use %c to print out a single character as a
character, but %d to print out its decimal value. The C++ programmer
needs to know the behaviors of the classes and operators used as well as
the C language promotion rules. My bizarre example below is the same
result in C or C++ because of the promotion rules. So, as you indicate,
a "char' is a signed 8-bit integer, and behaves that way,

On 12/15/2012 03:39 PM, Mark Woodward wrote:
> This whole discussion exemplifies one of my pet peeves about C++
> practitioners. C++ is a superset of C, thus, all the constructs
> available in C are also available in C++.
>
> A simple printf or snprintf would have truncated this discussion quickly.
>
> printf("byte->%02x<-byte int->%02x<-int", (unsigned int) mybyte,  myint);
>
> In both C and C++ you need to take care that "char" is a signed byte,
> i.e. anything over 0x7F will be a negative number and any conversion
> to an integer will sign extend and make it 0xffffff - something
>
> On 12/15/2012 11:19 AM, Jerry Feldman wrote:
>> On 12/13/2012 06:33 PM, Greg London wrote:
>>> Hmm. C++ is NOT helping here.
>>>
>>>    char mybyte=0x52;
>>>    int myint=0x52;
>>>
>>>    cout<<"byte->"<<std::hex<<mybyte<<"<-byte"
>>>        <<"int->"<<std::hex<<myint<<"<-int"<<endl;
>>>
>>>    output is:: byte->R<-byte  int->52<-int
>>>
>>>
>>> It wants to print any 8 bit type, signed or unsigned, as a character,
>>> even if there's a std::hex in front of it.
>>>
>>> Is there an easy fix for this?
>>>
>>> I could fake it out by converting the byte to an int,
>>> and then masking the upper bits I suppose.
>>>
>>> But it makes dealing with 8 bit data a bit of a pain.
>>>
>>> Greg
>>>
>>>
>>>
>>>> That's it.
>>>> Thanks!
>>>>
>>>>
>>>>> Try this:
>>>>> ss << "actual=0x" << std::hex << actual << " expected=0x" << std::hex <<
>>>>> expected << " " << msg;
>>>>>
>> cout<<"byte->"<<std::hex<<mybyte<<"<-byte"
>>        <<"int->"<<std::hex<<myint<<"<-int"<<endl;
>>
>> In this case, mbyte is a char, and the value of 0x52 is 'R'.
>> Solution cast the byte to an int.
>>
>> cout<<"byte->"<<std::hex<<|static_cast<int>(|mybyte)<<"<-byte"
>>        <<"int->"<<std::hex<<myint<<"<-int"<<endl;
>>
>> Here is an interesting one for you:
>> #include <iostream>
>> using namespace std;
>>
>> int main()
>> {
>>  
>>   int a = -2;
>>   unsigned b = 1;
>>   long result;
>>   result = a * b;
>>   cout << "Result is " << result << " or 0x" << std::hex << result << "\n";
>>   return 0;
>> }
>> In the above example, the result is -2 if compiled on a 32-bit system,
>> and 4294967294 if compiled on a 64-bit system. The issue is that the
>> expression, a + b, becomes an unsigned 32-bit integer expression, so
>> when the result of the expression is assigned to result, there is no
>> sign extension. In the 32-bit environment, result would be 32-bits, with
>> the high order bit set.
>> The hex result in both is the same: 0xfffffffe.
>>
>>
>>
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>


-- 
Jerry Feldman <gaf at blu.org>
Boston Linux and Unix
PGP key id:3BC1EB90 
PGP Key fingerprint: 49E2 C52A FC5A A31F 8D66  C0AF 7CEA 30FC 3BC1 EB90


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