[HH] c++ strings?

[Mark Woodward]

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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