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convert to hexadecimal

Posted on 2013-11-09
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Last Modified: 2013-12-02
I am trying to understand how the runtime converts this to hex

	std::stringstream s;
        s << std::hex;    
	int i = 0;
        s << std::setw(2) << std::setfill('0') << (unsigned int)(unsigned char)a[i];      // WORKS
        s << std::setw(2) << std::setfill('0') << (unsigned int)a[i];     // DOES NOT WORK
        

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1.  Why do we need this line
      s << std::hex;    

2) Why does it not work without (unsigned char)
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Question by:perlperl
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6 Comments
 
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Expert Comment

by:käµfm³d 👽
ID: 39635806
It's a stream manipulator. It coerces the stream to read its data as hexadecimal (where as the default is decimal). You could certainly read the data as a string first, and then do the conversion to hexadecimal value, but this would take a bit more code.

As to why it does not work without (unsigned char), what is the type of your a array?
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Author Comment

by:perlperl
ID: 39635832
The array was void* and passed to this function as const char*
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Expert Comment

by:jkr
ID: 39635841
You could use a 'unsigned int*' cast since it is an array, e.g.

        s << std::setw(2) << std::setfill('0') << ((unsigned int*)a)[i];

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Author Comment

by:perlperl
ID: 39636407
Thanks jkr. that works too
but I am trying to understand my solution as why
this works   (unsigned int)(unsigned char)
and why this not    (unsigned int)a
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Accepted Solution

by:
jkr earned 250 total points
ID: 39636429
>>but I am trying to understand my solution as why
>>this works   (unsigned int)(unsigned char)

This only works because you are printing 2-digit hexadecimal expressions that consist of one byte only. That will fail for values that consist of more than one byte..

Basically, you are casting 'void a[x]' to 'unsigned char' first. OK, that's fine. The next cast makes it clear to the stringstream that you want an unsigned int to be printed, so the single byte taken from 'a[ix]' (as a char) is taken and converted.
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Assisted Solution

by:sarabande
sarabande earned 250 total points
ID: 39639402
2) Why does it not work without (unsigned char)
there are two reasons:
- a void pointer points to unknown type. in order to calculate the address where a[ i ] is  located by "a + (i * sizeof(a[ 0 ])"  the compiler needs to know the size of a single element of the array.

 - an unsigned int is 32-bit while unsigned char is 8-bit. so omitting the unsigned char would cause 4 bytes to be converted to an integer, what gives wrong numbers.

note, my compiler does not accept the cast you posted as "works" cause a[ i ]  still has an undefined size (the cast only determines the wished output of the a[ i ] but not the type of the array) and the address of a[ i ] cannot correctly calculated, because of that.

to solve the issue correctly you need to cast the void pointer to a byte pointer (see also code of jkr above). when the compiler knows the type of the pointer, operations like + or [] will compile properly.


unsigned char x[] = { 33, 34, 35, 36 };
void * a = x;
std::cout << (unsigned int)(*(unsigned char*)a+i);

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the final cast to unsigned int is necessary to get the right operator<< selected which takes an int or unsigned int. otherwise the operator<<(std::ostream&, unsigned char) was used, which would not evaluate the std::hex manipulator, but would print the character code associated to the byte code.

to make all the code simpler you may use a byte buffer (pointer to unsigned char) instead of a void pointer as early as possible.

unsigned char * pbuf = (unsigned char *)getData();
...
std::cout << std::hex << std::setw(2) << std::right << std::setfill('0') << (int)pbuf[i];

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it also needs two casts, but is much better readable.

Sara
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