How to convert ’QString 1’ to ’unsigned char’ (solved)

[JKSH]

Are you trying to assign the whole string, or just one character? (Remember that a string is an array of characters)

Asked another way: Do you want to use "unsigned char" or "unsigned char *"?

[houmingc]

Trying to assigned one character
Trying to replace Array[6] with QString value & Array[7] with char value.

Initially, I tried append to variable Brightness0[] index6 & index7, but not successful.

[JKSH]

[quote author="houmingc" date="1420706407"]Trying to assigned one character[/quote]Think of a QString as an array of QChar.

@
Brightness[6] = value[6].toLatin1();
@

[houmingc]

Value of ui->BrightnessCombo is from 0 to 100
QString value=ui->BrightnessCombo->currentText();

I need to convert QString to one QChar and insert into index 6, but this is not successful
@
Brightness[6] = value.toLatin1();
@

[andre]

If value is a string that represents a number, then you'll first want to convert the string to an actual number: use one of the functions that create an unsigned number for you, like toUShort or toUInt. Then, it's a simple cast to put that value into a char (which is now used as a one-byte number, not as a character), which can then be assigned to your array.

So, you end up with something like this:
@
bool ok;
ushort sValue = value.toUShort(&ok);
if (!ok) {
//handle error
}
char cValue = static_cast<char>(sValue);

Brightness[6] = cValue;
@

However, I think it would be simpler if you replace your combo box with something that actually returns a numeric value to begin with, like a QSpinBox and/or a QSlider.

[houmingc]

QString value=ui->BrightnessCombo->currentText();

is change to

QString value=ui->BrightnessCombo->currentindex();

to solve the problem

[andre]

[quote author="houmingc" date="1420710723"]QString value=ui->BrightnessCombo->currentText();

is change to

QString value=ui->BrightnessCombo->currentindex();

to solve the problem[/quote]
That's just wrong too. The last variant returns an integer, and then you put it into a string again. Why?

[houmingc]

i need to put it into a unsigned char array bit 6; my code is in the first post :>

[andre]

But... A char is a number, not a string. Why convert a number to a string, and then jump through hoops to get a number again?

[houmingc]

Need to add hex value in an array[] and minus the total with 0x100.
Below is the code i scribbled.
Is there an easier way to parse it. Hope to get some feedback before bedtime.

@
unsigned char Brightness0[]= {0x08,0x11,0x01,0x00,0x00,0x00,0x00,0x64};
unsigned char EndHexValue =0x100;

   //convert Hex value to int
   int EndIntValue=EndHexValue.toUInt(0,16);
   //Array below is integer Equal of Brightness0
   int EqualArray[];
   //IntSum to contain total value of integer  
   int IntSum=0;
   unsigned char HexSum=0;
   for (int a=0; a<sizeof(Brightness0); a++)
  {     
         //Convert hex to int and set it into Integer array              
         Equalarray[a]= Brightness0[a].toUInt(0,16); 
         //Integer total value
         IntSum += EqualArray[a];

 }

        //Integer total value

        EndIntValue -= intSum; 

        HexSum=HexSum.Num(EndIntValue,16);

@

Edit: merged with your previous thread on the same issue. Please don't start multiple threads on the same problem. Andre

[ckakman]

Hi,

You can replace the for loop with std::accumulate().

By the way I don't get why there are so many conversions there. The computer doesn't know number bases.