Hello,
I am having troubles with char data type I am using 128 to intiallize it to my variable --> char x = 128; and printing it as an unsigned type (%u) I understand that when printing it as a %u it is printed as unsigned decimal integer. I have tried printf("%u",(unsigned char)x); and it does print 128 but I want to be able to print it without casting in the printf statement. Any help/feedback/thoughts would be very much appreciated.
If it's defined as a char, it will be passed to printf() as a char. There is no way to avoid the cast in this case. The problem is that you're trying to store 128 into a signed char, when a signed char is not capable of holding such a value.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
Thanks brewbuck I knew that signed char had a range -128 to 127, was trying to see if there is workaround out there. But thank you very much for the reply.
whats the point of a signed char anyway?
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What's the point of a signed int anyway?Originally Posted by ಠ_ಠ
Quzah.
Hope is the first step on the road to disappointment.
oh, they are visible characters... I didn't know that
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You are aware that characters are just tiny ints, right? You aren't forced to interpret them as displayable characters. For example, this is perfectly valid:You are confusing a data type, with a character set.Code:char x; for( x = 0; x < 10; x++ ) printf( "Yay!\n" );
Quzah.
Hope is the first step on the road to disappointment.
