CODEWITHSUNDEEP
c
user2467545
user2467545

Reputation:

Confusion in the output

I have just started my Data types revised chapter. I am currently studying the concept of signed and unsigned character. My doubt is that the signed char has a range from -128 to 127, then why the below code is still running ? Also, the below code is giving the infinite o/p which is not understandable to me. 

main( )
{
char ch ;
for ( ch = 0 ; ch <= 255 ; ch++ )
printf ( "\n%d %c", ch, ch ) ; 
}

I am currently using GCC 32-bit compiler. Can anyone please help me in explaining the o/p of the above code ?

Upvotes: 0

Views: 234

Answers (3)

Vrashabh Irde
Vrashabh Irde

Reputation: 14367

You are probably confused with the output. I think in the o/p you are seeing something like this.

0

1

2

3 ...

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32...

33 ! 34 " .... 125 } 126 ~ 127

255 � 256 257

... 511 � 512 513 .. and so on

0 to 32 are all flags(unprintable codes)(hence you dont see the output , but only the numbers for the first 33), followed by characters till 127. As you can see it wraps around at every 255 characters to give you the same result but it actually stops printing characters after multiples of 127( this is the 127 char list - http://web.cs.mun.ca/~michael/c/ascii-table.html) . It just resets after 127 to -128, so the program continues to print numbers to infinity even though it is resetting the char. This is because when you do printf("%d",ch) for -127 it prints 128, and so on until ch = 255 and then it flips again and starts printing 256 onwards and so on but the actual ch value never goes above 127 and hence it goes to infinity

A signed char c should you be giving you the above output. A char is essentially an integer 8 bits wide, but by default probably signed on your compiler.

Upvotes: 0

guangxyou
guangxyou

Reputation: 121

Because signed character is from -128 to 127, its binary number are 10000000 and 01111111, When 'ch' run to 127, next increment 'ch' will become -128, always less than 255, so it will infinite o/p.

Upvotes: 1

Adam Liss
Adam Liss

Reputation: 48330

for ( ch = 0 ; ch <= 255 ; ch++ )

If ch is a signed character, it will start at 0 and increment to 127. Then, at the next increment, it will "wrap around" and become -128. Using an unsigned char:

127 = 0x7F
128 = 0x80

But, using a signed char, 0x80 becomes -128.

So now ch will run from -128 through 127. And since all of those values are less than 255, this will repeat until you stop the program..

Upvotes: 4

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