confused about %c and ASCII codes

Question

Why would printf("%c ", 2293552); print 0?

ASCII values are from 0 to 127 I know this must be some cyclic thing but I want a clear explanation. Thank you

score 3 · Accepted Answer · answered Sep 20 '12 at 16:10

3

The number 2293552 corresponds to 0x22ff30. When printf interprets it as ASCII, it ignores all bits beyond the last eight bits containing 0x30, which is the code for '0'.

From the C99 standard:

7.19.1.6.8 -- %c: If no l length modifier is present, the int argument is converted to an unsigned char, and the resulting character is written.

answered Sep 20 '12 at 16:10

Sergey Kalinichenko

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Hah! You beat me to the answer, but I beat you to the standard quote :D (10k only) – Daniel Fischer Sep 20 '12 at 16:15
btw, what is the l length modifier? – bad_keypoints Sep 20 '12 at 16:59
And where can i find more or rather whole of the C99 standard document? Is this it? http://www.open-std.org/jtc1/sc22/WG14/www/docs/n1256.pdf – bad_keypoints Sep 20 '12 at 17:01
haha yes that link is it. L for long. thank you the documentation helped. – bad_keypoints Sep 20 '12 at 17:05
@ronnieaka Yes, that's the pdf from which I quoted. And yes, the `l` is lowercase `L`, for "long". – Sergey Kalinichenko Sep 20 '12 at 17:13

score 0 · Answer 2 · answered Sep 20 '12 at 16:11

0

Likely %c is using only the low-order byte of your argument, which is 2293552 & 255 = 48 = '0'.

answered Sep 20 '12 at 16:11

Keith Randall

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confused about %c and ASCII codes

2 Answers2