Some confusion of the character constant

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Some confusion of the character constant

AuthorQuestion

Ken H

Some confusion of the character constant
Posted on: 12 Mar 2012

hi all,
Consider the following problem:
first:

#include<stdio.h>
void main()
{
char ch='\48';
printf("%c\n",ch);
}
//Why the above code output is 8?

second:

#include<stdio.h>
void main()
{
char *s="\ta\018bc";
for(;*s!='\0';s++) printf("*");
}
//Why the above code output is 6 *?

thank very much.

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AuthorReply

Vulpes

Re: Some confusion of the character constant
Posted on: 12 Mar 2012

Multi-character constants are implementation defined in C++ though (FWIW) the documentation for Visual C++ is here:

http://msdn.microsoft.com/en-us/library/6aw8xdf2.aspx

With the first expression '\48', what seems to happen is that \4 is read as the octal literal 4 but 8 is read literally as it's not an octal digit. However, \4 is then ignored (frankly I've no idea why) and only 8 is printed.

With the second expression "\ta\018bc" we now have a string constant rather than a character constant. Ignoring the terminating \0, this expression has 6 characters namely \t, a, \01 (an octal constant), 8, b and c. Consequently, 6 asterisks get printed as you iterate through them.

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