Halo to all,
I think I've increased my frequency of asking questions. But when I see a problem like this I've no choice but to ask.
So, here's a piece of code.
How does the output to this question comes out to be 3?Code:#include<stdio.h> int main(void) { char p[] = "\123\xAB" ; printf("%d",sizeof(p)); }
On a similar note, I just changed the code to
How does the output of this one is 2?Code:#include<stdio.h> int main(void) { char p[] = "\0" ; printf("%d",sizeof(p)); }
Thanks
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
'\123' is legal as a character literal as is '\xAB' or the familiar '\0'. There is nothing unusual about the size since you are using escaped characters. sizeof is also not the same as strlen and the actual terminating zero will be counted.
>> "\123\xAB"
The '\' character is an escape sequence. It should only be followed by certain specific values (eg: "\n"). In your example, '\1' is meaningless, so the compiler is free to generates whatever it pleases. In the second case, you're instructing the compiler to insert a 0 into the buffer, which is already going to be null terminated so the compiler generates 2 total.
EDIT:
>> '\123' is legal as a character literal as is '\xAB' or the familiar '\0'.
Ok, I've never seen that before. My mistake.
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
Then what does '\123' signify? I mean for what it is used. Is it a special character? Also Isn't this '\0' a NULL character which means the string should terminate here and the size should be 1byte.
HOPE YOU UNDERSTAND.......
By associating with wise people you will become wise yourself
It's fine to celebrate success but it is more important to heed the lessons of failure
We've got to put a lot of money into changing behavior
PC specifications- 512MB RAM, Windows XP sp3, 2.79 GHz pentium D.
IDE- Microsoft Visual Studio 2008 Express Edition
>> Then what does '\123' signify?
I looked it up, and it's used for octal numbers, so that's equivalent to '\x53'.
>> Also Isn't this '\0' a NULL character which means the string should terminate here and the size should be 1byte.
No, it inserts however many you specify and then terminates the whole thing with another zero for good measure.In other words, "\0\0\0\0" would be 5 bytes total.
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
And just for good measure, the standard has very little to say on what is or is not a char literal:
6.4.4.4p10: "The value of an integer character constant
containing more than one character (e.g., 'ab'), or containing a
character or escape sequence that does not map to a single-byte
execution character, is implementation-defined."
Yes, but '\123' does map to a single-byte character. So no issue here.
char p[] = "\123\xAB" ;
means... assign the first three elements as follows:
p[0] = 83; // octal 0123 or hex 0x53
p[1] = 171; // octal 0253 or hex 0xAB
p[2] = 0;
Length of array is three bytes.
