Recently in my Institute I was given a code and was asked to find out the answer. The code looks like this.
#include <stdio.h>
int main()
{
char ch=500;
printf("%d\n",ch);
}
The output will come as -12
My question is: How can I calculate the value for this kind of code? Is there any formula or process for finding the values?
The code is syntactically correct but not doing what you think. As Paul Hankin is suggesting your compiler should give you a warning as you are trying to set a number that does not fit into a char in a char variable.
A char is 1 byte so it can store a number up to 127 if signed or 255 if unsigned. The value overflows and only the lower 8 bits are taken into account.
500 = 0b111110100
Take only the lower 8 bits: 0b11110100
The MSB is 1 so it's negative number.
1-complement plus 1 is 0b00001100 which is 12
That's why you get -12.
Replacing it by a short or an int should correctly print 500.
You can't know, unless you know the specifics of the given system.
char is typically only 8 bits wide and can't hold the value of 500. Furthermore, the char type is unsuitable for storing integer values, since it has implementation-defined signedness.
Meaning you can't know if it can contain values from 0 to 255 or from -128 to 127 (two's complement). It can even in theory have other constrains and other signedness formats.
Also, the conversion from a large unsigned integer to a smaller signed one is implementation-defined.
I would guess that your specific system has signed 8 bit char type and two's complement signedness. The raw value of 500 is 0x1F4. Upon initialization, your particular compiler will truncate this to fit an 8 bit variable, meaning you end up with only the least significant byte, 0xF4. Since you have an 8 bits signed variable in two's complement format, 0xF4 equals -12.
(The implicit type promotion done by printf preserves the sign.)
None of this behavior is guaranteed across different systems.
Needless to say, the code is completely non-portable. You should not write code like this, which heavily relies on numerous forms of poorly defined behavior.
here in this code ch is signed character type and seems its 8-bit ascii code... so its range is from -128 to +127 (total 256)...
now lets cycle over this ascii range starting from 0 to 127 then -128 to 0 and so... total value of integer number is 500
from 0 to 127 and -128 to 0 (whole circle), left integer value is 500-256 = 244.
from 0 to 127, left integer value is 244-128=116. from -128 to -12, left integer value is 116-116=0.
so after counting all 500 we are at -12 of ascii range....
"sorry for long and non-mathematical explanation....effort is to make you understand"
When using char data types to encode integers, the char values go from -128 to 127. The value is overflowing.
500 - 127 = 373
373 - 128 = 245
245 - 127 = 118
118 - 128 = -10
Since we are passing over 0 twice and that counts, it give us -12. If you try to print the character you will see that it is invalid there is no -12 literal. You can also check man ascii
warning: implicit conversion from 'int' to 'char' changes value from 500 to -12 [-Wconstant-conversion]CHAR_BIT == 8in<limits.h>), then characters can hold 256 different values. The plainchartype can have the same range asunsigned char(0..255) or the same range assigned char(-128..+127, assuming two's-complement). On your machine, it appears that plaincharis signed. When the value 500 is stored, the extra (more significant) bit(s) are removed; the least significant 8 bits are stored in the variable — as either +244 or -12 (-12 for your system). When passed toprintf(), the value is converted to (signed)int.