#include <stdio.h>
int main() {
unsigned char c = (int) 0.54;
for(; c ; printf("%d", c));
printf("%d", c);
return 0;
}
And when I run the program the output is displayed as 1
How can the output be 1? Thanks in advance
CodePudding user response:
In the first line- unsigned char c=(int)0.54;
char actually stores the ASCII code of the character so in another way it's an integer which stores the data in this way :
The compiler converts the number stored into the character from decimal number system to binary number system and takes into consideration only the first 8 bits from the right of that number represented in binary.(we don't need to consider the case of negative numbers since you use an unsigned char)-so as a result the variable c
takes 0 as value at the end of this line.
For the second line of your code- for(;c ;printf("%d",c));
:
so we have a for loop
( for (INITIALIZATION; CONDITION; AFTERTHOUGHT))
** In the INITIALIZATION part , Leaving this empty is fine, it's just equivalent that you have already initialized the variable that you are going to use as loop variable .
** In the CONDITION part , c is equivalent to c !=0. It keeps the for loop running until c ==0.(in your case c is initialized by 0 , so we will have c =0 and the program immediately exits the for loop).
** In the AFTERTHOUGHT part , The printf is run at the end of each iteration but doesn't change the value of c.(since the condition c !=0 is not verified the code will not write the iteration, so it will not pass to printf)
printf("%d",c);
will display 1 ( c in your loop is used as condition and in the same time to increment variable c by 1 ; c post-increment operator uses the principle 'use-then-change' so , c is incremented by 1 exactly after exiting the for loop , and c becomes equal to 1).