Why the output is 'No' in the given code ?

Question

unsigned int i = 23;
signed char c = -23;
if(i > c)
printf("Yes");
else
printf("No");

How the comparison between variable i & c is done here ?

Answer 1

A char can be either signed or unsigned and are signed by default having a range of -127 to 127. The issue is that i is unsigned, so when the comparison is made in the if statement the signed char c is implicitly cast to an unsigned int which causes an overflow resulting in the int value of c wrapping around to the upper bounds creating the overflow value to be higher than i. Remove the unsigned keyword and it will work as expected.

You can also cast i to signed and it will work.
Note, this could also cause issues if the unsigned value of i is greater than or less than the upper or lower bounds of c. Resulting in another overflow issue.
if((signed)i > c)

You can run the following code to see the overflow value of c when it is cast to an unsigned int.

signed char c = -23;
printf("%u", (unsigned)c);

Answer 2

RenDev you can sign type char.
A char is just an 8-bit integer number used to store...numbers. When you use %c (i.e. in printf )
the ascii symbol is shown... The only difference...

https://code.sololearn.com/cFBC6ny0u19L/?ref=app

Answer 3

i think you cant sign char’s. chars are stored as hex and ranges from 0x00-0xFF, (or 0-255 if your working with base 10). Since there is no way a letter can be “negitive” per say, the negitve sign on the char is dropped and so -23 becomes 23 meaning that i=23
and c = 23 and 23 is not larger than 23 so it returns “no”

Answer 4

RenDev I executed this code with condition i >= c, still the output was no.