why a negative signed char is bigger than a signed Int?

Question

logically 23 is bigger than - 23 so why here the result is different?
#include <stdio.h>
int main() {
unsigned int a=23;
signed char c=-23;
if(a>c)
printf("%d",a);
if(a<c)
printf("%d",c);
return 0;}
// output - 23

Answer 1

{ SorousH } I suspect the signed char might be getting converted to an unsigned char in the conditional statements. This could be a compiler specific behavior on comparison operators with mixed signed operands like this.
If this is what is occurring, then the -23 signed char would be converted to an unsigned char which still supports 256 total values. However, the supported range changes from (-128 to 127) to (0 to 255).
Since -23 isn't supported, the value would become: 233 => 256 - 23.
Here, we see (23 < 233) or (a < c).

Answer 2

Negative numbers have their first bit set to 1 so they are usually larger than positive numbers. C++ should be more strict with this, comparing unsigned and signed values is usually nonsense.
https://code.sololearn.com/cUxBYa65ULzb/?ref=app

Answer 3

David Carroll I suspected the same but was hesitant to answer.
Also the signed char has top limit of 127 so if we add even 1, it gives -128 again, so 127 + 106 = -23 or 233 as you mentioned.