Problems when signed and unsigned numbers are involved in operations

trap topic：
下面的代码输出是（ ）

#include<stdio.h>

int main(int argc, const char *argv[])
{
    
	unsigned int a=6;
	int b=-20;
	(a+b>6)?printf(">6\n"):printf("<=6\n");
	return 0;
}

A > 6
B <= 6
C 无
The final result after execution isA
So what is this asking？

b=-20;相加时,先转换成(unsigned int)-20 + 2^32-1=4,294,967,275;在与aThe result of the addition must be greater than 6.

总结：When signed and unsigned numbers are involved in operations,Converts signed numbers to unsigned numbers for operation.
举个例子：

#include <stdio.h>

int main(int argc, const char *argv[])
{
    
	unsigned int a = 10;
	int b = -20;

	if(a+b > 0){
    
		printf("yes\n");//会打印yes
	}else{
    
		printf("no\n");
	}

	return 0;
}

The final result is printed“yes”Prove that the above is correct.