Why this program is output Yes, while it should either have thrown error or NO

Question

Here a + b is -14 which should have been lesser than 'a' and hence should have printed NO but it printed yes.

unsigned int a = 6;
int b = -20;

if((a+b)  > a){
    printf("Yes");
} else {
    printf("NO");
}

return 1;

Answer 1

According to the C Standard (6.3.1.8 Usual arithmetic conversions)

1 Many operators that expect operands of arithmetic type cause conversions and yield result types in a similar way. The purpose is to determine a common real type for the operands and result. For the specified operands, each operand is converted, without change of type domain, to a type whose corresponding real type is the common real type. Unless explicitly stated otherwise, the common real type is also the corresponding real type of the result, whose type domain is the type domain of the operands if they are the same, and complex otherwise. This pattern is called the usual arithmetic conversions:

...

Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.

Types unsigned int and int have the same rank.

Thus the value of the variable b from your example is interpreted as an unsigned value.

If to use the correct conversion specifier %u for the expression ( a + b ) in a function call of printf then you might get

#include <stdio.h>

int main(void) 
{
    unsigned int a = 6;
    int b = -20;
    
    printf( "a + b = %u\n", a + b );
    
    return 0;
}


a + b = 4294967282

Answer 2

You compare signed with unsigned. As all values of unsigned cannot be accommodated in the signed so the the signed value is converted to unsigned and is much larger than 6