How Do C \'s Usual Arithmetic Conversions Handle Mixed Signed and Unsigned Integer Operations?

Promotion Rules for Mixed Signed Binary Operators

Operators that take operands of arithmetic types perform automatic conversions and determine the result type according to predefined rules. When operands have differing signedness, C follows the usual arithmetic conversion process outlined in §5/9 of the C standard.

In the first example:

int max = std::numeric_limits<int>::max();
unsigned int one = 1;
unsigned int result = max + one;

The integral promotions are performed, converting both max and one to int, resulting in an unsigned int result. The signedness of max is ignored, leading to a result of 2147483648.

In the second example:

unsigned int us = 42;
int neg = -43;
int result = us + neg;

The conversions follow the rule that the unsigned operand takes precedence, causing neg to be converted to an unsigned type. This conversion results in an implementation-defined value for the int result since the value of us neg is unrepresentable as int.

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