How Do Usual Arithmetic Conversions Determine the Result Type of Binary \' \' Operators with Signed and Unsigned Operands?

Promotion Rules for Signed and Unsigned Binary Operators

Consider the following code snippets:

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

// Snippet 2
unsigned int us = 42;
int neg = -43;
int result = us + neg;

How does the " " operator determine the correct result type in these cases, given the differing signedness of the operands?

The operator follows the "usual arithmetic conversions" rule, which dictates the type conversion steps based on the operand types. According to this rule, if either operand is:

• long double, both operands are converted to long double.
• double, both operands are converted to double.
• float, both operands are converted to float.
• unsigned long, the other operand is converted to unsigned long.
• long int and the other operand unsigned int, both operands are converted to unsigned long int if the value of the unsigned int can be represented in a long int; otherwise, both are converted to long.
• long, the other operand is converted to long.
• unsigned, the other operand is converted to unsigned.

Since int and unsigned int are interchangeable in the rule, the operand with the wider type (unsigned int) is chosen as the result type.

This explains why in Snippet 1, the result is unsigned int (2147483648) and in Snippet 2, the result is int (-1). The signed operand (neg) is implicitly converted to unsigned int, resulting in an undefined value in the latter case.

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