Consider the following code snippet:
int a = INT_MAX; unsigned int b = UINT_MAX; bool c = false; if (a < b) // warning C4018: '<' : signed/unsigned mismatch c = true; ... // Additional comparisons if (a == b) // no warning c = true; ... // Additional comparisons
Understanding the Signed/Unsigned Mismatch
In C , signed and unsigned integers have distinct types, and when comparing them directly (e.g., using <, >), MSVC issues a warning (C4018). This is because the compiler attempts to promote both operands to a common type before performing the comparison.
The Promotion Rules
For integer comparisons, MSVC follows the "usual arithmetic conversions" rules as defined in the C standard. These rules determine which promotions occur before the comparison:
Why No Warning for a == b?
In the example, when comparing a and b using ==, MSVC does not issue a warning because both operands are converted to unsigned int before the comparison. Since the values of a and b are both positive, there is no signed/unsigned mismatch, and the result is correct.
Intuitive and Non-Intuitive Results
While the lack of a warning for a == b may seem intuitive, the same cannot be said for comparisons like a < b. In this case, the conversion of a to unsigned results in a larger value, which affects the outcome of the comparison. Thus, MSVC issues a warning to highlight potential issues with signed/unsigned comparisons.
Conclusion
MSVC does not issue a warning for a == b because the usual arithmetic conversions result in both operands being converted to the same unsigned type, eliminating any signed/unsigned mismatch. However, for comparisons using operators like < or >, the potential for unexpected results due to conversions warrants a warning from the compiler.
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