{"id":17828,"date":"2022-10-27T07:50:57","date_gmt":"2022-10-27T02:20:57","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-bad-math-with-value-multiplied-by-1\/"},"modified":"2022-10-27T07:50:57","modified_gmt":"2022-10-27T02:20:57","slug":"solved-bad-math-with-value-multiplied-by-1","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-bad-math-with-value-multiplied-by-1\/","title":{"rendered":"[Solved] Bad Math with value multiplied by -1?"},"content":{"rendered":"

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Although you believe your code is similar to this:<\/p>\n

#include <iostream>\n\ndouble f()\n{\n    double x = 3;\n    return x * -1;\n}\n\nint main()\n{\n    std::cout << f() << std::endl;\n}\n<\/code><\/pre>\n
The code you have actually takes the type of x<\/code> from the result of a vector’s size()<\/code> – this returns a std::size_t<\/code>:<\/p>\n
#include <cstdint>\n#include <iostream>\n\ndouble f()\n{\n    std::size_t x = 3;\n    return x * -1;\n}\n\nint main()\n{\n    std::cout << f() << std::endl;\n}\n<\/code><\/pre>\n
What happens here is that a std::size_t<\/code> is multiplied by int<\/code>.  The rules of promotion when one argument is signed and one unsigned say that (source: CppReference<\/a>)<\/p>\n
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if the unsigned operand’s conversion rank is greater or equal to the conversion rank of the signed operand, the signed operand is converted to the unsigned operand’s type.<\/p>\n<\/blockquote>\n
std::size_t<\/code> is usually higher rank than int<\/code>, so we are now multiplying 3 by (std::size_t)-1<\/code> – a large number.  Then when we return<\/code>, this large number is converted to double<\/code> to match the signature of f()<\/code>.<\/p>\n
The issue can be avoided a few ways:<\/p>\n