CODEWITHSUNDEEP
c++
S55
S55

Reputation: 53

How does c++ determine whether an integer literal fit int type or not?

I know the standard says if the integer literal does not fit the int, it tries unsigned int, and so forth, per section 2.14.2 Table 6 in the standard.

My question is: what's the criteria to determine it fits or not?

Why do both std::is_signed<decltype(0xFFFFFFFF)>::value std::is_signed<decltype(0x80000000)>::value gives false. Why don't they fit in int? 0x80000000 has the same bit representation as signed -1 signed -2147483648.

Upvotes: 2

Views: 181

Answers (3)

Aykhan Hagverdili
Aykhan Hagverdili

Reputation: 29975

0xFFFFFFFF is hex for 4'294'967'295. On platforms where sizeof(int) == 4, the range of int is -2'147'483'648 to 2'147'483'647. As you can see 4'294'967'295 isn't in that range. Simple as that.

Upvotes: 3

Acorn
Acorn

Reputation: 26066

what's the criteria to determine it fits or not?

Up to the platform and compiler. They define how big int is.

Why do both std::is_signed<decltype(0xFFFFFFFF)>::value std::is_signed<decltype(0x80000000)>::value gives false.

Because in most platforms 0x80000000 and 0xFFFFFFFF will be unsigned int because they do not fit in an int.

Why don't they fit in int?

Because in most platforms int is 32-bit, two's complement, which means 0x7FFFFFFF is the biggest int

Upvotes: 3

HolyBlackCat
HolyBlackCat

Reputation: 96116

You don't need to look at "bit representation" to check if the number fits or not.

Assuming sizeof(int) == 4, int can represent numbers from -231 to 231-1 inclusive.

0x80000000 is 231, which is 1 larger than the maximum value.

Upvotes: 3

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