CODEWITHSUNDEEP
c++macros
Anon
Anon

Reputation: 421

Size of macro type?

I wanted to know what is the maximum number I could type with in a macro expression, until it overflowed. I wrote the following program:

#define I (INT_MAX + 1)
#define J (LONG_MAX + 1)

int main()
{
    cout << INT_MAX << I  << endl;
    cout << LONG_MAX << J << endl;
    return 0;
}

I got the following output: 

2147483647-2147483648
9223372036854775807-9223372036854775808

I don't understand the overflow for the int case, if it can store a value of LONG_MAX. Also then I remembered that macros can also store decimals like 3.14 (PI), so it should be a float type. I am now confused. Please help.

Upvotes: 0

Views: 467

Answers (1)

eerorika
eerorika

Reputation: 238401

Macros do not have a type. They do not interact with the type system at all. Macros are part of a pre-processing step that is performed before compilation. Macros are textual replacement. After your source has been pre-processed, the result seen by the compiler is as follows:

int main()
{
    cout << INT_MAX << (INT_MAX + 1)  << endl;
    cout << LONG_MAX << (LONG_MAX + 1) << endl;
    return 0;
}

... except INT_MAX and LONG_MAX are themselves macros, so they would have been expanded to some integer literal as part of the pre-processing as well. The exact values are implementation specific and can differ between systems, so I did not expand them here in order for the answer remain general.

P.S. The behaviour of the program is undefined, because of signed overflow.

Upvotes: 1

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