What does GCC __attribute__((mode(XX)) actually do?

13,471

Solution 1

These allow you to explicitly specify a size for a type without depending on compiler or machine semantics, such as the size of 'long' or 'int'.

They are described fairly well on this page.

I quote from that page:

QI: An integer that is as wide as the smallest addressable unit, usually 8 bits.

HI: An integer, twice as wide as a QI mode integer, usually 16 bits.

SI: An integer, four times as wide as a QI mode integer, usually 32 bits.

DI: An integer, eight times as wide as a QI mode integer, usually 64 bits.

SF: A floating point value, as wide as a SI mode integer, usually 32 bits.

DF: A floating point value, as wide as a DI mode integer, usually 64 bits.

So DI is essentially sizeof(char) * 8.

Further explanation, including TI mode, can be found here (possibly better than the first link, but both provided for reference).

So TI is essentially sizeof(char) * 16 (128 bits).

Solution 2

@haelix Just read this question and I also tried to understand this thing. By my reading: you can find the definitions in the [gcc/gcc/machmode.def] in GCC source tree. For 'SD' it should be:

    /* Decimal floating point modes.  */ 
DECIMAL_FLOAT_MODE (SD, 4, decimal_single_format);

and 'DECIMAL_FLOAT_MODE' says:

     DECIMAL_FLOAT_MODE (MODE, BYTESIZE, FORMAT);
declares MODE to be of class DECIMAL_FLOAT and BYTESIZE bytes
wide.  All of the bits of its representation are significant.
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Updated on June 05, 2022

Comments

  • Admin
    Admin about 2 years

    This arose from a question earlier today on the subject of bignum libraries and gcc specific hacks to the C language. Specifically, these two declarations were used:

    typedef unsigned int dword_t __attribute__((mode(DI)));
    

    On 32 bit systems and

    typedef unsigned int dword_t __attribute__((mode(TI)));
    

    On 64-bit systems.

    I assume given this is an extension to the C language that there exists no way to achieve whatever it achieves in current (C99) standards.

    So my questions are simple: is that assumption correct? And what do these statements do to the underlying memory? I think the result is I have 2*sizeof(uint32_t) for a dword in 32-bit systems and 2*sizeof(uint64_t) for 64-bit systems, am I correct?