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0x0000000000400507 <main+28>: 74 0c je 0x400515 <main+42> 0x0000000000400509 <main+30>: bf 28 06 40 00 mov $0x400628,%edi .. 0x400507 <main+28>: 0x28bf0c74 I think shows the machine code is big-endian. Is my conclusion right?
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The x86 processors use little-endian byte ordering. The least significant byte (LSB) of an integer is stored at the lowest address of the integer. The most significant byte is stored at the highest address for data items in this processor.
If it is little-endian, it would be stored as “01 00 00 00”. The program checks the first byte by dereferencing the cptr pointer. If it equals to 0, it means the processor is big-endian(“00 00 00 01”), If it equals to 1, it means the processor is little-endian (“01 00 00 00”).
The following platforms are considered little endian: AXP/VMS, Digital UNIX, Intel ABI, OS/2, VAX/VMS, and Windows. On big endian platforms, the value 1 is stored in binary and is represented here in hexadecimal notation.
In the SAS System, the following platforms are considered big endian: IBM mainframe, HP-UX, AIX, Solaris, and Macintosh. The following platforms are considered little endian: VAX/VMS, AXP/VMS, Digital UNIX, Intel ABI, OS/2, and Windows.
No, Intel CPUs are little endian: http://en.wikipedia.org/wiki/Endianness
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