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Editorial Team
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Asked: 2026-06-11T11:05:34+00:00

In this example of x86_64 hex/disassembled code I see: 48B80000000000000000 mov rax, 0x0 Signed

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In this example of x86_64 hex/disassembled code I see:

48B80000000000000000 mov rax, 0x0

    Signed Byte 52
    Unsigned Byte   52
    Signed Short    14388
    Unsigned Short  14388
    Signed Int  943863860
    Unsigned Int    943863860
    Signed Int64    3472328296363079732
    Unsigned Int64  3472328296363079732
    Float   4.630555e-05
    Double  1.39804332763832e-76
    String  48B80000000000000000

which to me appears to have the same functionality as:

48C7C000000000 mov rax, 0x0

    48C7C000000000

    Signed Byte 52
    Unsigned Byte   52
    Signed Short    14388
    Unsigned Short  14388
    Signed Int  927152180
    Unsigned Int    927152180
    Signed Int64    3472328377950746676
    Unsigned Int64  3472328377950746676
    Float   1.163599e-05
    Double  1.39806836023098e-76
    String  48C7C000000000

How is the first example treated differently from the second example?

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1 Answer

  1. Editorial Team

    The C7 opcode moves a signed 32 bit number into a 64 bit register. It will sign-extend it, which means it will fill the high bits with whatever the sign bit of the source is, so that it will have the same value when interpreted as a signed number.

    Since it’s an immediate value, the only difference that you’ll see is that the top one will support an immediate value of up to 64 bits long, but the bottom one will only support up to 32 bits.

    Here are the relevant lines from Intel’s reference manual.

    REX.W + B8+ rd    MOV r64, imm64     OI     Valid     N.E.     Move imm64 to r64.
REX.W + C7 /0     MOV r/m64, imm32   MI     Valid     N.E.     Move imm32 sign extended to 64-bits to r/m64.
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