64位GCC混合32位和64位指针 [英] 64-bit GCC mixing 32-bit and 64-bit pointers

问题描述

尽管代码可以正常工作，但我对编译器似乎混合使用相同类型的32位和64位参数的决定感到困惑.具体来说，我有一个接收三个char指针的函数.查看汇编代码，将三个中的两个作为64位指针(按预期方式)传递，而将第三个本地常量(但仍然是字符串)作为32位指针传递.我看不到当第3个参数不是完全加载的64位指针时函数如何知道.显然，只要高端为0都没有关系，但我认为它并没有努力确保这一点.在此示例中，任何东西都可能在RDX的高端.我想念什么?顺便说一句，接收函数假定它是一个完整的64位指针，并在条目中包含以下代码:

Although the code works, I'm baffled by the compiler's decision to seemingly mix 32 and 64 bit parameters of the same type. Specifically, I have a function which receives three char pointers. Looking at the assembly code, two of the three are passed as 64-bit pointers (as expected), while the third, a local constant, but character string nonetheless, is being passed as a 32-bit pointer. I don't see how my function could ever know when the 3rd parameter isn't a fully loaded 64-bit pointer. Obviously it doesn't matter as long as the higher side is 0, but I don't see it making an effort to ensure that. Anything could be in the high side of RDX in this example. What am I missing? BTW, the receiving function assumes it's a full 64-bit pointer and includes this code on entry:

     movq    %rdx, -24(%rbp)

这是有问题的代码:

.LC4
    .string "My Silly String"

    .text
    .globl funky_funk
    .type  funky_funk, @function
    funky_funk:
        pushq     %rbp
            movq      %rsp, %rbp
            pushq     %rbx
            subq      $16, %rsp
            movq      %rdi, -16(%rbp)          ;char *dst 64-bit
            movl      %esi, -20(%rbp)          ;int len, 32 bits OK

            movl      $.LC4, %edx              ;<<<<---- why is it not RDX?

            movl      -20(%rbp), %ecx          ;int len 32-bits OK
            movq      -16(%rbp), %rbx          ;char *dst 64-bit
            movq      -16(%rbp), %rax          ;char *dst 64-bit
            movq      %rbx, %rsi               ;char *dst 64-bit
            movq      %rax, %rdi               ;char *dst 64-bit
            call      edc_function


    void funky_funk(char *dst, int len)
    {                                             //how will function know when 
         edc_function(dst, dst, STRING_LC4, len); //a str passed in 3rd parm
    }                                             //is 32-bit ptr vs 64-bit ptr?

    void edc_function(char *dst, char *src, char *key, int len)
    {
         //so, is key a 32-bit ptr? or is key a 64-bit ptr?
    }

推荐答案

将其添加为答案，因为它包含原始问题的部分难题":

Adding this as an answer, as it contains "part of the puzzle" for the original question:

只要编译器可以[通过例如指定满足此要求的内存模型]确定.LC4在前4GB之内，它就可以执行此操作. ％edx将被加载LC4地址的32位，并且高位被设置为零，因此，当edc_function()被调用时，它可以使用％rdx的完整64位，并且只要地址在该范围之内即可.较低的4GB，它将正常工作.

As long as the compiler can determine [by for example specifying a memorymodel that satisfies this] that .LC4 is within the first 4GB, it can do this. %edx will be loaded with 32 bits of the address of LC4, and upper bits set to zero, so when the edc_function() is called, it can use the full 64-bits of %rdx, and as long as the address is within the lower 4GB, it will work out fine.

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